mathsvol1ans

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Exercise - 1.1






(1) Write the following in roster form.
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(i) {x ∈ N : x < 121 and x is a prime }.
Solution: The required set = {2, 3, 5, 7}
2
(ii) the set of positive roots of the equation (x − 1)(x + 1)(x − 1) = 0.
Solution: The set of positive roots of the equation is = {1}
(iii) {x : x ∈ N, 4x + 9 < 52}.
Solution: The required set = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
x − 4
(iv) {x : = 3, x ∈ R − {−2}}.
x + 2
Solution: Given that
x 6= −2
x − 4 = 3(x + 2)

x − 4 = 3x + 6
2x = −10

x = −5
The required set = { - 5 }
Method 2
−6
1 + = 3
x + 2
−6
= 2
x + 2
−6 = 2x + 4
x = −5

The required set = { - 5 }
(2) Write the set {−1, 1} in set builder form.
Solution: Many solutions are possible. Three solutions are given below.
Sol(1):{x ∈ R : (x + 1)(x − 1) = 0}.
2
Sol(2):{x ∈ Z : x = 1}.
Sol(3):{x ∈ Q : x is a non-zero integer lies between −2 and 2}.
(3) State whether the following sets are finite or infinite.
(i) {x ∈ N : x is an even prime number}.
Solution: finite set
(ii) {x ∈ N : x is an odd prime number}.
Solution: infinite set
(iii) {x ∈ Z : x is even and less than10}.
Solution: Infinite set
(iv) {x ∈ R : x is a rational number}.
Solution: infinite set
(v) {x ∈ N : x is a rational number}. Solution: infinite set
Note to the teacher
For the below problem the student is expected to form three sets A,B,C and verify the results. The
Formal proof is given here.
(4) Verify the following results:

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(i) A × (B ∩ C) = (A × B) ∩ (A × C).
Solution:
Let (a, b) ∈ A × (B ∩ C) ⇔ a ∈ A, b ∈ (B ∩ C)

Since a ∈ A, b ∈ B ⇔ (a, b) ∈ A × B
and a ∈ A, b ∈ C ⇔ (a, b) ∈ A × C
We have (a, b) ∈ (A × B) ∩ (A × C).

Thus A × (B ∩ C) = (A × B) ∩ (A × C).
(ii) A × (B ∪ C) = (A × B) ∪ (A × C).
Solution:
Let (a, b) ∈ A × (B ∪ C) