Example: 36 If f (x) = sin[x] , [x] ≠ 0 , then lim f(x) equals [IIT 1985; Rajasthan PET 1995]
Solution: (d) [x]
x→0
Example: 37 0 ,
Solution: (a) [x] = 0
Example: 38
Solution: (d) (a) 1 (b) 0 (c) –1 (d) Does not exist
Example: 39 In closed interval of x = 0 at right hand side [x] =0 and at left hand side [x] = −1. Also [0] =0.
Solution: (a)
Example: 40 Therefore function is defined as f(x) = sin[x] , (−1 ≤ x < 0)
Solution: (d) [x]
0 , (0 ≤ x < 1)
∴ Left hand limit = lim f(x) = lim sin[x] = sin(−1) = sin 1c
[x] −1
x→0− x→0−
Right hand limit = 0, Hence, limit doesn’t exist.
lim tan x − sin x [IIT 1974; Rajasthan PET 2000]
x3
x→0 (d) None of these
(a) 1 (b) 1 (c) 2
2 −2 3
sin x 2 sin2 x sin 2 x
2 2
lim tan x − sin x = lim sin x − sin x cos x = lim = xli→m0 sin x . 2 . . 1 = 1
x3 x3 cos x x3 cos x x cos 2 4 2
x→0 x→0 x→0 x x
2
If f(x) = sin(e x− 2 − 1) , then lim f(x) is given by
log(x 1)
− x→2
(a) – 2 (b) –1 (c) 0 (d) 1
lim f(x) = lim sin(e x−2 − 1) = lim sin(et − 1) . (Putting x = 2 + t)
log(t + 1) log(t + 1)
x→2 x→2 t→0
lim sin(e t − 1) . e t − 1 . t lim sin(et − 1) 1 t ..... 1
et − 1 t log(1 et −1 1! 2!
= x→∞ + t) = t→0 + + 1 1 1
2 3
− t + t2 − .....
= 1.1.1 = 1 [ As t → 0, et −1→ 0, ∴ sin(et − 1) = 1]
(et − 1)
lim 2 acot x − acos x = [Kerala (Engg.) 2001]
cot x − cos x
x→π / (d) log x
(a) log a (b) log 2 (c) a
lim acot x − acos x = lim acos x acot x−cos x − 1
cot − cos x cot x − cos x
x→π / 2 x x→π / 2
= a cos(π / 2) x lim 2 a cot x−cos x − 1 = 1 log a = log a .
cot x − cos x
→π /
sin x cos x tan x f(x)
x2
If f(x) = x 3 x 2 x , then lim is [Karnataka CET 2002]
2x 1 1
x→0 (d) 1
(a) 3 (b) –1 (c) 0
f(x) = x(x − 1) sin x − (x 3 − 2x 2 )cos x − x 3 tan x
= x 2 sin x − x3 cos x − x3 tan x + 2x2 cos x − x sin x
Hence, lim f(x) = lim sin x − x cos x − x tan x + 2 cos x − sin x = 0−0−0+ 2−1 =1.
x2 x
x→0 x→0
Example: 41 If f (x) = cot −1 3x − x3 and g(x) = cos −1 1 − x 2 , then lim f(x) − f (a) , 0 <a < 1 is [Orissa JEE 2003]
Solution: (d) 1 − 3x2 1 + x 2 g(x) − g(a) 2
x→a
Example: 42
Solution: (c) (a) 3 (b) 3 (c) 3 (d) 3
Example: 43 2(1 + a 2 ) 2(1 + x 2 ) 2 −2
Solution: (d)
f(x) = cot −1 3x − x3 and g(x) = cos −1 1 − x2
1− 3x 2 1 + x2
Put x = tanθ in both equation
f(θ ) = cot −1 3 tan θ − tan 3 θ = cot −1{tan 3θ }
1− 3 tan 2 θ
f(θ ) = cot −1 cot π − 3θ = π − 3θ ⇒ f ′(θ ) = −3 ..….(i)
2 2
and g(θ ) = cos −1 1 − tan 2 θ = cos −1(cos 2θ ) = 2θ ⇒ g′(θ ) = 2 ….. (ii)
1 + tan 2 θ
Now xli→ma f(x) − f (a) = lim f (x) − f (a) 1 = f ′(x). 1 = −3 × 1 =− 3 .
g(x) − g(a) x →a x − a g(x) g ′(x) 2 2
lim − g(a)
x →a x − a
1 − tan x [1 − sin x]
2
lim is [AIEEE 2003]
x→π2 1 + x
tan 2 [π − 2x]3
(a) 1 (b) 0 (c) 1 (d) ∞
8 32
tan π − x (1 − sin x)
4 2
lim
π (π − 2x)3
x → 2
tan −y (1 − cos y) − tan y .2 sin 2 y 1 tan y sin y 2 1
2 2 2 32 2 . y 2 32
Let x= π + y, then y → 0 ⇒ lim = lim = lim = .
2 (−2y)3 (−8)y 3 y
y→0 y→0 y→0 2
2
If lim [(a − n) nx − tan x] sin nx = 0, where n is non-zero real number, then a is equal to [IIT Screening 2003]
x
x→0 2
(a) 0 (b) n+1 (c) n (d) n + 1
n n
lim n sin nx . lim (a − n)n − tan x = 0 ⇒ n [(a − n)n − 1] = 0 ⇒ (a − n)n = 1 ⇒ a = n + 1 .
nx x→0 x n
x→0
(3) Logarithmic limits : To evaluate the logarithmic limits we use following formulae
(i) log(1 + x) = x − x2 + x3 − ............to ∞ where −1 < x ≤ 1 and expansion is true only if base is e.
2 3
(ii) lim log(1 + x) = 1 (iii) lim log e x =1
x
x→0 x→e
(iv) lim log(1 − x) = −1 (v) lim loga(1 + x) = loga e,a > 0,≠ 1
x x
x→0 x→0
Example: 44 lim log e (1 + 2h) − 2 log e (1 + h) [IIT Screening 1997]
h2
Solution: (a) h→0
Example: 45 (a) –1 (b) 1 (c) 2 (d) –2
Solution: (c)
Example: 46 loge(1 + 2h) − 2 loge(1 + h) (2h) − (2h)2 + (2h)3 − .....∞ − 2 h − h2 + h3 − ......
Solution: (b) h2 2 3 h2 2 3
Example: 47 lim = lim
Solution: (c) h→0 x→a
= lim − h2 + 2h3 − .... = lim h2{−1 + 2h − ....} = lim {−1 + 2h + ....} = −1 .
h2 h2
h→0 h→0 h→0
lim log{1 + (x − a)} =
(x − a)
x→a
(a) –1 (b) 2 (c) 1 (d) –2
Let x – a = y, when x → a, y → 0,
∴ The given limit = lim log{1 + y} =1.
y
y→0
lim log10(1 + h) =
h
h→0
(a) 1 (b) log10 e (c) loge 10 (d) None of these
lim log e (1 + h) . 1 = log10 e .
h loge 10
h→0
If lim log(3 + x) − log(3 − x) = k, then the value of k is [AIEEE 2003]
x
x→0
(a) 0 (b) − 1 (c) 2 (d) − 2
3 3 3
lim log(3 + x) − log(3 − x) = lim log 3+ x = lim log 1 + (x / 3)
x 3− x 1 − (x / 3)
x→0 x→0 x x→0
x
= lim log(1 + (x / 3)) − lim log(1 − (x / 3)) = 1 − − 1 = 2 .
3 3 3
x→0 x x→0 x
(4) Exponential limits :
(i) Based on series expansion : We use ex =1+ x + x2 + x3 + .............∞
2! 3!
To evaluate the exponential limits we use the following results –
(a) lim ex − 1 = 1 (b) lim ax − 1 = log e a (c) lim eλx − 1 =λ (λ ≠ 0)
x x x
x→0 x→0 x→0
(ii) Based on the form 1∞ : To evaluate the exponential form 1∞ we use the following results.
(a) If lim f(x) = lim g(x) = 0 , then lim {1 + f(x)}1 / g(x) = e lim f(x) , or
x→a g(x)
x→a x→a x→a
when lim f(x) = 1 and lim g(x) = ∞ . Then lim { f (x)} g( x) = lim [1 + f(x) − 1]g(x) = e lim ( f ( x)−1)g(x)
x→a
x→a x→a x→a x→a
(b) lim (1 + x)1/ x = e (c) lim 1 + 1 x (d) lim (1 + λx)1/ x = e λ (e) lim 1 + λ x
x→0 x→∞ x x→0 x→∞ x
=e = eλ
Note : lim a x = ∞, if a >1 i.e., a∞ = ∞, if a >1 and a∞ = 0 if a < 1.
0 , if a <1
x→∞
Example: 48 lim eαx − e βx = [MP PET 1994]
Solution: (d) x
Example: 49 x→0
Solution: (b) (a) α + β (b) 1 + β (c) α 2 − β 2 (d) α − β
Example: 50 α
Solution: (a)
lim eαx − e βx = lim (eαx − 1) − (e βx − 1) = lim eαx − 1 − lim eβx − 1 = α−β .
Example: 51 x x x x
x→0 x→0 x→0 x→0
Solution: (d)
The value of lim e x − (1 + x) is [Karnataka CET 1995]
Example: 52 x2
Solution: (a) x→0
Example: 53
Solution: (b) (a) 0 (b) 1 (c) 1 (d) 1
2 4
(1 + x + x2 + .....) − (1 + x) x 2 1 + x x2 + .....
2! x2 2! 3! + 4!
lim ex − (1 + x) = lim = lim = 1 = 1 .
x2 x2 2! 2
x→0 x→0 x→0
lim ax − 1 is equal to
x→0 1 + x − 1
(a) 2 loge a (b) 1 loge a (c) a loge 2 (d) None of these
2
lim ax − 1 = lim ax − 1 . 1 + x + 1 ( ) ( )= (a x a x− 1
x→0 1 + x − 1 x→0 1 + x − 1 1 + x + 1 lim − 1) 1+ x +1 = lim x . 1+ x +1
1+ x −1
x→0 x→0
( )= ax − 1
lim x . lim 1 + x + 1 = (loge a).(2) = 2 loge a .
x→0 x→0
The value of lim x + 3 x + 2 is [UPSEAT 2003]
x→∞ x + 1
(a) e4 (b) 0 (c) 1 (d) e2
x 3 x +2 2 x2+1.( x+ 2).(x2+1) 2 x + 1 2. 1+ 2 2 lim 1+ 2 1+ 1
x 1 + + 2 1+ x x x
lim + lim 1 + lim 1 1 e x→∞ = e2.
x→∞ + = x→∞ = + x =
x 1 x →∞ x 1
lim x + 3 x +2 lim 1 2 x + 2 e lim 2 (x + 2) e lim 1+ 2
x→∞ x + 1 x →∞ + x→∞ x +1 x→∞ 2 1+ x
Alternative method : = 1 = e2
+ = = x
x 1
lim 1 + 1 c + dx
x→∞ + bx
If a, b, c, d are positive, then a [EAMCET 1992]
(a) ed / b (b) ec / a (c) e(c +d) /(a+b) (d) e
ed / b
lim 1 + 1 c + dx = lim 1 + 1 a+bx c + dx = lim 1 + 1 a+bx =e and lim c + dx = d
x→∞ a + bx a + bx a+bx + bx a + bx
x→∞ x→∞ a x→∞ b
e xl→im∞ 1 c + dx
+ bx 1
Alternative method : a = ed / b .
lim x x = [Roorkee 1987]
x→0 (d) None of these
(a) 0 (b) 1 (c) e
Let y = x x ⇒ log y = x log x ; ∴ lim log y = lim x log x = 0 = log 1 ⇒ lim x x = 1
y→0 x→0 x→0
Example: 54 The value of lim (1 + x)1 / x− e + 1 ex is
Solution: (a) x2 2
x→0 [DCE 2001]
Example: 55
Solution: (d) (a) 11e (b) −11e (c) e (d) None of these
24 24 24
Example: 56
Solution: (d) (1 + x)1 / x 1 log(1+ x) 1 x − x2 + x3 −.... e1− x + x2 −....... e− x + x2 −......
Example: 57 x x 2 3 2 3 2 3
Solution: (b) = e = e = = e
= − x + x2 − ..... + 1 − x + x2 − ..... 2 = e1 − x + 11 x2 − ............
e1 + 2 3 2! 2 3 + ..... 2 24
∴ lim (1 + x)1 / x − e + ex 11e
x2 2 24
x→0 =
lim (1 + x)1 / x − e equals
x
x→0 [UPSEAT 2001]
(a) π / 2 (b) 0 (c) 2 / e (d) – e / 2
1 1 [log(1+ x)] 1 x − x2 + x3 − x4 +.... e1 − x x2 − x3 + .... − x + x2 − x3 +....
x x 2 3 4 2+ 3 4 2 3 4
(1 + x) x = e = e = = e.e
= − x + x2 x3 + ..... − x + x2 x3 + ...2 = e − ex + 11e x2 − ..........
2 3− 4 2 3− 4 2 24
e 1 + +
1! 2! + ...
lim (1 + x)1 / x − e lim e − ex + 11e x 2.......... − e lim − e 11e x + ... e .
x 2 24 x →0 2 24 2
∴ x→0 = x→0 ⇒ − = −
x
lim cos x m = [AMU 2001]
m [AMU 2002]
m→∞
(a) 0 (b) e (c) 1/e (d) 1
lim cos x m = mli→m∞1 + cos x − 1m = mli→m∞1 − − cos x + 1 m
m→∞ m m m
= e 2 sin x 2 x2
2m 4m2
x m lim − x m lim − x m x2
2m 2 sin 2 2m 4m
mli→m∞1 − 2 sin2 em→∞ m→∞ − 2 lim = e0
= = 2m = e m→∞ = 1.
lim n2 − n + 1 n(n−1) =
n2 − n − 1
n→∞
(a) e (b) e2 (c) e −1 (d) 1
nl→im∞ n2 − n + 1 n(n−1) = nl→im∞ n(n − 1) + 1 n(n−1) = lim 1 + 1 1) n(n−1) = e = e2 .
n2 − n − 1 n(n − 1) − 1 n→∞ 1 − n(n − e −1
1 1) n(n−1)
n(n −
Alternative Method: lim 1 + n2 2 n(n−1) = e lim 2n(n−1) = e2 .
n→∞ −n n→∞ n2 −n−1
− 1
(5) L’ Hospital’s rule : If f(x) and g(x) be two functions of x such that
(i) lim f(x) = lim g(x) = 0
x→a x→a
(ii) Both are continuous at x = a
(iii) Both are differentiable at x = a .
(iv) f ' (x) and g' (x) are continuous at the point x = a , then lim f(x) = lim f ' (x) provided that g' (a) ≠ 0
g(x) g' (x)
x→a x→a
Note : The above rule is also applicable if lim f(x) = ∞ and lim g(x) = ∞ .
x→a x→a
If lim f ' (x) assumes the indeterminate form 0 or ∞ and f ' (x), g' (x) satisfy all the condition
g' (x) 0 ∞
x→a
embodied in L’ Hospital rule, we can repeat the application of this rule on f ' (x) to get, lim f ' (x) =
g' (x) g' (x)
x→a
lim f " (x) . Sometimes it may be necessary to repeat this process a number of times till our goal of
g" (x)
x→a
evaluating limit is achieved.
Example: 58 lim 1− cos mx = [Kerala (Engg.) 2002]
Solution: (c) 1− cos nx
x→0
Example: 59
Solution: (c) (a) m / n (b) n / m (c) m2 (d) n2
n2 m2
mx mx 2
lim 1 − cos mx 2 sin2 2 lim sin 2 m2 x 2 . 1 . 4 m2 ×1 m2
1 − cos nx xli→m0 2 sin2 nx 4 n2 x 2 n2 n2
x→0 = 2 = x→0 mx nx 2 = =
2
2 sin
nx
2
Trick : Apply L-Hospital rule ,
lim 1− cos mx = lim m sin mx = lim m2 cos mx = m2 .
1− cos nx n sin nx n2 cos nx n2
x→0 x→0 x→0
The integer n for which lim (cos x − 1) (cos x − ex ) is a finite non-zero number is [IIT Screening 2002]
xn
x→0
(a) 1 (b) 2 (c) 3 (d) 4
n cannot be negative integer for then the limit = 0
Limit = lim 2 sin2 x e x − cos x 1 lim e x − cos x (n≠1 for then the limit = 0)
2 xn−2 2 xn−2
x→0 22(x / 2)2 = x→0
= 1 lim ex + sin x . So, if n = 3 , the limit is 1 2) which is finite. If n = 4, the limit is infinite.
2 (n − 2)xn−3 2(n −
x→0
f (1 + x) 1
f (1)
Example: 60 Let f:R→R be such that f(1) = 3 and f′ (1) = 6 . Then lim x equals [IIT Screening 2002]
Solution: (c)
Example: 61 x→0 (d) e3
Solution: (a)
Example: 62 (a) 1 (b) e1 / 2 (c) e2 [IIT Screening 1997; AMU 1997]
Solution: (d)
Example: 63 f (1 + x) 1 lim 1 [log f(1+ x)−log f (1)] lim f ′(1+ x) / f(1+ x) f ′(1)
Solution: (a) f (1) x 1
xli→m0 x = e x→0 = e x →0 = e f(1) = e6 / 3 = e2 .
Example: 64
Solution: (d)
Example: 65
Solution: (c) lim sin α − cos α =
α −π / 4
α →π / 4
(a) 2 (b) 1 / 2 (c) 1 (d) None of these
(By ‘L’ Hospital rule)
lim sin α − cos α ( 0 form) = lim cosα + sinα
α −π / 4 0 1
α →π / 4 α →π / 4
= 1 + 1 = 2 = 2.
2 2 2
lim x3 − a3 =
x2 − a2
x→a
(a) 0 (b) Not defined (c) 2a (d) 3a
2
x3 − a3 ( )0 = 3x2 (By ‘L’ Hospital rule) = 3a2 3a .
x2 − a2 0 2x 2a 2
lim form lim =
x→a x→a
lim x +h − x= [Roorkee 1983]
h
h→0
(a) 1 / 2 x (b) 1/ 2 h (c) Zero (d) None of these
lim x +h − x = lim x +h − x x +h + x = lim (x + h) − x = lim h =1
h h × x +h + x) 2 x
h→0 h→0 x h→0 h( x + h + x) h→0 h( x + h +
Trick : Applying ‘L’ Hospital’s rule, [Differentiating Nr and Dr with respect to h]
We get, lim 2 1 −0 1
2x
h→0 x+h = .
1
lim sin2 α − sin 2 β [MP PET 2001]
α2 − β2
α→β
(a) 0 (b) 1 (c) sin β (d) sin 2β
β 2β
lim sin2 α − sin2 β = lim sin(α + β )sin(α − β) = lim sin(α − β ) lim sin(α + β ) = lim sin(α + β ) = sin 2β .
α2 − β2 (α + β ) (α − β) (α − β ) (α + β ) (α + β ) 2β
α→β α −β →0 α −β →0 α −β →0 α→β
Trick : By L’ Hospital’s rule, lim 2 sinα cosα = sin 2β .
2α 2β
α→β
lim tan 2x − x equals [IIT 1971]
3x − sin x
x→0
(a) 2/3 (b) 1/3 (c) 1/2 (d) 0
tan 2x − x 2 tan 2x − 1 1
3x − sin x x 2
lim = lim 2x sin = .
x
x→0 x→0 3−
Example: 66 If G(x) = − 25 − x2 , then lim G(x) − G(1) equals [IIT 1983]
Solution: (d) x − 1
x →1
Example: 67
Solution: (c) (a) 1/24 (b) 1/5 (c) − 24 (d) None of these
Example: 68
Solution: (a) lim G(x) − G(1) = lim − 25 − x2 + 24 [Multiply both numerator and denominator by ( 24 + 25 − x2 )]
Example: 69 x −1 x −1
Solution: (c) x →1 x →1
Example: 70 = lim x +1 = 1
24 + 25 − x2 24
x →1
Alternative method: By L'-Hospital rule, lim G′(x) = lim −1(−2x) = 1
1 2 25 − x2 24
x →1 x →1
If f(a) = 2, f ' (a) = 1, g(a) = 1, g' (a) = 2, then lim g(x)f(a) − g(a) f (x) equals
x− a
x→a
[IIT 1983; Rajasthan PET 1990; MP PET 1995; DCE 1999; Karnataka CET 1999, 2003]
(a) –3 (b) 1 (c) 3 (d) 1
3 −3
Applying L – Hospital's rule, we get, lim g(x) f(a) − g(a) f(x) = lim g′(x) f(a) − g(a) f′(x)
x−a 1
x→a x→a
= g′(a) f(a) − g(a) f′ (a) = 2 × 2 − 1× (1) = 3.
lim (1 + x)n −1 = [Kurukshetra CEE 2002]
x
x→0
(a) n (b) 1 (c) –1 (d) None of these
lim (1 + nx + nC2x 2 + ......higher powers of x to xn) − 1 = n
x
x→0
Trick : Apply L- Hospital rule.
lim sin x + log(1 − x) is equal to [Roorkee 1995]
x2
x→0
(a) 0 (b) 1 (c) 1 (d) None of these
2 −2
cos x−1 1 x − sin x 1 1
2x − − (1 − x)2 =−2
Apply L- Hospital rule, we get, lim = lim 2
x→0 x→0
sin x log(1 x) x − x3 x5 − ...... − x − x2 − x3 x4 − ......
x2 3! + 5! 2 3− 4
Alternative method : lim + − = lim x2 + lim x2
x→0 x→0 x→0
sin x = x− x3 + x5 − ...... and log (1 − x) = −x − x2 − x3 ........
3! 5! 2 3
− x2 − x 3 1 + 1 − x4 .... 1
2 3! 3 4 2
Hence, lim = − .
x2
x→0
lim xe x − log(1 + x) equals [Rajasthan PET 1996]
x2
x→0
(a) 2 (b) 1 (c) 1 (d) 3
3 3 2 2
Solution: (d) Let y = lim xe x − log(1 + x) 0 form
x2 0
Example: 71 x→0
Solution: (d)
Applying L–Hospital's rule, y lim ex + xe x − 1 1 x 0 form
Example: 72 2x + 0
Solution: (d) = x→0
Example: 73
Solution: (a) y = lim 1 e x + ex + xe x + 1 = lim 1 [1 + 1 + 0 + 1] = 3
Example: 74 2 2 2
Solution: (a) x→0 (1 + x)2 x→0
lim sin−1 x − tan−1 x is equal to [Rajasthan PET 2000]
x3
x→0
(a) 0 (b) 1 (c) – 1 (d) 1
2
lim sin−1 x − tan−1 x 0 form
x3 0
x→0
Applying L-Hospital’s rule,
= lim 11 0 form
1− x2 − 1+ x2 0
x→0
3x 2
−1 × (1 −2x / 2 2x 1 1 2 1
2 − x 2)3 + (1 + x 2 )2 6 x 2)3/ x 2
= lim = lim + = .
6x (1 2 (1 2 )2
x→0 x→0 − +
lim 1 + log x −x = [Karnataka CET 2000]
x→1 1 − 2x + x2
(a) 1 (b) – 1 (c) 0 (d) − 1
2
Applying L-Hospital’s rule, lim 1 + log x −x lim 1 −1 lim 1− x
1 − 2x + x2 x 2x(x − 1)
x →1 = x →1 − 2 + 2x = x →1
Again applying L-Hospital’s rule, we get lim 4 −1 2 1
x− =−2
x →1
lim 4x − 9x = [EAMCET 2002]
x(4 x + 9x )
x→0
(a) log 2 (b) 1 log 3 (c) 1 log 3 (d) log 3
3 2 2 2 2 2
y = lim 4x − 9x 0 form
x(4 x + 9 x ) 0
x→0
Using L-Hospital’s rule, y = lim 4 x log 4 − 9 x log 9 ⇒ y = log 4 − log 9 ⇒ y = log 2 2 = log 2 .
9 x ) + x(4 x log 4+ 9 2 3 3
x→0
(4 x + x log 9) 2
If f(a) = 2 , f ′(a) = 1 , g(a) = −3 , g′(a) = −1 , then lim f(a) g(x) − f(x) g(a) = [Karnataka CET 2003]
x− a
x→a
(a) 1 (b) 6 (c) – 5 (d) – 1
lim f(a) g(x) − f(x) g(a) 0 form
x− a 0
x→a
Using L-Hospital’s rule, lim f(a) g′(x) − f′(x) g(a) = f(a) × g′(a) − f ′(a) × g(a) = 2 × (−1) − 1 × (−3) = 1.
1−0
x→a
Example: 75 The value of lim 2− x−3 is [MP PET 2003]
Solution: (d) x2 − 49
Example: 76 x→7
Solution: (a)
(a) 2 (b) − 2 (c) 1 (d) − 1
Example: 77 9 49 56 56
Solution: (c)
0− 1 −1 −1 −1
Example: 78 2 x−3 x− 7− 56
Applying L-Hospital’s rule, lim 2x = lim 4x 3 = 4.7 3 = .
Solution: (b)
Example: 79 x →7 x →7
Solution: (d)
Let f(a) = g(a) = k and their nth derivatives f n(a), gn(a) exist and are not equal for some n. If
lim f(a) g(x) − f(a) − g(a) f(x) + g(a) = 4, then the value of k is [AIEEE 2003]
g(x) − f(x)
x→a
(a) 4 (b) 2 (c) 1 (d) 0
lim k g(x) − k f(x) = 4
g(x) − f(x)
x→a
By L-Hospital’ rule, lim k g ′(x) − f ′(x) = 4 , ∴ k = 4.
g ′(x) −
x→a f ′(x)
The value of lim ∫ x2 sec2 t dt 0 is [AIEEE 2003]
x→0 x sin x
(a) 3 (b) 2 (c) 1 (d) 0
∫lim d x2 sec 2 t dt = lim sec 2 x 2 .2x (By L' –Hospital's rule)
dx sin x + x cos x
x→0 0 x→0
d
dx (x sin x)
= lim 2 sec 2 x 2 = 2 × 1 =1.
1 + 1
x→0 sin x + cos x
x
lim 3 sin x− 3 cos x [EAMCET 2003]
6x π
x→π / 6 −
(a) 3 (b) 1 (c) − 3 (d) − 1
3 3
Using L–Hospital’s rule, lim 3 cos x + 3 sin x = 3. 3 3 . 1 1.
6 2+ 2 3
x→π / =
6 6
Given that f ' (2) = 6 and f ′(1) = 4 , then lim f(2h + 2 + h2 ) − f(2) = [IIT Screening 2003]
f(h − h2 + 1) − f(1)
h→0 (d) 3
(a) Does not exist (b) − 3 (c) 3
2 2
lim f(2h + 2 + h2) − f(2) = lim f′(2h + 2 + h2)(2 + 2h) = 6×2 = 3.
f(h − h2 + 1) − f(1) f ′(h − h2 + 1)(1 − 2h) 4 ×1
h→0 h→0
Funtion limit continuity & differentiability
Introduction .
The word ‘Continuous’ means without any break or gap. If the graph of a function has no break, or gap or
jump, then it is said to be continuous.
A function which is not continuous is called a discontinuous function.
While studying graphs of functions, we see that graphs of functions sin x , x, cos x , ex etc. are continuous but
greatest integer function [x] has break at every integral point, so it is not continuous. Similarly tan x, cot x, sec x ,
1 etc. are also discontinuous function.
x
Y Continuous function Discontinuous function Y
Y Y 3
2
(0, 1) f (x) = x 1
–π/2 π 2π X XX
X′
–2π –π O π/2 Y′ 0 X X’ –3 –2 –1O 1 X
f(x)= 1/x –1 23
(0,–1) –2
y = sin x y = [x]
–3
Y’
Continuity of a Function at a Point.
A function f(x) is said to be continuous at a point x = a of its domain iff lim f(x) = f(a) . i.e. a function f(x) is
x→a
continuous at x = a if and only if it satisfies the following three conditions :
(1) f(a) exists. (‘a’ lies in the domain of f)
(2) lim f(x) exist i.e. lim f(x) = lim f(x) or R.H.L. = L.H.L.
x→a+ x→a−
x→a
(3) lim f(x) = f(a) (limit equals the value of function).
x→a
Cauchy’s definition of continuity : A function f is said to be continuous at a point a of its domain D if for
every ε > 0 there exists δ > 0 (dependent on ε) such that | x − a|< δ ⇒|f(x) − f(a)|< ε.
Comparing this definition with the definition of limit we find that f(x) is continuous at x = a if lim f(x) exists
x→a
and is equal to f(a) i.e., if lim f(x) = f(a) = lim f(x) .
x→a− x→a+
Heine’s definition of continuity : A function f is said to be continuous at a point a of its domain D,
converging to a, the sequence < an > of the points in D converging to a, the sequence < f(an ) > converges to
f(a)i.e. lim an = a ⇒ lim f(an ) = f(a). This definition is mainly used to prove the discontinuity to a function.
Note : Continuity of a function at a point, we find its limit and value at that point, if these two exist and
are equal, then function is continuous at that point.
Formal definition of continuity : The function f(x) is said to be continuous at x = a, in its domain if for
any arbitrary chosen positive number ∈ > 0 , we can find a corresponding number δ depending on ∈ such
that| f (x) − f(a)| < ∈ ∀x for which 0 <| x − a|< δ .
Continuity from Left and Right.
Function f(x) is said to be
(1) Left continuous at x = a if lim f(x) = f(a)
x→a−0
(2) Right continuous at x = a if Lim f(x) = f(a) .
x→a+0
Thus a function f(x) is continuous at a point x = a if it is left continuous as well as right continuous at x = a.
x + λ, x < 3
Example: 1 If f(x) = 4, x = 3 is continuous at x = 3, then λ = [MP PET 1994, 2001; Rajasthan PET 1999]
Solution: (d)
3x − 5, x > 3 (d) 1
Example: 2 …..(i)
Solution: (c) (a) 4 (b) 3 (c) 2 …..(ii)
…..(iii)
Example: 3 L.H.L. at x = 3, lim f(x) = lim (x + λ) = lim (3 − h + λ) = 3 + λ
Solution: (c) x→3− x→3− h→0
Example: 4 R.H.L. at x = 3, lim f(x) = lim (3x − 5) = lim {3(3 + h) − 5} = 4
Solution: (d) x→3+ x→3+ h→0
Value of function f(3) = 4
For continuity at x = 3
Limit of function = value of function 3 + λ = 4 ⇒ λ = 1.
If f(x) = x sin 1 , x ≠ 0 is continuous at x = 0, then the value of k is [MP PET 1999; AMU 1999; Rajasthan PET 2003]
x
k, x = 0
(a) 1 (b) –1 (c) 0 (d) 2
If function is continuous at x = 0, then by the definition of continuity f(0)= lim f(x)
x→0
since f(0) = k . Hence, f(0) = k = lim (x) sin 1
x
x→0
⇒ k = 0 (a finite quantity lies between –1 to 1) ⇒ k = 0.
2x + 1 when x < 1
If f(x) = k when x = 1 is continuous at x =1, then the value of k is [Rajasthan PET 2001]
5x − 2 when x > 1 (d) 4
(a) 1 (b) 2 (c) 3
Since f(x) is continuous at x = 1,
⇒ lim f(x) = lim f(x) = f(1) …..(i)
x →1− x →1+
Now lim f(x) = lim f(1 − h) = lim 2(1 − h) + 1 = 3 i.e., lim f(x) = 3
x →1− h→0 h→0 x →1−
Similarly, lim f(x) = lim f(1 + h) = lim 5(1 + h) − 2 i.e., lim f(x) = 3
x →1+ h→0 h→0 x →1+
So according to equation (i), we have k = 3.
The value of k which makes f(x) = sin 1 , x≠0 continuous at x = 0 is [Rajasthan PET 1993; UPSEAT 1995]
x
k , x = 0
(a) 8 (b) 1 (c) –1 (d) None of these
We have lim f(x) = lim sin 1 = An oscillating number which oscillates between –1 and 1.
x
x→0 x→0
Hence, lim f(x) does not exist. Consequently f(x) cannot be continuous at x = 0 for any value of k.
x→0
Example: 5 The value of m for which the function f(x) = mx2, x ≤ 1 is continuous at x = 1 , is [MP PET 1998]
Solution: (c) 2x, x > 1
Example: 6 (a) 0 (b) 1 (c) 2 (d) Does not exist
Solution: (a)
LHL = lim f(x) = lim m(1 − h)2 = m
x →1− h→0
RHL = lim f(x) = lim 2(1 + h) = 2 and f(1) = m
x →1+ h→0
Function is continuous at x = 1 , ∴ LHL = RHL = f(1)
Therefore m = 2 .
If the function f (x) (cos x)1 / x,x ≠ 0 is continuous at x = 0 , then the value of k is [Haryana CEE 1996]
,x = 0
= k
(a) 1 (b) –1 (c) 0 (d) e
lim (cos x)1 / x = k ⇒ lim 1 log(cos x) = log k ⇒ lim 1 lim log cos x = log k ⇒ lim 1 × 0 = log e k ⇒ k =1.
x x x
x→0 x→0 x→0 x→0 x→0
Continuity of a Function in Open and Closed Interval.
Open interval : A function f(x) is said to be continuous in an open interval (a, b) iff it is continuous at every
point in that interval.
Note : This definition implies the non-breakable behavior of the function f(x) in the interval (a, b).
Closed interval : A function f(x) is said to be continuous in a closed interval [a, b] iff,
(1) f is continuous in (a, b)
(2) f is continuous from the right at ‘a’ i.e. lim f(x) = f(a)
x→a+
(3) f is continuous from the left at ‘b’ i.e. lim f(x) = f(b) .
x →b −
x + a2 2 sin x , 0 ≤ x < π
4
If the function f(x) = x cot x + b
Example: 7 , π ≤ x < π , is continuous in the interval [0, π] then the values of (a, b) are
Solution: (b) 4 2
π
2
b sin 2x − a cos 2x , ≤ x ≤ π
[Roorkee 1998]
(a) (–1, –1) (b) (0, 0) (c) (–1, 1) (d) (1, –1)
Since f is continuous at x = π ; ∴ f π = f π + h = f π − h ⇒ π (1) + b = π − 0 + a2 2 sin π − 0
4 4 h→0 4 h→0 4 4 4 4
⇒ π + b = π + a2 2 sin π ⇒ b = a2 2. 1 ⇒ b = a2
4 4 4 2
Also as f is continuous at x = π ; ∴ f π = lim f(x) = lim f π − h
2 2 2
x→π2 − 0 h→0
⇒ b sin 2 π − a cos 2 π = hli→m0(π2 − h) cot(π2 − h) + b ⇒ b . 0 − a(−1) = 0 + b ⇒ a = b .
2 2
Hence (0, 0) satisfy the above relations.
1 + sin πx for − ∞ < x ≤ 1
+b 2 for 1 < x < 3
Example: 8 If the function f (x) = ax is continuous in the interval (−∞, 6) then the values of a and b are
Solution: (c) xπ
6 tan 12 for 3≤ x <6
Example: 9
Solution: (a) respectively [MP PET 1998]
Example: 10
(a) 0, 2 (b) 1, 1 (c) 2, 0 (d) 2, 1
The turning points for f(x) are x = 1, 3.
So, lim f(x) = lim f(1 − h) = lim 1 + sin π (1 − h) = 1 + sin π − 0 =2
2 2
x →1− h→0 h→0
Similarly, lim f(x) = lim f(1 + h) = lim a(1 + h) + b = a + b
x →1+ h→0 h→0
f(x) is continuous at x = 1, so lim f(x) = lim f(x) = f(1)
x →1− x →1+
⇒ 2=a+b ..…..(i)
Again, lim f(x) = lim f(3 − h) = lim a(3 − h) + b = 3a + b and lim f(x) = lim f(3 + h) = lim 6 tan π (3 + h) =6
x→3− h→0 12
h→0 x→3+ h→0 h→0
f(x) is continuous in (−∞, 6) , so it is continuous at x = 3 also, so lim f(x) = lim f(x) = f(3)
x→3− x→3+
⇒ 3a + b = 6 ..….(ii)
Solving (i) and (ii) a = 2, b = 0.
Trick : In above type of questions first find out the turning points. For example in above question they are x = 1,3. Now
find out the values of the function at these points and if they are same then the function is continuous i.e., in above
problem.
1 + sin π x ; − ∞ < x ≤ 1, f(1) = 2
2 ; 1< x <3 f(1) = a + b, f(3) = 3a + b
f(x) = ax +b f(3) = 6
πx
6 12
tan ; 3≤ x <6
Which gives 2 = a + b and 6 = 3a + b after solving above linear equations we get a = 2, b = 0 .
If f(x) x sin x, when 0 < x ≤ π then
2
= π2 sin(π [IIT 1991]
π
+ x), when 2 < x < π
(a) f(x) is discontinuous at x = π (b) f(x) is continuous at x = π
2 2
(c) f(x) is continuous at x = 0 (d) None of these
lim f(x) = π , lim + f(x) = − π and f π = π . `
π− 2 2 2 2
x → 2 x → π
2
Since lim ≠ lim f(x) , ∴ Function is discontinuous at x = π
− + 2
x π x π
+ 2 + 2
1 − cos 4 x , when x < 0
, when x = 0 is continuous at x = 0, then the value of ‘a’ will be
x2
a
If f(x) = [IIT 1990; AMU 2000]
x , when x > 0
(16 + x ) − 4
(a) 8 (b) –8 (c) 4 (d) None of these
Solution: (a) lim f (x) lim 2 sin2 2x 4 = 8 and lim f(x) = lim [( 16 + x ) + 4] = 8
(2x)2 x→0+ x→0+
x→0− x→0−
Hence a = 8 .
Continuous Function.
(1) A list of continuous functions :
Function f(x) Interval in which f(x) is continuous
(–∞, ∞)
(i) Constant K (–∞, ∞)
(ii) xn, (n is a positive integer) (–∞, ∞) – {0}
(–∞, ∞)
(iii) x–n (n is a positive integer) (–∞, ∞)
(iv) |x – a| (–∞, ∞) – {x : q(x) = 0}
(v) p(x) = a0 xn + a1xn−1 + a2 xn−2 + ........ + an (–∞, ∞)
(–∞, ∞)
(vi) p(x) , where p(x) and q(x) are polynomial in x (–∞, ∞) – {(2n + 1)π/2 : n ∈ I}
q(x) (–∞, ∞) – {nπ : n ∈ I}
(–∞, ∞) – {(2n + 1)π /2 : n ∈ I}
(vii) sin x (–∞, ∞) – {nπ : n ∈ I}
(–∞, ∞)
(viii) cos x (0, ∞)
(ix) tan x
(x) cot x
(xi) sec x
(xii) cosec x
(xiii) e x
(xiv) loge x
(2) Properties of continuous functions : Let f(x) and g(x) be two continuous functions at x = a.Then
(i) cf(x) is continuous at x = a, where c is any constant
(ii) f(x) ± g(x) is continuous at x = a.
(iii) f(x) . g(x) is continuous at x = a.
(iv) f(x) / g(x) is continuous at x = a , provided g(a) ≠ 0 .
Important Tips
• A function f(x) is said to be continuous if it is continuous at each point of its domain.
• A function f(x) is said to be everywhere continuous if it is continuous on the entire real line R i.e. (−∞, ∞) . eg. polynomial function
ex, sin x, cos x, constant, xn, | x − a| etc.
• Integral function of a continuous function is a continuous function.
• If g(x) is continuous at x = a and f(x) is continuous at x = g(a) then (fog) (x) is continuous at x = a .
• If f(x) is continuous in a closed interval [a, b] then it is bounded on this interval.
• If f(x) is a continuous function defined on [a, b] such that f(a) and f(b) are of opposite signs, then there is atleast one value of x for which
f(x) vanishes. i.e. if f(a)>0, f(b) < 0 ⇒ ∃c ∈ (a, b) such that f(c) = 0.
• If f(x) is continuous on [a, b] and maps [a, b] into [a, b] then for some x ∈ [a, b] we have f(x) = x.
(3) Continuity of composite function : If the function u = f(x) is continuous at the point x = a, and the
function y = g(u) is continuous at the point u = f(a) , then the composite function y = (gof)(x) = g(f(x)) is
continuous at the point x = a.
Discontinuous Function.
(1) Discontinuous function : A function ‘f’ which is not continuous at a point x = a in its domain is said to
be discontinuous there at. The point ‘a’ is called a point of discontinuity of the function.
The discontinuity may arise due to any of the following situations.
(i) lim f(x) or lim f(x) or both may not exist
x→a+ x→a−
(ii) lim f(x) as well as lim f(x) may exist, but are unequal.
x→a+ x→a−
(iii) lim f(x) as well as lim f(x) both may exist, but either of the two or both may not be equal to f(a) .
x→a+ x→a−
Important Tips
• A function f is said to have removable discontinuity at x = a if lim f(x) = lim f(x) but their common value is not equal to f(a).
x+a+ x+a−
Such a discontinuity can be removed by assigning a suitable value to the function f at x = a.
• If lim f(x) does not exist, then we can not remove this discontinuity. So this become a non-removable discontinuity or essential
x→a
discontinuity.
• If f is continuous at x = c and g is discontinuous at x = c, then
(a) f +g and f – g are discontinuous (b) f.g may be continuous
• If f and g are discontinuous at x = c, then f + g, f – g and fg may still be continuous.
• Point functions (domain and range consists one value only) is not a continuous function.
Example: 11 The points of discontinuity of y= 1 where u= 1 is
Solution: (a) u2 + u − 2 x −1
Example: 12 (a) 1 ,1, 2 (b) −1 , 1, − 2 (c) 1 , − 1, 2 (d) None of these
Solution: (b) 2 2 2
The function u= f(x) = 1 is discontinuous at the point x = 1. The function y = g(x) = 1 = 1 is
x −1 u2 + u − 2 (u + 2)(u − 1)
discontinuous at u = −2 and u = 1
when u= −2⇒ 1 = −2 ⇒ x = 1 , when u =1⇒ 1 =1⇒ x = 2.
x −1 2 x −1
Hence, the composite y = g( f (x)) is discontinuous at three points = 1 , 1, 2.
2
The function f(x) = log(1 + ax) − log(1 − bx) is not defined at x = 0 . The value which should be assigned to f at x = 0 so
x
that it is continuous at x = 0, is [IIT 1983; MP PET 1995; Karnataka CET 1999; Haryana CEE 2002; AMU 2002]
(a) a − b (b) a + b (c) log a + log b (d) log a − log b
Since limit of a function is a + b as x → 0 , therefore to be continuous at x = 0 , its value must be a + b at
x = 0 ⇒ f(0) = a + b .
x 2 −4x + 3 , for x 1
x2 −1 , for x 1
Example: 13 If f(x) = ≠ , then [IIT 1972]
Solution: (c) = [Roorkee 1988]
Example: 14 2
Solution: (c)
(a) lim f(x) = 2 (b) lim f(x) = 3
x →1+ x →1−
(c) f(x) is discontinuous at x = 1 (d) None of these
f(1) = 2, f(1+) = lim x2 − 4x + 3 = lim (x − 3) = −1
x2 −1 (x + 1)
x →1+ x →1+
f(1−) = lim x2 − 4x + 3 = −1 ⇒ f(1) ≠ f(1−) . Hence the function is discontinuous at x =1.
x2 −1
x →1−
x − 1, x < 0
1
If f(x) = 4 , x = 0 , then
x2, x > 0
(a) lim f(x) = 1 (b) lim f(x) = 1
x→0+ x→0−
(c) f(x) is discontinuous at x = 0 (d) None of these
Clearly from curve drawn of the given function f(x) , it is discontinuous at x = 0 .
y y = x2, x > 0
x
(0,1/4)
x'
O
(0,–1)
y = x-1, x < 0
y'
a π
(1+ | sin x |)|sin x| , 6 x 0
− < <
Example: 15 Let f(x) = b, x=0 , then the values of a and b if f is continuous at x = 0, are respectively
Solution: (b)
tan 2x
π
e tan 3x , 0 < x < 6
(a) 2 , 3 (b) 2 , e 2 / 3 (c) 3 , e 3 / 2 (d) None of these
3 2 3 2
a x| π x 0
(1+ sin x |)|sin − 6 < <
| b ;
;
f(x) = tan 2x ; x=0
e tan 3x 0 < x < π
6
For f(x) to be continuous at x = 0
⇒ lim f(x) = f(0) = lim f(x) ⇒ lim(1+| sin a = ex lim |sin x||sina x| = ea
→0−
x |)|sin x|
x →0− x →0+ x→0
e tan 2 x .2 x tan 3 x .3 x
2x 3x
Now, lim e tan 2x / tan 3x = lim = lim e2 / 3 = e2 / 3.
x→0+ x→0+ x→0+
∴ ea = b = e2/ 3 ⇒ a= 2 and b = e2/ 3 .
3
Example: 16 Let f(x) be defined for all x > 0 and be continuous. Let f(x) satisfy f x = f(x) − f(y) for all x, y and f(e) = 1, then
Solution: (a) y
Example: 17
Solution: (c) [IIT 1995]
Example: 18 (a) f(x) = In x (b) f(x) is bounded (c) f 1 → 0 as x→0 (d) xf(x) → 1 as x → 0
Solution: (b) x
Let f(x) = In (x), x > 0 f(x) = In (x) is a continuous function of x for every positive value of x.
f x = In x = In (x) − In (y) = f(x) − f(y).
y y
Let f(x) = [x] sin π 1] , where [.] denotes the greatest integer function. The domain of f is ….. and the points of
[x +
discontinuity of f in the domain are [IIT 1996]
(a) {x ∈ R| x ∈ [−1, 0)}, I − {0} (b) {x ∈ R|x ∉ [1, 0)}, I − {0}
(c) {x ∈ R| x ∉ [−1, 0)}, I − {0} (d) None of these
Note that [x + 1] = 0 if 0 ≤ x + 1 < 1
i.e. [x + 1] − 0 if −1 ≤ x < 0.
Thus domain of f is R − [−1, 0) = {x ∉ [−1, 0)}
We have sin π 1] is continuous at all points of R − [−1, 0) and [x] is continuous on R − I, where I denotes the set of
[x +
integers.
Thus the points where f can possibly be discontinuous are……, −3, − 2, − 1, 0 1, 2, ........... But for 0 ≤ x < 1,[x] = 0 and
sin π 1] is defined.
[x +
Therefore f(x) = 0 for 0 ≤ x < 1.
Also f(x) is not defined on −1 ≤ x < 0.
Therefore, continuity of f at 0 means continuity of f from right at 0. Since f is continuous from right at 0, f is
continuous at 0. Hence set of points of discontinuities of f is I − {0}.
If the function f(x) = 2x − sin −1 x ,(x ≠ 0) is continuous at each point of its domain, then the value of f (0) is
2x + tan −1 x
[Rajasthan PET 2000]
(a) 2 (b) 1/3 (c) 2/3 (d) – 1/3
f(x) = lim 2x − sin −1 x = f (0) , 0 form
2x + tan −1 x 0
x→0
2 − 1
1− x2 2−1 1
Applying L-Hospital’s rule, f (0) = lim = 2+1 = 3
2 + 1
x→0 +x
1 2
Trick : f (0) = lim 2x − sin−1 x ⇒ lim 2− sin−1 x = 2 − 1 = 1 .
2x + tan−1 x 2+ x x 2 + 1 3
x→0 x→0
tan−1
x
− 2 sin x, x ≤ − π
2
Example: 19 The values of A and B such that the function f(x) = A sin x + B, − π < x < π , is continuous everywhere are
2 2
Solution: (c) π
2
Example: 20 cos x, x ≥
Solution: (a)
Example: 21 [Pb. CET 2000]
Solution: (c)
Example: 22 (a) A = 0, B = 1 (b) A = 1, B = 1 (c) A = −1, B = 1 (d) A = −1, B = 0
Solution: (a)
For continuity at all x ∈ R, we must have f − π = lim (−2 sin x) = lim (A sin x + B)
2
x→(−π / 2)− x→(−π / 2)+
⇒ 2 = −A + B …..(i)
and f π = lim (A sin x + B) = lim (cos x)
2
x→(π / 2)− x→(π / 2)+
⇒ 0= A+B ….(ii)
From (i) and (ii), A = −1 and B = 1 .
If f(x) = x 2 − 10x + 25 for x ≠ 5 and f is continuous at x = 5, then f(5) = [EAMCET 2001]
x 2 − 7x + 10
(a) 0 (b) 5 (c) 10 (d) 25
f (5) = lim f(x) = lim x2 − 10x + 25 = lim (x (x − 5)2 = 5−5 = 0 .
x 2 − 7x + 10 − 2)(x − 5) 5−2
x→5 x→5 x→5
In order that the function f(x) = (x + 1)cot x is continuous at x = 0 , f(0) must be defined as
(a) f (0) = 1 (b) f(0) = 0 (c) f(0) = e [UPSEAT 2000; Haryana CEE 2001]
e
(d) None of these
For continuity at 0, we must have f(0) = lim f(x)
x→0
xli→m0(1 + 1 x cot x xli→m0(1 + 1 lim x x
x tan
+ 1)cot x x) x x→0 = e1 = e .
= lim (x = = x)
x→0
The function f(x) = sin | x | is [DCE 2002]
(a) Continuous for all x (b) Continuous only at certain points
(c) Differentiable at all points (d) None of these
It is obvious.
1 − sin x , x ≠ π
2
Example: 23 If f(x) = π − 2x π be continuous at x = π , then value of λ is [Rajasthan PET 2002]
Solution: (c) 2 2
Example: 24 λ , x = (d) 2
Solution: (d)
Example: 25 (a) –1 (b) 1 (c) 0
Solution: (c)
Example: 26 f(x) is continuous at x = π , then lim f(x) = f (0) or λ = lim 2 1 − sin x , 0 form
Solution: (b) 2 π − 2x 0
x→π / 2 x→π /
Example: 27
Applying L-Hospital’s rule, λ = lim − cos x ⇒ λ = lim cos x = 0.
−2 2
x→π / 2 x→π / 2
If f (x) = 2− x+4 ; (x ≠ 0), is continuous function at x = 0 , then f (0) equals [MP PET 2002]
sin 2x
(a) 1 (b) − 1 (c) 1 (d) − 1
4 4 8 8
If f(x) is continuous at x = 0, then, f(0) = lim f(x) = lim 2 − x+ 4 , 0 form
sin 2x 0
x→0 x→0
Using L–Hospital’s rule, f (0) = lim 1 = − 1 .
− 8
x→0
2 x+4
2 cos 2x
If function f(x) = x x if x is rational , then f(x) is continuous at ...... number of points
1 − if x is irrational
[UPSEAT 2002]
(a) ∞ (b) 1 (c) 0 (d) None of these
At no point, function is continuous.
The function defined by f(x) = x 2 + e 1 −1 , x ≠ 2 , is continuous from right at the point x = 2, then k is equal to
2− x
k , x=2
[Orissa JEE 2002]
(a) 0 (b) 1/4 (c) –1/4 (d) None of these
1 −1
f(x) = x 2 + e 2−x and f(2) = k
If f(x) is continuous from right at x = 2 then lim f(x) = f(2) = k
x→2+
1 −1 1 −1
⇒ lim x2 + e 2−x (2 h)2 2−(2+h)
=k ⇒ k = lim f(2 + h) ⇒ k = lim + + e
x 2+
→ h→0 h→0
[ ]⇒ k = lim 4 + h2 + 4h + e −1/ h −1 ⇒ k = [4 + 0 + 0 + e −∞ ] −1 ⇒ k= 1 .
h→0 4
The function f(x) = 1− sin x + cos x is not defined at x =π. The value of f(π ), so that f(x) is continuous at x = π , is
1+ sin x + cos x
[Orissa JEE 2003]
(a) − 1 (b) 1 (c) – 1 (d) 1
2 2
[EAMCET 2003]
Solution: (c) lim f(x) = lim 2 cos2 x − 2 sin x cos x lim cos x − sin x lim tan π x
2 cos2 2 + 2 sin 2 cos 2 cos 2 2 4 2 (d) – 1
x →π x →π x x x = x →π x x = x →π −
2 2 2 2 2 [Karnataka CET 2003]
+ sin
∴ At x =π, f (π ) = − tan π = −1 .
4
1 + kx − 1 − kx , for − 1 ≤ x < 0 is continuous at x = 0, then k =
x , for 0 ≤ x ≤ 1
Example: 28 If f(x) =
2x 2 + 3x − 2
(a) – 4 (b) – 3 (c) – 2
Solution: (c) L.H.L. = lim 1 + kx − 1 − kx = k
Example: 29 x
x→0−
Solution: (c)
R.H.L. = lim (2x2 + 3x − 2) = −2
x→0+
Since it is continuous, hence L.H.L = R.H.L ⇒ k = −2 .
The function f(x) =| x|+ | x| is
x
(a) Continuous at the origin
(b) Discontinuous at the origin because |x| is discontinuous there
(c) Discontinuous at the origin because |x| is discontinuous there
x
(d) Discontinuous at the origin because both |x| and |x| are discontinuous there
x
| x | is continuous at x =0 and |x| is discontinuous at x=0
x
∴ f(x) =| x | + | x | is discontinuous at x=0.
x
Funtion limit continuity & differentiability
Differentiability of a Function at a Point.
(1) Meaning of differentiability at a point : Consider the [a + h, f (a+h)]
function f(x) defined on an open interval (b, c) let P(a, f(a)) be a R
point on the curve y = f(x) and let Q (a − h, f(a − h)) and Y
[a–h, f (a–h)]
R(a + h, f(a + h)) be two neighbouring points on the left hand side and Q P [a, f (a)]
right hand side respectively of the point P.
O
Then slope of chord PQ = f(a − h) − f(a) = f(a − h) − f(a)
(a − h) − a −h
X
and, slope of chord PR = f(a + h) − f (a) = f (a + h) − f (a) .
a+h − a h
As h → 0, point Q and R both tends to P from left hand and right hand respectively. Consequently, chords
PQ and PR becomes tangent at point P.
Thus, lim f(a − h) − f(a) = lim (slope of chord PQ)= lim (slope of chord PQ)
−h h→0
h→0 Q→P
Slope of the tangent at point P, which is limiting position of the chords drawn on the left hand side of point P
and lim f(a + h) − f(a) = lim (slope of chord PR) = lim (slope of chord PR).
h
h→0 h→0 R→P
⇒ Slope of the tangent at point P, which is the limiting position of the chords drawn on the right hand side of
point P.
Now, f(x) is differentiable at x = a ⇔ lim f(a − h) − f (a) = lim f(a + h) − f (a)
−h h
h→0 h→0
⇔ There is a unique tangent at point P.
Thus, f(x) is differentiable at point P, iff there exists a unique tangent at point P. In other words, f(x) is
differentiable at a point P iff the curve does not have P as a corner Y
point. i.e., "the function is not differentiable at those points on which
function has jumps (or holes) and sharp edges.”
Let us consider the function f(x) =| x − 1|, which can be f (x)=–x+1 f (x) = x – 1
graphically shown, f ' (x)= –1 f ' (x) = 1
Which show f(x) is not differentiable at x = 1 . Since, f(x) has O 1 23 X
sharp edge at x = 1 .
Mathematically : The right hand derivative at x = 1 is 1 and
left-hand derivative at x = 1 is –1. Thus, f(x) is not differentiable at
x =1.
(2) Right hand derivative : Right hand derivative of f(x) at x = a , denoted by f ' (a + 0) or f ' (a+) , is the
lim f(a + h) − f(a) .
h
h→0
(3) Left hand derivative : Left hand derivative of f(x) at x = a, denoted by f ' (a − 0) or f ' (a−), is the
lim f(a − h) − f(a) .
−h
h→0
(4) A function f(x) is said to be differentiable (finitely) at x = a if f ' (a + 0) = f ' (a − 0) = finite
i.e., lim f(a + h) − f (a) = lim f(a − h) − f (a) = finite and the common limit is called the derivative of f(x) at
h −h
h→0 h→0
x = a , denoted by f ' (a) . Clearly, f ' (a) = lim f(x) − f(a) {x → a from the left as well as from the right}.
x→a x−a
Example: 1 Consider f(x) = x2 , x ≠ 0 [EAMCET 1994]
Solution: (b) |x|
Example: 2 0 , x = 0
Solution: (a)
(a) f(x) is discontinuous everywhere
Example: 3
Solution: (b) (b) f(x) is continuous everywhere but not differentiable at x = 0
(c) f '(x) exists in (–1, 1)
(d) f '(x) exists in (–2, 2)
We have, f(x) = |xx2|, x ≠ 0 = x2 = x, x > 0
x ,x =0
0
0 , x = 0 x2
= −x, x < 0
− x
⇒ lim f(x) = lim − x = 0 , lim f(x) = lim x = 0 and f(0) = 0.
x→0− x→0− x→0+ x→0+
So f(x) is continuous at x = 0 . Also f(x) is continuous for all other values of x. Hence, f(x) is everywhere continuous.
Also, Rf ' (0) = f ' (0 + 0) = lim f(h) − f(0) = lim h − 0 =1
h−0 h
h→0 h→0
i.e. Rf ' (0) = 1 and Lf ' (0) = f ' (0 − 0) = lim f(−h) − f(0) = lim h = −1
−h −h
h→0 h→0
i.e. Lf ' (0) = −1 So, Lf ' (0) ≠ Rf ' (0) i.e., f(x) is not differentiable at x = 0 .
If the function f is defined by f (x) = 1+ x x | , then at what points f is differentiable [EAMCET 1992]
|
(a) Everywhere (b) Except at x = ±1 (c) Except at x = 0 (d) Except at x = 0 or ± 1
x , x >0 −h
1 , x =0; 1+h
x + x , f(−h) − f (0) − 0
| 0 −h
We have, f(x) = 1+ x | = 1 x Lf ' (0) = lim = lim −h =1
− x
h→0 h→0
x<0
f(h) − f(0) lim 1 h h − 0 1
h + +
Rf ' (0) = lim = h→0 = lim =1
h→0 h h→0 1 h
So, Lf ' (0) = Rf ' (0) = 1
So, f(x) is differentiable at x = 0 ; Also f(x) is differentiable at all other points.
Hence, f(x) is everywhere differentiable.
The value of the derivative of | x − 1| + | x − 3| at x = 2 is [UPSEAT 1993]
(a) –2 (b) 0 (c) 2 (d) Not defined
− (x − 1) − (x − 3) , x < 1 − 2x + 4 , x < 1
Let f(x) =| x − 1| + | x − 3| = (x − 1) − (x − 3)
, 1≤ x <3 = 2 , 1≤ x <3
(x − 1) + (x − 3) , x ≥ 3 2x − 4 , x ≥ 3
Since, f(x) = 2 for 1 ≤ x < 3 . Therefore f ' (x) = 0 for all x ∈(1, 3) .
Hence, f ' (x) = 0 at x = 2 .
sin x 2 , x≠0
x
Example: 4 The function f defined by f(x) =
Solution: (a)
Example: 5 0 , x = 0
Solution: (b) (a) Continuous and derivable at x = 0 (b) Neither continuous nor derivable at x = 0
Example: 6 (c) Continuous but not derivable at x = 0 (d) None of these
Solution: (a)
We have, lim f(x) lim sin x 2 xli→m0 sin x 2 x 1× 0 0 f (0)
Example: 7 x x
Solution: (a) x→0 = x→0 = 2 = = =
So, f(x) is continuous at x = 0 , f(x) is also derivable at x = 0 ,
because lim f(x) − f (0) = lim sin x 2 = 1 exists finitely. Y
x− 0 x2
x→0 x→0
If f(x) =|log| x||, then
(a) f(x) is continuous and differentiable for all x in its domain f(x)=|log|x|
(b) f(x) is continuous for all x in its domain but not differentiable X' (–1, 0) O (1, 0) X
at x = ±1 .
(c) f(x) is neither continuous nor differentiable at x = ±1 Y'
(d) None of these
It is evident from the graph of f(x) = | log | x || that f(x) is everywhere continuous but not differentiable at x = ±1. .
The left hand derivative of f(x) = [x] sin(πx) at x = k (k is an integer), is
(a) (−1)k (k − 1)π (b) (−1)k−1(k − 1)π (c) (−1)k kπ (d) (−1)k−1 kπ
f(x) = [x] sin(π x)
If x is just less than k, [x] = k – 1. ∴ f(x) = (k − 1)sin(πx) , when x < k ∀k ∈ I
Now L.H.D. at x = k ,
= lim (k − 1) sin(π x) − k sin(π k) = lim (k − 1)sin(π x) [as sin(π k) = 0, k ∈ integer]
x −k (x − k)
x→k x→k
= lim (k − 1) sin(π (k − h)) [Let x = (k − h) ]
−h
h→0
= lim (k − 1)(−1)k −1 sin hπ = lim (k − 1)(−1)k −1 sin hπ × (−π ) = (k − 1)(−1)k π = (−1)k (k − 1)π .
−h hπ
h→0 h→0
The function f(x) =| x|+| x − 1| is [Haryana CEE 2002]
(a) Continuous at x = 1 , but not differentiable (b) Both continuous and differentiable at x = 1
(c) Not continuous at x = 1 (d) None of these
− 2x + 1, x < 0
We have, f(x) =| x |+ | x − 1|= 1 , 0≤ x <1 Y
2x − 1, x ≥ 1
Since, lim f(x) = lim 1 = 1, lim f(x) = lim (2x − 1) = 1 and f(1) = 2 × 1 − 1 = 1 y=2x–1
x →1− x →1− x →1+ x →1+
y = –2x+1 y=1
∴ lim f(x) = lim f(x) = f(1) . So, f(x) is continuous at x = 1.
x →1− x →1+
Now, lim f(x) − f (1) lim f(1 − h) − f (1) lim 1−1 0 , x =1/2 x =1 X
x − 1 −h −h
x →1− = h→0 = h→0 =
and lim f(x) − f (1) = lim f(1 + h) − f (1) = lim 2(1 + h) − 1 − 1 = 2 .
x − 1 h h
x →1+ h→0 h→0
∴ (LHD at x = 1 ) ≠ (RHD at x = 1 ). So, f(x) is not differentiable at x = 1.
− 2x + 1 , x < 0
Trick : The graph of f(x) =| x | +| x − 1| i.e. f(x) = 1 , 0 ≤ x < 1 is
2x − 1 , x ≥ 1
By graph, it is clear that the function is not differentiable at x = 0 , 1 as there it has sharp edges.
Example: 8 Let f(x) =| x − 1| + | x + 1|, then the function is
Solution: (c)
(a) Continuous (b) Differentiable except x = ±1
Example: 9 (c) Both (a) and (b) (d) None of these
Solution: (a)
2x , when x > 1
Here f(x) = | x − 1|+| x + 1| ⇒ f(x) = 2 , when − 1 ≤ x ≤ 1
− 2x , when x < −1
Graphical solution : The graph of the function is shown alongside,
From the graph it is clear that the function is continuous at all real x, also differentiable at all real x except at x = ±1; Since
sharp edges at x = −1 and x = 1 .
At x = 1 we see that the slope from the right i.e., R.H.D. = 2, while slope from the left i.e., L.H.D.= 0
Similarly, at x = −1 it is clear that R.H.D. = 0 while L.H.D. = – 2
Trick : In this method, first of all, we differentiate the function and on the Y
derivative equality sign should be removed from doubtful points.
− 2 , x < −1
Here, f' (x) = 0 , − 1 < x < 1 (No equality on –1 and +1) 2
2 , x > 1
Now, at x = 1, f ' (1+) = 2 while f ' (1−) = 0 and X’ –1 O 1X
at x = −1, f ' (−1+) = 0 while f ' (−1−) = −2
Thus, f(x) is not differentiable at x = ±1 . Y’
Note : This method is not applicable when function is discontinuous.
If the derivative of the function f(x) = ax 2 +b + 4 , x < −1 is everywhere continuous and differentiable at x = 1 then
bx 2 + ax , x ≥ −1
(a) a = 2, b = 3 (b) a = 3, b = 2 (c) a = –2, b = – 3 (d) a = – 3, b = – 2
f(x) = ax 2 + b + , x < −1
bx 2 + ax 4, x ≥ −1
∴ f ' (x) = 2ax , x < −1
, x ≥ −1
2bx + a
To find a, b we must have two equations in a, b
Since f(x) is differentiable, it must be continuous at x = −1 .
∴ R = L = V at x = −1 for f(x) ⇒ b − a + 4 = a + b
∴ 2a = 4 i.e., a = 2
Again f ' (x) is continuous, it must be continuous at x = −1 .
∴R = L = V at x = −1 for f ' (x)
−2b + a = −2a. Putting a = 2, we get −2b + 2 = −4
Example: 10 ∴ 2b = 6 or b = 3.
Solution: (b)
Let f be twice differentiable function such that f " (x) = − f(x) and f ' (x) = g(x), h(x) = {f(x)}2 + {g(x)}2 . If h(5) = 11, then
Example: 11 h(10) is equal to
Solution: (c)
(a) 22 (b) 11 (c) 0 (d) None of these
Differentiating the given relation h(x) = [ f(x)]2 + [g(x)]2 w.r.t x, we get h'(x) = 2 f(x)f '(x) + 2g(x)g'(x) .......(i)
But we are given f ' ' (x) = − f(x) and f ' (x) = g(x) so that f ' ' (x) = g' (x).
Then (1) may be re-written as h' (x) = −2 f ' ' (x)f ' (x) + 2 f ' (x)f ' ' (x) = 0. Thus h′(x) = 0
Whence by intergrating, we get h(x) = constant = c (say). Hence h(x) = c, for all x.
In particular, h(5) = c. But we are given h(5) = 11.
It follows that c = 11 and we have h(x) = 11 for all x. Therefore, h(10) = 11.
|si2nx − 3|[x] , x ≥1
, x <1
The function f(x) = πx
2
(a) Is continuous at x = 2 (b) Is differentiable at x = 1
(c) Is continuous but not differentiable at x = 1 (d) None of these
[2 + h] = 2,[2 − h] = 1,[1 + h] = 1,[1 − h] = 0
At x = 2, we will check R = L = V
R = lim |4 + 2h − 3|[2 + h] = 2, V = 1. 2 = 2
h→0
L = lim | 4 − 2h − 3 |[2 − h] = 1, R ≠ L , ∴ not continuous
h→0
At x = 1, R = lim| 2 + 2h − 3|[1 + h] = 1.1 = 1,
V =|−1|[1] = 1
L = lim sin π (1 − h) = 1
2
h→0
Since R = L = V ∴ continuous at x = 1.
R.H.D. = lim | 2 + 2h − 3|[1 + h] − 1 = lim | −1|.1 − 1 = lim 1 − 1 = 0
h h h
h→0 h→0 h→0
L.H.D. = lim | 2 − 2h − 3|[1 − h] − 1 = lim 1.0 − 1 = lim 1 =∞
−h − h h
h→0 h→0 h→0
Since R.H.D. ≠ L.H.D. ∴ not differentiable. at x = 1 .
Differentiability in an Open Interval.
A function f(x) defined in an open interval (a, b) is said to be differentiable or derivable in open interval (a, b)
if it is differentiable at each point of (a, b).
Differentiability in a closed interval : A function f : [a, b] → R is said to be differentiable in [a, b] if
(1) f ' (x) exists for every x such that a < x < b i.e. f is differentiable in (a, b).
(2) Right hand derivative of f at x = a exists.
(3) Left hand derivative of f at x = b exists.
Everywhere differentiable function : If a function is differentiable at each x ∈ R , then it is said to be
everywhere differentiable. e.g., A constant function, a polynomial function, sin x, cos x etc. are everywhere
differentiable.
Some standard results on differentiability
(1) Every polynomial function is differentiable at each x ∈ R .
(2) The exponential function ax, a > 0 is differentiable at each x ∈ R .
(3) Every constant function is differentiable at each x ∈ R .
(4) The logarithmic function is differentiable at each point in its domain.
(5) Trigonometric and inverse trigonometric functions are differentiable in their domains.
(6) The sum, difference, product and quotient of two differentiable functions is differentiable.
(7) The composition of differentiable function is a differentiable function.
Important Tips
• If f is derivable in the open interval (a, b) and also at the end points ‘a’ and ‘b’, then f is said to be derivable in the closed interval [a, b].
• A function f is said to be a differentiable function if it is differentiable at every point of its domain.
• If a function is differentiable at a point, then it is continuous also at that point.
i.e. Differentiability ⇒ Continuity, but the converse need not be true.
• If a function ‘f’ is not differentiable but is continuous at x = a, it geometrically implies a sharp corner or kink at x = a.
• If f(x) is differentiable at x = a and g(x) is not differentiable at x =a, then the product function f(x).g(x) can still be differentiable at x = a.
• If f(x) and g(x) both are not differentiable at x = a then the product function f(x).g(x) can still be differentiable at x = a.
• If f(x) is differentiable at x = a and g(x) is not differentiable at x = a then the sum function f(x) + g(x) is also not differentiable at x = a
• If f(x) and g(x) both are not differentiable at x = a, then the sum function may be a differentiable function.
Example: 12 The set of points where the function f(x) = 1 − e −x2 is differentiable
Solution: (b)
(a) (−∞, ∞) (b) (−∞, 0) ∪ (0, ∞) (c) (−1, ∞) (d) None of these
Example: 13
Solution: (a) Clearly, f(x) is differentiable for all non-zero values of x, For x ≠ 0 , we have f'(x) = xe −x2
Now, (L.H.D. at x = 0) 1 − e−x2
= lim f(x) − f (0) = lim f(0 − h) − f(0) = lim 1 − e −h2 = lim − 1 − e −h2 = − lim eh2 − 1 × 1 = −1
x − 0 −h −h h h2 e h2
x→0− h→0 h→0 h→0 h→0
and, (RHD at x = 0) = lim f(x) − f (0) = lim 1 − e −h2 −0 = lim e h2 − 1 × 1 =1.
x − 0 h h2 e h2
x→0+ h→0 h→0
So, f(x) is not differentiable at x = 0 , Hence, the points of differentiability of f(x) are (−∞,0) ∪ (0, ∞) .
The function f(x) = e −|x| is [AMU 2000]
(a) Continuous everywhere but not differentiable at x = 0
(b) Continuous and differentiable everywhere
(c) Not continuous at x = 0
(d) None of these
We have, f (x) = e −x , x ≥ 0
e x , x < 0
Clearly, f(x) is continuous and differentiable for all non-zero x.
Now, lim f(x) = lim ex = 1 and lim f(x) = lim e−x = 1
x→0− x→0 x→0+ x→0
Y
Also, f(0) = e 0 = 1
f(x)=e–|x|
So, f(x) is continuous for all x. (0, 1)
(LHD at x =0)= d (e x ) = [e x ]x=0 = e0 =1 X' O y X
dx Y'
x=0
(RHD at x =0)= d (e −x ) = [−e −x ]x=0 = −1
dx x=0
So, f(x) is not differentiable at x = 0 .
Hence, f(x) = e −|x| is everywhere continuous but not differentiable at x = 0. This fact is also evident from the graph of the
function.
Example: 14 If f(x) = 1 − 1 − x 2 , then f(x) is (b) Continuous on [–1,1] and differentiable on (–1, 0) ∪ (0, 1)
Solution: (b) (d) None of these
(a) Continuous on [–1, 1] and differentiable on (–1, 1)
Example: 15 (c) Continuous and differentiable on [–1, 1]
Solution: (d)
Example: 16 We have, f(x) = 1 − 1 − x 2 . The domain of definition of f(x) is [–1, 1].
Solution: (d)
For x ≠ 0, x ≠ 1, x ≠ −1 we have f '(x) = 1 × x
Note : 1− x2
1− 1− x2
Since f(x) is not defined on the right side of x = 1 and on the left side of x = −1 . Also, f ' (x) → ∞ when x → −1+ or
x → 1− . So, we check the differentiability at x = 0.
Now, (LHD at x = 0)= lim f(x) − f(0) = lim f(0 − h) − f (0)
x−0 −h
x→0 h→0
= lim 1− 1 − h2 −0 = − lim 1 − {1 − (1 / 2)h2 + (3 / 8)h4 + ....} = − lim 1 − 3 h2 + ..... = − 1
−h h 2 8 2
h→0 h→0 h→0
Similarly, (RHD at x = 0) = 1
2
Hence, f(x) is not differentiable at x = 0.
Let f(x) be a function differentiable at x = c . Then lim f(x) equals
x→c
(a) f '(c) (b) f " (c) (c) 1 (d) None of these
f (c)
Since f(x) is differentiable at x = c, therefore it is continuous at x = c . Hence, lim f(x) = f(c) .
x→c
The function f(x) = (x 2 − 1)| x 2 − 3x + 2|+ cos(| x|) is not differentiable at [IIT Screening 1999]
(a) – 1 (b) 0 (c) 1 (d) 2
(x 2 − 3x + 2) = (x − 1)(x − 2) = +ive
When x < 1 or > 2, –ive when 1 ≤ x ≤ 2
Also cos| x|= cos x (since cos(−x) = cos x )
∴ f(x) = −(x 2 − 1)(x 2 − 3x + 2) + cos x, 1 ≤ x ≤ 2
∴ f(x) = (x 2 − 1)(x 2 − 3x + 2) + cos x, x > 2 .........(i)
Evidently f(x) is not differentiable at x = 2 as L′ ≠ R′
For all other values like x < 0, 0 ≤ x < 1, f(x) is same as given by (i).
Example: 17 If f (x) xe− |1x| + 1 x ≠ 0 , then f(x) is
Solution: (b) x x=0
= , [AIEEE 2003]
Example: 18
Solution: (d) 0 , (b) Continuous for all x but not differentiable at x = 0
(d) Discontinuous every where
Example: 19
Solution: (b) (a) Continuous as well as differentiable for all x
Example: 20
(c) Neither differentiable nor continuous at x = 0
f(0) = 0 and f(x) = xe − |1x|+ 1
x
R.H.L. = lim (0 + h)e−2 / h = lim h =0
e2 / h
h→0 h→0
L.H.L. = lim (0 h)e − 1 − 1 =0
h h
−
h→0
∴ f(x) is continuous.
Rf ′(x) at (x = 0) = lim f(0 + h) − f(0) = lim he −2 / h = e−∞ =0
h h
h→0 h→0
f(0 − h) − f (0) he − 1 − 1
−h h h
Lf ′(x) at (x = 0) = lim = lim − = +1 ⇒ Lf ′(x) ≠ Rf ′(x)
h→0 h→0 −h
f(x) is not differentiable at x = 0.
The function f(x) = x 2 sin 1 , x ≠ 0, f(0) = 0 at x=0 [MP PET 2003]
x
(a) Is continuous but not differentiable (b) Is discontinuous
(c) Is having continuous derivative (d) Is continuous and differentiable
lim f(x) = x 2 sin 1 but − 1 ≤ sin 1 ≤ 1 and x→0
x x
x→0
∴ lim f(x) = 0 = lim f(x) = f(0)
x→0+ x→0−
Therefore f(x) is continuous at x = 0 . Also, the function f(x) = x 2 sin 1 is differentiable because
x
h2 sin 1 −0 0, h2 sin 1 .
h −h
Rf ′(x) = lim = Lf ′(x) = lim = 0
h −h
h→0 h→0
Which of the following is not true [Kerala (Engg.) 2002]
(a) A polynomial function is always continuous (b) A continuous function is always differentiable
(c) A differentiable function is always continuous (d) ex is continuous for all x
A continuous function may or may not be differentiable. So (b) is not true.
If f(x) = sgn (x 3 ), then [DCE 2001]
(a) f is continuous but not derivable at x = 0 (b) f '(0+) = 2
(c) f '(0 − ) = 1 (d) f is not derivable at x = 0
x 3 | for x3 ≠ 0
| x 3
0 for x3 = 0
|
Solution: (d) Here, f(x) = sgn x3 = x | for x≠0 . Thus, f(x) = sgn x 3 = sgn x, which is neither continuous nor derivable at 0.
x
Note : 0 for x = 0
− 1, x < 0
Example: 21
Solution: (c) 0, x=0
Example: 22 1, x > 0
Solution: (c)
f ′(0 + ) = lim f(0 + h) − f(0) = lim 1 − 0 → ∞ and f ′(0 − ) = lim f(0 − h) − f(0) = lim −1 − 0 → ∞ .
h h h h
h→0+ h→0+ h→0− h→0−
∴ f ′(0+ ) ≠ f ′(0− ) , ∴ f is not derivable at x = 0 .
A function f(x) = 1 + x, x≤2 is [AMU 2001]
5 − x, x>2 [IIT Screening 2000]
(a) Not continuous at x = 2 (b) Differentiable at x = 2
(c) Continuous but not differentiable at x = 2 (d) None of the above
lim 1 + (2 − h) = 3 , lim 5 − (2 + h) = 3 , f(2) = 3
h→0− h→0+
Hence, f is continuous at x = 2
Now Rf ′(x) = lim 5 − (2 + h) − 3 = −1
h
h→0
Lf ′(x) = lim 1 + (2 − h) − 3 = 1
−h
h→0
Rf ′(x) ≠ Lf ′(x)
∴ f is not differentiable at x = 2 .
Let f : R → R be a function. Define g : R → R by g(x) =| f(x)| for all x. Then g is
(a) Onto if f is onto (b) One-one if f is one-one
(c) Continuous if f is continuous (d) Differentiable if f is differentiable
g(x) =| f(x)| ≥ 0 . So g(x) cannot be onto. If f(x) is one-one and f(x1) = − f(x 2 ) then g(x1) = g(x 2 ) . So, ‘ f(x) is one-
one’ does not ensure that g(x) is one-one.
If f(x) is continuous for x ∈ R,| f(x)| is also Y y = f (x) y = | f (x)| Y y = | f (x)|
continuous for x ∈ R . This is obvious from the
following graphical consideration. PO X PO X
y = f (x)
So the answer (c) is correct. The fourth answer
(d) is not correct from the above graphs y = f(x)
is differentiable at P while y =| f(x)| has two
tangents at P, i.e. not differentiable at P.
Determinants and matrices
Definition .
(1) Consider two equations, a1 x + b1y = 0 .....(i) and a2 x + b2y = 0 .....(ii)
Multiplying (i) by b2 and (ii) by b1 and subtracting, dividing by x, we get, a1b2 − a2b1 = 0
The result a1b2 − a2b1 is represented by a1 b1
a2 b2
Which is known as determinant of order two and a1b2 − a2b1 is the expansion of this determinant. The
horizontal lines are called rows and vertical lines are called columns.
Now let us consider three homogeneous linear equations
a1 x + b1y + c1z = 0 , a2 x + b2y + c 2 z = 0 and a3 x + b3y + c3 z = 0
Eliminated x, y, z from above three equations we obtain
a1(b2c 3 − b3c 2 ) − b1(a2c 3 − a3c 2 ) + c1(a2b3 − a3b2 ) = 0 .....(iii)
a1 b1 c1
The L.H.S. of (iii) is represented by a2 b2 c2
a3 b3 c3
Its contains three rows and three columns, it is called a determinant of third order.
Note : The number of elements in a second order is 22 = 4 and the number of elements in a third
order determinant is 32 = 9 .
(2) Rows and columns of a determinant : In a determinant horizontal lines counting from top 1st, 2nd,
3rd,….. respectively known as rows and denoted by R1, R2, R3 , ...... and vertical lines counting left to right, 1st, 2nd,
3rd,….. respectively known as columns and denoted by C1, C2, C3 ,.....
(3) Shape and constituents of a determinant : Shape of every determinant is square. If a determinant of
n order then it contains n rows and n columns.
i.e., Number of constituents in determinants = n2
(4) Sign system for expansion of determinant : Sign system for order 2, order 3, order 4,….. are given
by + − , +−+ +−+−
−+ − + −, − + − + ,.....
+−+−
+−+ −+−+
Expansion of Determinants .
Unlike a matrix, determinant is not just a table of numerical data but (quite differently) a short hand way of
writing algebraic expression, whose value can be computed when the values of terms or elements are known.
(1) The 4 numbers a1,b1, a2 ,b2 arranged as a1 b1 is a determinant of second order. These numbers are
a2 b2
called elements of the determinant. The value of the determinant is defined as a1 b1 = a1b2 − a2b1 .
a2 b2
The expanded form of determinant has 2! terms.
a1 b1 c1
(2) The 9 numbers ar , br , cr (r = 1, 2, 3)arranged as a2 b2 c2 is a determinant of third order. Take any
a3 b3 c3
row (or column); the value of the determinant is the sum of products of the elements of the row (or column) and the
corresponding determinant obtained by omitting the row and the column of the element with a proper sign, given
by the rule (−1)i+ j , where i and j are the number of rows and the number of columns respectively of the element of
the row (or the column) chosen. Thus a1 b1 c1 = a1 b2 c2 − b1 a2 c2 + c1 a2 b2
a2 b2 c2 b3 c3 a3 c3 a3 b3
a3 b3 c3
The diagonal through the left-hand top corner which contains the element a1, b2, c3 is called the leading diagonal
or principal diagonal and the terms are called the leading terms. The expanded form of determinant has 3! terms.
Short cut method or Sarrus diagram method : To find the value of third order determinant, following
method is also useful
a1 b1 c1 a1 b1
a2 c2 a2 b2
b2 c3 a3 b3
a3 b3
Taking product of R.H.S. diagonal elements positive and L.H.S. diagonal elements negative and adding them.
We get the value of determinant as = a1b2c3 + b1c 2a3 + c1a2b3 − c1b2a3 − a1c 2b3 − b1a2c3
Note : This method does not work for determinants of order greater than three.
Evaluation of Determinants .
If A is a square matrix of order 2, then its determinant can be easily found. But to evaluate determinants of
square matrices of higher orders, we should always try to introduce zeros at maximum number of places in a
particular row (column) by using the properties and then we should expand the determinant along that row
(column).
We shall be using the following notations to evaluate a determinant :
(1) Ri to denote ith row.
(2) Ri ↔ R j to denote the interchange of ith and j th rows.
(3) Ri → Ri + λR j to denote the addition of λ times the elements of j th row to the corresponding elements
of ith row.
(4) Ri (λ) to denote the multiplication of all element of ith row by λ .
Similar notations are used to denote column operations if R is replaced by C.
Properties of Determinants .
P-1 : The value of determinant remains unchanged, if the rows and the columns are interchanged.
a1 b1 c1 a1 a2 a3
If D = a2 b2 c2 and D' = b1 b2 b3 . Then D' = D, D and D' are transpose of each other.
a3 b3 c3 c1 c2 c3
Note : Since the determinant remains unchanged when rows and columns are interchanged, it is
obvious that any theorem which is true for ‘rows’ must also be true for ‘columns’.
P-2 : If any two rows (or columns) of a determinant be interchanged, the determinant is unaltered in
numerical value but is changed in sign only.
a1 b1 c1 a2 b2 c2
Let D = a2 b2 c 2 and D' = a1 b1 c1 . Then D' = −D
a3 b3 c3 a3 b3 c3
P-3 : If a determinant has two rows (or columns) identical, then its value is zero.
a1 b1 c1
Let D = a1 b1 c1 . Then, D = 0
a2 b2 c2
P-4 : If all the elements of any row (or column) be multiplied by the same number, then the value of
determinant is multiplied by that number.
a1 b1 c1 ka1 kb1 kc1
Let D = a2 b2 c 2 and D′ = a2 b2 c 2 . Then D' = kD
a3 b3 c3 a3 b3 c3
P-5 : If each element of any row (or column) can be expressed as a sum of two terms, then the determinant
can be expressed as the sum of the determinants.
a1 + x b1 + y c1 + z a1 b1 c1 xyz
e.g., a2 b2 c2 = a2 b2 c2 + a2 b2 c2
a3 b3 c3 a3 b3 c3 a3 b3 c3
P-6 : The value of a determinant is not altered by adding to the elements of any row (or column) the same
multiples of the corresponding elements of any other row (or column)
a1 b1 c1 a1 + ma2 b1 + mb2 c1 + mc 2
e.g., D = a2 b2 c 2 and D' = a2 b2 c 2 . Then D' = D
a3 b3 c3 a3 − na1 b3 − nb1 c 3 − nc1
Note : It should be noted that while applying P-6 at least one row (or column) must remain
unchanged.
P-7 : If all elements below leading diagonal or above leading diagonal or except leading diagonal elements
are zero then the value of the determinant equal to multiplied of all leading diagonal elements.
a1 b1 c1 a1 0 0 a1 0 0
e.g., 0 b2 c2 = a2 b2 0 = 0 b2 0 = a1b2c3
0 0 c3 a3 b3 c3 0 0 c3
P-8 : If a determinant D becomes zero on putting x = α , then we say that (x − α) is factor of determinant.
x 52
e.g., if D = x 2 9 4 . At x = 2, D = 0 (because C1 and C2 are identical at x = 2 )
x 3 16 8
Hence (x − 2) is a factor of D.
Note : It should be noted that while applying operations on determinants then at least one row (or
column) must remain unchanged.or, Maximum number of operations = order or determinant –1
It should be noted that if the row (or column) which is changed by multiplied a non zero number,
then the determinant will be divided by that number.
Example: 1 1 ω n ω 2n
Solution: (a) If n ≠ 3k and 1, ω , ω 2 are the cube roots of unity, then ∆ = ω 2n 1 ωn has the value
Example: 2
ω n ω 2n 1
Solution: (a)
(a) 0 (b) ω (c) ω 2 [Pb. CET 1991; Rajasthan PET 2001; AIEEE 2003]
(d) 1
Applying C1 → C1 + C2 + C3 , we get
1 + ω n + ω 2n ω n ω 2n 0 ω n ω 2n
∆ = 1 + ω n + ω 2n 1 ω n = 0 1 ω n = 0 ( 1 + ω n + ω 2n = 0 if n is not multiple of 3)
1 + ω n + ω 2n ω 2n 1 0 ω 2n 1
1+ x 1 1 [Rajasthan PET 1992; Kerala (Engg.) 2002]
1 1+y 1 =
1 1 1+z
(a) xyz1 + 1 + 1 + 1 (b) xyz (c) 1+ 1 + 1 + 1 (d) 1 + 1 + 1
x y z x y z x y z
1 + 1 1 1 11 1
x x x
1 1 1 xyz1 + 1 1 1 1 1 1
∆ = xyz y 1 + y = x + y + z y 1+ y (by R1 → R1 + R2 + R3 )
1 y 1 y
z 1 1 z 1 1
1 + z 1 + z
z z
1 1 1 1 00
xyz1 + x y z 1
= + + y 1 0 (by C2 → C2 − C1 and C3 → C3 − C1 )
1 0 1
z
= xyz1 + 1 + 1 + 1 1 0 = xyz1 + 1 + 1 + 1
x y z 0 1 x y z
211
Trick : Put x = 1, y = 2 and z = 3 , then 1 3 1 = 2(11) − 1(3) + 1(1 − 3) = 17 .
114
option (a) gives 1× 2× 31 + 1 + 1 + 1 = 17
1 2 3
Example: 3 10 C4 10 C5 11 Cm is equal to zero, where m is
The value of ∆ = 11 C6 11 C7 12 Cm+2
12 C9 13 Cm+4
12 C8
(a) 6 (b) 4 (c) 5 (d) None of these
Solution: (c) 10 C4 10 C5 11 Cm =0
∆ = 11 C6 11 C7 12 Cm+2
Example: 4 12 C9 13 Cm+4
Solution: (d) 12 C8
Example: 5 Applying C2 → C1 + C2
Solution: (a)
Example: 6 10 C4 10 C4 +10 C5 11 Cm =0⇒ ∆= 10 C4 11 C5 11 Cm =0
Solution: (c) ∆ = 11 C6 11 C6 +11 C7 12 Cm+2 11 C6 12 C7 12 Cm+2
12 C8 +12 C9 13 Cm+4 12 C8 13 C9 13 Cm+4
12 C8
Clearly m = 5 satisfies the above result [ C2, C3 will be identical]
If a1, a2 , a3 ,......., an , ...... are in G.P. then the value of the determinant log an log an+1 log an+2 is
log an+3 log an+4 log an+5
log an+6 log an+7 log an+8
[AIEEE 2004; IIT 1993]
(a) –2 (b) 1 (c) 2 (d) 0
a1, a2 , a3 ,......., an , ...... are in G.P.
∴ an2+1 = an .an+2 ⇒ 2 log an+1 = log an + log an+2
a 2 4 = an+3 .an+5 ⇒ 2 log an+4 = log an+3 + log an+5
n+
a 2 = an+6 .an+8 ⇒ 2 log an+7 = log an+6 + log an+8
n+7
Putting these values in the second column of the given determinant, we get
∆ = 1 log an log an + log an+2 log an+2
2 log an+3 log an+3 + log an+5 log an+5
log an+6 log an+6 + log an+8 log an+8
= 1 (0) = 0 [C2 is the sum of two elements, first identical with C1 and second with C3 ]
2
The value of 1 1 1
(2x + 2−x )2 (3x + 3−x )2 (5x + 5−x )2
(2x − 2−x )2 (3x − 3−x )2 (5x − 5−x )2
(a) 0 (b) 30 x (c) 30 −x (d) None of these
Applying R2 → R2 − R3
1 1 1 1 1 1 = 0 [ R1 and R2 are identical]
∆ = 2.2 x.2.2−x 2.3 x.2.3−x 2.5 x.2.5 −x
(3 x − 3 −x )2 (5 x − 5 −x )2 =4 1 1 1
(2 x − 2−x )2 (2 x − 2−x )2 (3 x − 3 −x )2 (5 x − 5 −x )2
Trick : Putting x = 0 , we get option (a) is correct
If x, y, z are integers in A.P. lying between 1 and 9 and x51, y41 and z31 are three digit numbers then the value of
543
x 51 y 41 z 31 is
xyz
(a) x + y + z (b) x − y + z (c) 0 (d) None of these
x 51 = 100x + 50 + 1 ,
y 41 = 100y + 40 + 1
z 31 = 100z + 30 + 1
543
∴ ∆ = 100x + 50 + 1 100y + 40 + 1 100z + 30 + 1
xyz
Applying R2 → R2 − 100R3 − 10R1
543
∆ = 1 1 1 = x − 2y + z
xyz
x, y, z are in A.P. , ∴ x − 2y + z = 0 , ∴ ∆ = 0
Example: 7 If a ≠ b ≠ c , the value of x which satisfies the equation 0 x−a x −b
Solution: (a) x+a 0 x − c = 0 is
x+b
Example: 8 x+c 0
Solution: (c)
[EAMCET 1988; Karnataka CET 1991; MNR 1980; MP PET 1988, 99, 2001; DCE 2001]
Example: 9
Solution: (b) (a) x = 0 (b) x = a (c) x = b (d) x = c
Example: 10
Expanding determinant, we get, ∆ = −(x − a)[−(x + b)(x − c)] + (x + b)[(x + a)(x + c)] = 0 ⇒ 2x3 − (2 Σab)x = 0
⇒ Either x = 0 or x2 = Σab . Since x = 0 satisfies the given equation.
0 −a −b
Trick : On putting x = 0 , we observe that the determinant becomes ∆ x=0 = a 0 − c = 0
bc 0
∴ x = 0 is a root of the given equation.
sin x cos x cos x π π
cos x sin x 4 4
The number of distinct real roots of cos x cos x cos x =0 in the interval − ≤ x ≤ is [IIT Screening 2001]
sin x
(a) 0 (b) 2 (c) 1 (d) 3
1 cos x cos x
(2 cos x + sin x) 1 sin x cos x = 0
cos x sin x
1
Applying, R2 → R2 − R1 and R3 → R3 − R1
1 cos x cos x
(2 cos x + sin x) 0 sin x − cos x 0 = 0 ⇒ (2 cos x + sin x)(sin x − cos x)2 = 0
0 0 sin x − cos x
∴ tan x = −2,1 But tan x ≠ −2 in − π , π . Hence tan x = 1 ⇒ x= π
4 4 4
λ2 + 3λ λ −1 λ+3 [IIT 1981]
If pλ4 + qλ3 + rλ2 + sλ + t = λ + 1 2−λ λ − 4 , then value of t is
λ+4
λ−3 3λ
(a) 16 (b) 18 (c) 17 (d) 19
Since it is an identity in λ so satisfied by every value of λ . Now put λ = 0 in the given equation, we have
0 −1 3
t = 1 2 − 4 = −12 + 30 = 18
−3 4 0
1 x (x + 1)
2x
If f(x) = 3x(x − 1) x(x − 1) (x + 1)x , then f(100) is equal to [IIT 1999, MP PET 2000]
(a) 0
x(x − 1)(x − 2) (x + 1)x(x − 1) (d) –100
(b) 1 (c) 100
Solution: (a) 1 x (x + 1)
f(x) = 2x x(x − 1) (x + 1)x
Example: 11 x(x − 1)(x − 2) (x + 1)x(x − 1)
Solution: (b) 3x(x − 1)
Example: 12
Applying C3 → C3 − C2 , we get
Solution: (b)
1x 1
2x = 0 . Hence
f(x) = 2x x(x − 1) 3x(x − 1) f(100) = 0
3x(x − 1) x(x − 1)(x − 2)
1 0 0 00 is
2 20 00
The value of 4 4 3 0 0
5 5 5 40
6666 5
(a) 6! (b) 5! (c) 1.22. 3.4 3. 5 4.6 4 (d) None of these
The elements in the leading diagonal are 1, 2, 3, 4, 5. On one side of the leading diagonal all the elements are zero.
∴ The value of the determinant
= The product of the elements in the leading diagonal = 1.2.3.4.5 = 5!
The determinant a b aα + b [DCE 2000, 2001]
b c bα + c = 0 if
aα + b bα + c 0
(a) a, b, c are in A.P.
(b) a, b, c are in G.P. or (x − α) is a factor of ax 2 + 2bx + c = 0
(c) a, b, c are in H.P.
(d) α is a root of the equation
ab aα + b
Applying R3 → R3 − αR1 − R2 , we get b c bα + c =0
0 0 − aα 2 − bα − bα − c
⇒ − (aα 2 + 2bα + c)(ac − b 2 ) = 0 ⇒ aα 2 + 2bα + c = 0 or b 2 = ac
⇒ x = α is a root of ax2 + 2bx + c = 0 or a, b, c are in G.P.
⇒ (x − α) is a factor of ax2 + 2bx + c = 0 or a, b, c are in G.P.
Minors and Cofactors .
(1) Minor of an element : If we take the element of the determinant and delete (remove) the row and
column containing that element, the determinant left is called the minor of that element. It is denoted by M ij
a11 a12 a13 M11 M12 M13
Consider the determinant ∆ = a21 a22 a23 , then determinant of minors M = M 21 M 22 M 23 ,
a31 a32 a33 M 31 M 32 M 33
where M11 = minor of a11 = a 22 a 23 , M12 = minor of a12 = a21 a23
a32 a33 a31 a33
M13 = minor of a13 = a 21 a 22
a31 a32
Similarly, we can find the minors of other elements . Using this concept the value of determinant can be
∆ = a11 M11 − a12 M12 + a13 M13
or, ∆ = −a21 M 21 + a22 M 22 − a23 M 23 or, ∆ = a31 M 31 − a32 M 32 + a33 M 33 .
(2) Cofactor of an element : The cofactor of an element aij (i.e. the element in the ith row and jth column)
is defined as (−1)i+ j times the minor of that element. It is denoted by Cij or Aij or Fij . Cij = (−1)i+ j Mij
a11 a12 a13 C11 C12 C13
If ∆ = a21 a22 a23 , then determinant of cofactors is C = C21 C22 C23 , where
a31 a32 a33 C31 C32 C33
C11 = (−1)1+1 M11 = +M11 , C12 = (−1)1+2 M12 = −M12 and C13 = (−1)1+3 M13 = +M13
Similarly, we can find the cofactors of other elements.
Note : The sum of products of the element of any row with their corresponding cofactor is equal to
the value of determinant i.e. ∆ = a11C11 + a12C12 + a13C13 = a11C11 + a21C21 + a31C31
where the capital letters C11, C12, C13 etc. denote the cofactors of a11, a12, a13 etc.
In general, it should be noted
ai1C j1 + ai2C j2 + ai3C j3 = 0, if i ≠ j or a1iC1 j + a2iC2 j + a3iC3 j = 0, if i ≠ j
If ∆' is the determinant formed by replacing the elements of a determinant ∆ by their
corresponding cofactors, then if ∆ = 0 , then ∆C = 0 , ∆' = ∆n−1 , where n is the order of the
determinant.
1351
Example: 13 The cofactor of the element 4 in the determinant 2 3 4 2 is [MP PET 1987]
Solution: (b) 8 0 1 1
Example: 14
0211
Solution: (b)
Example: 15 (a) 4 (b) 10 (c) –10 (d) –4
Solution: (d)
1 31
The cofactor of element 4, in the 2nd row and 3rd column is (−1)2+3 8 0 1 = −{1(−2) − 3(8 − 0) + 1.16} = 10
0 21
a1 b1 c1
If ∆ = a2 b2 c 2 and A1, B1, C1 denote the cofactors of a1, b1, c1 respectively, then the value of the determinant
a3 b3 c3
A1 B1 C1 [MP PET 1989]
A2 B2 C2 is
A3 B3 C3
(a) ∆ (b) ∆2 (c) ∆3 (d) 0
a1 b1 c1 A1 B1 C1 Σa1 A1 0 0 ∆00
We know that ∆.∆' = a2 b2 c 2 . A2 B2 C2 = 0 Σa2 A2 0 = 0 ∆ 0 = ∆3
a3 b3 c3 A3 B3 C3 0 0 Σa3 A3 0 0 ∆
⇒ ∆' = ∆2
Trick : According to property of cofactors ∆' = ∆n−1 = ∆2 ( Hence n = 3)
If the value of a third order determinant is 11, then the value of the square of the determinant formed by the cofactors will
be [Roorkee 1990; DCE 2000]
(a) 11 (b) 121 (c) 1331 (d) 14641
∆' = ∆n−1 = ∆3−1 = ∆2 = (11)2 = 121 . But we have to find the value of the square of the determinant, so required value is
(121)2 = 14641.
Product of two Determinants .
Let the two determinants of third order be,
a1 b1 c1 α1 β1 γ 1
D1 = a2 b2 c2 and D2 = α 2 β 2 γ 2 . Let D be their product.
a3 b3 c3 α3 β3 γ 3
(1) Method of multiplying (Row by row) : Take the first row of D1 and the first row of D2 i.e. a1, b1, c1
and α1, β1, γ 1 multiplying the corresponding elements and add. The result is a1α1 + b1β1 + c1γ 1 is the first element
of first row of D.
Now similar product first row of D1 and second row of D2 gives a1α 2 + b1β 2 + c1γ 2 is the second element of
first row of D, and the product of first row D1 and third row of D2 gives a1α 3 + b1β3 + c1γ 3 is the third element of
first row of D. The second row and third row of D is obtained by multiplying second row and third row of D1 with
1st , 2nd , 3rd row of D2 , in the above manner.
a1 b1 c1 α1 β1 γ 1 a1α1 + b1β1 + c1γ 1 a1α 2 + b1β 2 + c1γ 2 a1α 3 + b1β 3 + c1γ 3
Hence, D = a2 b2 c2 × α 2 β 2 γ 2 = a2α1 + b2 β1 + c2γ 1 a2α 2 + b2 β 2 + c2γ 2 a2α 3 + b2 β 3 + c2γ 3
a3 b3 c3 α3 β3 γ 3 a3α1 + b3 β1 + c3γ 1 a3α 2 + b3 β 2 + c3γ 2 a3α 3 + b3 β 3 + c3γ 3
Note : We can also multiply rows by columns or columns by rows or columns by columns.
Example: 16 For all values of A, B, C and P, Q, R the value of cos(A − P) cos(A − Q) cos(A − R) [IIT 1994]
cos(B − P) cos(B − Q) cos(B − R) is
cos(C − P) cos(C − Q) cos(C − R)
(a) 0 (b) cosA cos B cos C (c) sin A sin B sin C (d) cos P cos Q cos R
Solution: (a) The determinant can be expanded as
cos A cos P + sin A sin P cos A cos Q + sin A sin Q cos A cos R + sin A sin R cos A sin A 0 cos P sin P 0
cos B cos P + sin B sin P cos B cos Q + sin B sin Q cos B cos R + sin B sin R = cos B sin B 0 × cos Q sin Q 0 =0
cos C cos P + sin C sin P cos C cos Q + sin C sin Q cos C cos R + sin C sin R sin C 0 cos R sin R 0
cos C
Example: 17 log 3 512 log 4 3 × log 2 3 log 8 3 = [Tamilnadu (Engg.) 2002]
Solution: (b) log 3 8 log 4 9 log 3 4 log 3 4
(a) 7 (b) 10 (c) 13 (d) 17
log 3 512 log 4 3 × log 2 3 log 8 3 = log 512 × log 9 − log 3 × log 8 × log 3 × log 4 − log 3 × log 4
log 3 8 log 4 9 log 3 4 log 3 4 log 3 log 4 log 4 log 3 log 2 log 3 log 8 log 3
= log 29 log 32 − log 23 × log 22 − log 22 = 9 × 2 − 3 2 − 2 = 10
log 3 × log 22 log 22 log 2 log 23 2 2 3
Summation of Determinants .
f(r) a l
Let ∆r = g(r) b m , where a, b, c, l, m and n are constants, independent of r.
h(r) c n
n
∑ f(r) a l
r =1
nn
∑ ∑Then, ∆r = g(r) b m . Here function of r can be the elements of only one row or one column.
r =1 r=1
n
∑ h(r) c n
r =1
Example: 18 p 15 8 [Kurukshetra CEE 1998]
Solution: (d) If Dp = p 2
35 9 , then D1 + D2 + D3 + D4 + D5 =
p3 25 10
(a) 0 (b) 25 (c) 625 (d) None of these
1 15 8 2 15 8 3 15 8 4 15 8 5 15 8
D1 = 1 35 9 , D2 = 4 35 9 , D3 = 9 35 9 , D4 = 16 35 9 , D5 = 25 35 9
1 25 10 8 25 10 27 25 10 64 25 10 125 25 10
15 75 40
⇒ D1 + D2 + D3 + D4 + D5 = 55 175 45
225 125 50
= 15(3125) − 75(−7375) + 40(−32500) = 46875 + 553125 − 1300000 = −700000
Example: 19 Nn 1 5 [MNR 1994]
2N + 1 is
∑The value of Un , if Un = n2 2N + 1 (d) None of these
n=1 n 3 3N 2 3N
(a) 0 (b) 1 (c) –1
N(N + 1) 1 5
2N + 1
N 2 3N 2 N(N + 1) 6 1 5
N(N + 1)(2N + 1) 12 2N + 1
Solution: (a) ∑Un = 2N + 1 = 4N + 2 2N + 1
6 3N(N + 1) 3N 2 3N
n=1 N(N + 1)2
2 3N
N(N + 1) 6 1 6
12 4N + 2
Applying C3 → C3 + C2 = 4N + 2 2N +1 3N(N + 1) =0 [ C1 and C3 are identical]
3N(N + 1) 3N2
Differentiation and Integration of Determinants.
(1) Differentiation of a determinant : (i) Let ∆(x) be a determinant of order two. If we write
∆(x) =|C1 C2 |, where C1 and C2 denote the 1st and 2nd columns, then
∆' (x) = C'1 C2 + C1 C′2
where C'i denotes the column which contains the derivative of all the functions in the ith column Ci .
In a similar fashion, if we write ∆(x) = R1 , then ∆′(x) = R'1 + R1
R2 R2 R2′
(ii) Let ∆(x) be a determinant of order three. If we write ∆(x) = C1 C2 C3 , then
∆' (x) = C'1 C2 C3 + C1 C'2 C3 + C1 C2 C'3
R1 R'1 R1 R1
and similarly if we consider ∆(x) = R2 , then ∆' (x) = R2 + R'2 + R2
R3 R3 R3 R'3
(iii) If only one row (or column) consists functions of x and other rows (or columns) are constant, viz.
f1(x) f2(x) f3(x)
Let ∆(x) = b1 b2 b3 ,
c1 c2 c3
then ∆' (x) = f '1 (x) f'2 (x) f'3 (x) and in general ∆n(x) = f1n(x) f2n(x) f3n(x)
b1 b2 b3 b1 b2 b3
c1 c2 c3 c1 c2 c3
where n is any positive integer and f n (x) denotes the nth derivative of f(x) .
Example: 20 x b b and ∆ 2 = x b are the given determinants, then
x b a x
If ∆1 = a a x
a
[MNR 1986; Kurukshetra CEE 1998; UPSEAT 2000]
(a) ∆1 = 3(∆ 2 )2 (b) d (∆1 ) = 3∆ 2 (c) d (∆1 ) = 2(∆ 2 )2 (d) ∆1 = 3∆32/ 2
dx dx
Solution: (b) x b b = x3 − 3abx ⇒ d (∆1 ) = 3(x 2 − ab) and ∆ 2 = x b = x 2 − ab
x b dx a x
∆1 = a a x
a
∴ d (∆1 ) = 3(x 2 − ab) = 3∆ 2
dx
Example: 21 If y = sin mx , then the value of the determinant y y1 y2 , where yn = dny is
y4 y5 dx n
y3 y7 y8
y6
(a) m9 (b) m2 (c) m3 (d) None of these
y y1 y2 sin mx m cos mx − m2 sin mx
Solution: (d) y3 y4 y5 = − m3 cos mx m4 sin mx m5 cos mx
y6 y7 y8 − m6 sin mx − m7 cos mx m8 sin mx
Taking − m6 common from R3 , R1 and R3 becomes identical. Hence the value of determinant is zero.
f(x) g(x) h(x)
(2) Integration of a determinant : Let ∆(x) = a b c , where a, b, c, l, m and n are constants.
l mn
∫⇒ b ∆(x)dx = bf (x)dx ∫bg(x)dx ∫bh(x)dx
∫a a a
a b c
m n
a
l
Note : If the elements of more than one column or rows are functions of x then the integration can be
done only after evaluation/expansion of the determinant.
1 cos x 1 − cos x π /2
If ∆(x) = 1 + sin x cos x
Example: 22 sin x ∫1 + sin x − cos x , then ∆(x) dx is equal to
sin x 10
(a) 1/4 (b) 1/2 (c) 0 (d) –1/2
Solution: (d) Applying C3 → C3 + C2 − C1
1 cos x 0
∆(x) = 1 + sin x cos x 0 = cos x − cos x(1 + sin x) = − sin x cos x
sin x sin x 1
π /2 1 π /2 1 − cos 2x π / 2 1 (cos − cos 0) = 1
2 2 2 0 4 2
∆(x)dx sin 2xdx
0 0
∫ ∫∴ = − = − = π −
Application of Determinants in solving a system of Linear Equations.
a1 x + b1y + c1z = d1
.....(i)
Consider a system of simultaneous linear equations is given by a2x + b2y + c2z = d2
a3 x + b3 y + c3 z = d3
A set of values of the variables x, y, z which simultaneously satisfy these three equations is called a solution.
A system of linear equations may have a unique solution or many solutions, or no solution at all, if it has a solution
(whether unique or not) the system is said to be consistent. If it has no solution, it is called an inconsistent system.
If d1 = d2 = d3 = 0 in (i) then the system of equations is said to be a homogeneous system. Otherwise it is
called a non-homogeneous system of equations.
Theorem 1 : (Cramer’s rule) The solution of the system of simultaneous linear equations
a1 x + b1y = c1 .....(i) and a2 x + b2y = c2 .....(ii)
is given by x = D1 , y = D2 , where D= a1 b1 , D1 = c1 b1 and D2 = a1 c1 , provided that D≠0
D D a2 b2 c2 b2 a2 c2
Note : Here D= a1 b1 is the determinant of the coefficient matrix a1 b1 .
a2 b2 b2
a 2
The determinant D1 is obtained by replacing first column in D by the column of the right hand side
of the given equations. The determinant D2 is obtained by replacing the second column in D by
the right most column in the given system of equations.
(1) Solution of system of linear equations in three variables by Cramer’s rule :
Theorem 2 : (Cramer’s Rule) The solution of the system of linear equations
a1 x + b1y + c1z = d1 .....(i)
a2x + b2y + c2z = d2 .....(ii)
a3 x + b3y + c3z = d3 .....(iii)
is given by x = D1 , y = D2 and z = D3 , where D = a1 b1 c1
D D D a2 b2 c2 ,
a3 b3 c3
d1 b1 c1 a1 d1 c1 a1 b1 d1
D1 = d2 b2 c 2 , D2 = a2 d2 c 2 , and D3 = a2 b2 d2 , Provided that D ≠ 0
d3 b3 c3 a3 d3 c3 a3 b3 d3
Note : Here D is the determinant of the coefficient matrix. The determinant D1 is obtained by replacing
the elements in first column of D by d1, d2, d3 . D2 is obtained by replacing the element in the
second column of D by d1, d2, d3 and to obtain D3 , replace elements in the third column of D
by d1 , d2 , d3 .
Theorem 3 : (Cramer’s Rule) Let there be a system of n simultaneous linear equation n unknown given by
a11 x1 + a12 x 2 + .... + a1n xn = b1
a21 x1 + a22 x 2 + .... + a2n xn = b2
an1 x1 + an2 x 2 + .... + ann xn = bn
a11 a12 a1n
Let D = a 21 a 22 a 2n and let D j , be the determinant obtained from D after replacing the j th column
an1 an2 anm
b1
by b2 . Then, x1 = D1 , x2 = D2 ,....., xn = Dn , Provided that D ≠ 0
D D D
bn
(2) Conditions for consistency
Case 1 : For a system of 2 simultaneous linear equations with 2 unknowns
(i) If D ≠ 0 , then the given system of equations is consistent and has a unique solution given by
x = D1 ,y = D2 .
D D
(ii) If D = 0 and D1 = D2 = 0 , then the system is consistent and has infinitely many solutions.
(iii) If D = 0 and one of D1 and D2 is non-zero, then the system is inconsistent.
Case 2 : For a system of 3 simultaneous linear equations in three unknowns
(i) If D ≠ 0 , then the given system of equations is consistent and has a unique solution given by
x = D1 , y= D2 and z = D3
D D D
(ii) If D = 0 and D1 = D2 = D3 = 0 , then the given system of equations is consistent with infinitely many
solutions.
(iii) If D = 0 and at least one of the determinants D1, D2, D3 is non-zero, then given of equations is
inconsistent.
(3) Algorithm for solving a system of simultaneous linear equations by Cramer’s rule
(Determinant method)
Step 1 : Obtain D, D1, D2 and D3
Step 2 : Find the value of D. If D ≠ 0 , then the system of the equations is consistent has a unique solution.
To find the solution, obtain the values of D1, D2 and D3 . The solutions is given by x = D1 , y = D2 and z = D3 .
D D D
If D = 0 go to step 3.
Step 3 : Find the values of D1, D2, D3 . If at least one of these determinants is non-zero, then the system is
inconsistent. If D1 = D2 = D3 = 0, then go to step 4
Step 4 : Take any two equations out of three given equations and shift one of the variables, say z on the right
hand side to obtain two equations in x, y. Solve these two equations by Cramer’s rule to obtain x, y, in terms of z.
Note: The system of following homogeneous equations a1 x + b1y + c1z = 0 , a2 x + b2y + c2z = 0 ,
a3 x + b3y + c3 z = 0 is always consistent.
a1 b1 c1
If ∆ = a2 b2 c2 ≠ 0 , then this system has the unique solution x = y = z = 0 known as trivial
a3 b3 c3
solution. But if ∆ = 0 , then this system has an infinite number of solutions. Hence for non-trivial
solution ∆ = 0 .
Example: 23 If the system of linear equations x + 2ay + az = 0, x + 3 by + bz = 0, x + 4cy + cz = 0 has a non-zero solution, then a, b, c
Solution: (c)
(a) Are in A.P. (b) Are in G.P. (c) Are in H.P. [AIEEE 2003]
(d) Satisfy a + 2b + 3c = 0
1 2a a
System of linear equations has a non-zero solution, then 1 3b b = 0
1 4c c
10 a
Applying C2 → C2 − 2C3 ; 1 b b = 0
1 2c c
Applying R3 → R3 − R2 and R2 → R2 − R1
10 a 2 1 1
b a c
0 b b−a =0 ⇒ b(c − b) − (b − a)(2c − b) = 0 . On simplification = + ; ∴ a, b, c are in H.P.
0 2c − b c−b
Example: 24 If the system of equations x + ay = 0, az + y = 0 and ax + z = 0 has infinite solutions, then the value of a is
Solution: (a)
Example: 25 (a) –1 (b) 1 (c) 0 [IIT Screening 2003]
Solution: (c)
(d) No real values
Example: 26
1a0
0 1 a = 0 ⇒ 1 + a(a)2 = 0 ⇒ a 3 = −1⇒ a = −1
a01
If the system of equations ax + y + z = 0, x + by + z = 0 and x + y + cz = 0 , where a, b, c ≠ 1 has a non-trivial solution,
then the value of 1 + 1 + 1 is
1−a 1−b 1−c
(a) –1 (b) 0 (c) 1 (d) None of these
a11
As the system of the equations has a non-trivial solution 1 b 1 = 0
11c
Applying R2 → R2 − R1 and R3 → R3 − R1
a11
1 − a b − 1 0 = 0 ⇒ a(b − 1)(c − 1) − 1(1 − a)(c − 1) − 1(1 − a)(b − 1) = 0
1−a 0 c −1
⇒ a 1 1 =0 ⇒ 1 − 1 + 1 1 b 1 =0 ⇒ 1 1 1 =1
1−a + 1−b + 1−c 1−a − + 1−c 1−a + 1−b + 1−c
If the system of equations x + 2y − 3z = 1, (k + 3)z = 3, (2k + 1)x + z = 0 is inconsistent, then the value of k is
[Roorkee 2000]
(a) –3 (b) 1 (c) 0 (d) 2
2
Solution: (a) For the equations to be inconsistent D = 0
1 2 −3 1 2 −3
∴ D = 0 0 k + 3 = 0 ⇒ k = −3 and D1 = 3 0 0 ≠ 0 , Hence system is inconsistent for k = −3 .
2k + 1 0 1 00 1
Example: 27. The equations x + y + z = 6, x + 2y + 3z = 10, x + 2y + mz = n give infinite number of values of the triplet (x, y, z) if
(a) m = 3, n ∈ R (b) m = 3, n ≠ 10 (c) m = 3, n = 10 (d) None of these
Solution: (c) Each of the first three options contains m = 3 . When m = 3 , the last two equations become x + 2y + 3z = 10 and
x + 2y + 3z = n .
Example: 28 Obviously, when n = 10 these equations become the same. So we are left with only two independent equations to find the
values of the three unknowns.
Consequently, there will be infinite solutions.
The value of λ for which the system of equations 2x − y − z = 12 , x − 2y + z = −4, x + y + λz = 4 has no solution is
(a) 3 (b) –3 (c) 2 [IIT Screening 2004]
(d) –2
Solution: (d) 2 −1 −1
D = 1 − 2 1 = −3λ − 6 . For no solution the necessary condition is −3λ − 6 = 0 ⇒ λ = −2 . It can be see that for
11 λ
λ = −2 , there is no solution for the given system of equations.
Application of Determinants in Co-ordinate Geometry.
(1) Area of triangle whose vertices are (xr , yr ); r = 1, 2, 3 is
∆= 1 [ x 1 (y 2 − y3) + x 2 (y3 − y1) + x 3 (y1 − y2 )] = 1 x1 y1 1
2 2 x2 y2 1
x3 y3 1
(2) If ar x + br y + cr = 0, (r = 1, 2, 3) are the sides of a triangle, then the area of the triangle is given by
1 a1 b1 c1 2 where C1 = a2b3 − a3b2, C2 = a3b1 − a1b3 , C3 = a1b2 − a2b1 are the
∆ = 2C1C2C3 a2 b2 c2 ,
a3 b3 c3
a1 b1 c1
cofactors of the elements c1,c2, c3 respectively in the determinant a2 b2 c2 .
a3 b3 c3
x y1
(3) The equation of a straight line passing through two points (x1, y1) and (x2, y2) is x1 y1 1 = 0
x2 y2 1
a1 b1 c1
(4) If three lines ar x + br y + cr = 0; (r = 1, 2, 3) are concurrent if a2 b2 c2 = 0
a3 b3 c3
(5) If ax 2 + 2hxy + by2 + 2gx + 2 fy + c = 0 represents a pair of straight lines then
ahg
abc + 2 fgh − af 2 − bg 2 − ch2 = 0 = h b f
gfc
(6) The equation of circle through three non-collinear points A(x1, y1), B(x2, y2), C(x3, y3 ) is
x2 + y2 x y 1
x12 + y12 x1 y1 1 =0
x2 y2 1
x 2 + y 22
2
x 2 + y32 x3 y3 1
3
Example: 29 The three lines ax + by + c = 0, bx + cy + a = 0, cx + ay + b = 0 are concurrent only when [DCE 2000]
(a) a + b + c = 0 (b) a 2 + b 2 + c 2 = ab + bc + ca
(c) a3 + b3 + c3 = ab + bc + ca (d) None of these
abc
Solution: (a, b) Three lines are concurrent if b c a = 0 or, 3abc − a 3 − b 3 − c 3 = 0 ⇒ a 3 + b 3 + c 3 = 3abc
cab
Also, a 3 + b 3 + c 3 − 3abc = 0 ⇒ (a + b + c)(a 2 + b 2 + c 2 − ab − bc − ca) = 0
⇒ (a + b + c) = 0 or a 2 + b 2 + c 2 = ab + bc + ca .
Some Special Determinants.
(1) Symmetric determinant : A determinant is called symmetric determinant if for its every element
ahg
aij = a ji ∀ i, j e.g., h b f
gfc
(2) Skew-symmetric determinant : A determinant is called skew symmetric determinant if for its every
0 3 −1
element aij = − a ji ∀ i, j e.g., − 3 0 5
1 −5 0
Note : Every diagonal element of a skew symmetric determinant is always zero.
The value of a skew symmetric determinant of even order is always a perfect square and that
of odd order is always zero.
0 c −b
For example (i) − c 0 a = −c(0 − ab) − b(ac − 0) = abc − abc = 0
b −a 0
(ii) 0 a = 0 + a2 = a2 (Perfect square) 0 a−b e− f
−a 0 (iii) b − a 0 l − m = 0
f −e m−l 0
(3) Cyclic order : If elements of the rows (or columns) are in cyclic order.
1 a a2
i.e. (i) 1 b b2 = (a − b)(b − c)(c − a)
1 c c2
abc 111
(ii) a2 b2 c 2 = a2 b2 c 2 = (a − b)(b − c)(c − a)(ab + bc + ca)
bc ca ab a3 b3 c 3
a bc abc a a 2 a3
(iii) b ca abc = b b2 b3 = abc(a − b)(b − c)(c − a)
c ab abc c c 2 c 3
111 abc
(iv) a b c = (a − b)(b − c)(c − a)(a + b + c) (v) b c a = −(a3 + b3 + c 3 − 3abc)
a3 b3 c3 cab
Note : These results direct applicable in lengthy questions (As behavior of standard results)
0 i − 100 i − 500
Example: 30 ∆ = 100 − i 0 1000 − i is equal to
Solution: (d)
Example: 31 500 − i i − 1000 0
Solution: (c)
(a) 100 (b) 500 (c) 1000 (d) 0
Example: 32 This determinant of skew symmetric of odd order, hence is equal to 0.
[Pb. CET 1997; DCE 2002]
1 a a2
1 b b2 =
1 c c2
(a) a 2 + b 2 + c 2 (b) (a + b)(b + c)(c + a) (c) (a − b)(b − c)(c − a) (d) None of these
1 a a2
∆ = 1 b b2
1 c c2
Applying R1 → R1 − R2 & R2 → R2 − R3
0 a − b a2 − b2 0 1 a+b
= 0 b − c b 2 − c 2 = (a − b)(b − c) 0 1 b + c ;
1c c2 1 c c2
Applying R1 → R1 − R2 (a − c) 00 1
00
b+c = (a − b)(b − c)(a − c) 0 1 b + c = (a − b)(b − c)(a − c)(−1) = (a − b)(b − c)(c − a)
= (a − b)(b − c) 0 1 c2 1 c c2
1c
111
If a b c = (a − b)(b − c)(c − a)(a + b + c) Where a, b, c are all different, then the determinant
a3 b3 c3
1 1 1 vanishes when
(x − a)2 (x − b)2 (x − c)2
(x − b)(x − c) (x − c)(x − a) (x − a)(x − b)
(a) a + b + c = 0 (b) x = 1 (a +b + c) (c) x = 1 (a + b + c) (d) x = a + b + c
3 2
Solution: (b) 111 …… (i)
Example: 33 a b c = (a − b)(b − c)(c − a)(a + b + c)
Solution: (d) a3 b3 c3
Example: 34
Solution: (b) 111
Now, (x − a)2 (x − b)2 (x − c)2 = 0
(x − b)(x − c) (x − c)(x − a) (x − a)(x − b)
1 (x − a) (x − b) (x − c)
(x − a)(x − b)(x − c) (x − a)3 (x − b)3
⇒ (x − c)3 =0
(x − a)(x − b)(x − c) (x − a)(x − b)(x − c)
(x − a)(x − b)(x − c)
Applying C1 → C1(x − a), C2 → C2 (x − b), C3 → C3 (x − c)
(x − a) (x − b) (x − c)
(x − a)3 (x − b)3 (x − c)3 = 0
111
[(x − a) − (x − b)][(x − b) − (x − c)][(x − c) − (x − a)](x − a + x − b + x − c) = 0
(b − a)(c − b)(a − c)[3x − (a + b + c)] = 0 or x = 1 (a + b + c) [ a ≠ b ≠ c]
3
αβγ
If α, β and γ are the roots of the equations x 3 + px + q = 0 then value of the determinant β γ α is [AMU 1990]
γαβ
(a) p (b) q (c) p 2 − 2q (d) 0
Since α, β, γ are the roots of x 3 + px + q = 0 , ∴ α + β + γ = 0
αβγ
∆= β γ α
γαβ
α +β +γ α +β +γ α +β +γ 000
Applying R1 → R1 + R2 + R3 , We get, β γ α = β γ α =0
γ α β γαβ
x +1 x2 + 2 x(x + 1) = p0 x 6 + p1 x 5 + p2 x 4 + p3 x 3 + p4 x 2 + p5 x + p6 ,then (p5 , p6 ) =
x +1 x(x 2 + 2)
If ∆ = x(x + 1) x(x + 1)
x2 + 2 x +1
(a) (–3, –9) (b) (–5, –9) (c) (–3, –5) (d) (3, –9)
12 0
0 = p6 ⇒ p6 = 9 by expansion.
Putting x = 0 in both sides, we get, 0 1 1
20
p5 is the coefficient of x or constant term in the differentiation of determinant.
Differentiate both sides,
1 2x 2x +1 x +1 x2 + 2 x(x + 1) x + 1 x2 + 2 x(x + 1)
x +1 1 3x2 + 2 + x(x + 1) x +1 x(x2 + 2)
x(x + 1) x(x + 1) x(x2 + 2) + 2x + 1 2x +1
x2 + 2 x(x + 1) x +1 2x 1
x +1 x2 + 2
= 6p0 x 5 + 5p1 x 4 + 4 p2 x 3 + 3p3 x 2 + 2p4 x + p5
Putting x = 0 both sides, we get p5 = −5 ; ∴ (p5 , p6 ) = (−5, 9) .
Determinants-and-matrices
Definition .
A rectangular arrangement of numbers (which may be real or complex numbers) in rows and columns, is
called a matrix. This arrangement is enclosed by small ( ) or big [ ] brackets. The numbers are called the elements of
the matrix or entries in the matrix. A matrix is represented by capital letters A, B, C etc. and its elements by small
letters a,b,c,x,y etc. The following are some examples of matrices:
A = 1 4 , B = 2 + i −3 2 , C = [1, 4, 9] , a E = [l]
2 3 −3+i − 5 D = g,
1
h
Order of a Matrix.
A matrix having m rows and n columns is called a matrix of order m×n or simply m×n matrix (read as 'an m
by n matrix). A matrix A of order m×n is usually written in the following manner
a11 a12 a13 ...a1 j ...a1n
a 21 a 22 a 23 ...a2 j ...a 2n
A = ..... ..... ..... ..... ..... or A = [aij ]m×n , where i = 1, 2,.....m
ai2 ai3 ...aij ...ain j = 1, 2,.....n
ai1
..... ..... ..... ..... .....
am1
am2 am3 ...amj ...amn
Here aij denotes the element of ith row and jth column. Example : order of matrix 3 −1 5 is 2×3
6 2 − 7
Note : A matrix of order m×n contains mn elements. Every row of such a matrix contains n elements
and every column contains m elements.
Equality of Matrices .
Two matrix A and B are said to be equal matrix if they are of same order and their corresponding elements
are equal Example: If A = 1 6 3 and B= a1 a2 a3 are equal matrices.
5 2 1 b1 b2 b3
Then a1 = 1, a2 = 6, a3 = 3,b1 = 5,b2 = 2,b3 = 1
Types of Matrices.
(1) Row matrix : A matrix is said to be a row matrix or row vector if it has only one row and any number of
columns. Example : [5 0 3] is a row matrix of order 1× 3 and [2] is a row matrix of order 1×1.
(2) Column matrix : A matrix is said to be a column matrix or column vector if it has only one column and
2
any number of rows. Example : 3 is a column matrix of order 3×1 and [2] is a column matrix of order 1×1.
− 6
Observe that [2] is both a row matrix as well as a column matrix.
(3) Singleton matrix : If in a matrix there is only one element then it is called singleton matrix.
Thus, A = [aij]m×n is a singleton matrix if m = n = 1 Example : [2], [3], [a], [–3] are singleton matrices.
(4) Null or zero matrix : If in a matrix all the elements are zero then it is called a zero matrix and it is
generally denoted by O. Thus A = [aij ]m×n is a zero matrix if aij = 0 for all i and j.
Example : [0], 0 00, 0 0 00,[0 0] are all zero matrices, but of different orders.
0 0 0
(5) Square matrix : If number of rows and number of columns in a matrix are equal, then it is called a square
a11 a12 a13
matrix. Thus A = [aij ]m×n is a square matrix if m = n . Example : a21
a 22 a 23 is a square matrix of order 3×3
a31 a32 a33
(i) If m ≠ n then matrix is called a rectangular matrix.
(ii) The elements of a square matrix A for which i = j,i.e. a11,a22,a33,....ann are called diagonal elements and
the line joining these elements is called the principal diagonal or leading diagonal of matrix A.
(iii) Trace of a matrix : The sum of diagonal elements of a square matrix. A is called the trace of matrix A ,
n
∑which is denoted by tr A. tr A = aii = a11 + a22 + ...ann
i=1
Properties of trace of a matrix : Let A = [aii ]n×n and B = [bij ]n×n and λ be a scalar
(i) tr(λA) = λ tr(A) (ii) tr(A − B) = tr(A) − tr (B) (iii) tr(AB) = tr(BA)
(iv) tr (A)= tr (A') or tr (AT ) (v) tr (In) = n (vi) tr (0)= 0
(vii) tr (AB) ≠ tr A. tr B
(6) Diagonal matrix : If all elements except the principal diagonal in a square matrix are zero, it is called a
diagonal matrix. Thus a square matrix A = [aij ] is a diagonal matrix if aij = 0, when i ≠ j .
2 0 0
Example : 0 3 0 is a diagonal matrix of order 3×3, which can be denoted by diag [2, 3, 4]
0 0 4
Note : No element of principal diagonal in a diagonal matrix is zero.
Number of zeros in a diagonal matrix is given by n2 − n where n is the order of the matrix.
A diagonal matrix of order n × n having d1,d2,....., dn as diagonal elements is denoted by
diag [d1, d2 ,..., dn ] .
(7) Identity matrix : A square matrix in which elements in the main diagonal are all '1' and rest are all zero is
called an identity matrix or unit matrix. Thus, the square matrix A = [aij ] is an identity matrix, if aij = 1, if i= j
0, if i≠ j
We denote the identity matrix of order n by In .
Example : [1], 1 0 , 1 0 0
0 1 0 1 0 are identity matrices of order 1, 2 and 3 respectively.
0 0 1
(8) Scalar matrix : A square matrix whose all non diagonal elements are zero and diagonal elements are
α, if i j
equal is called a scalar matrix. Thus, if A = [aij ] is a square matrix and aij = 0, if i = j , then A is a scalar matrix.
≠
Example : [2], 1 01, 5 0 0
0 0 5 0 are scalar matrices of order 1, 2 and 3 respectively.
0 0 5
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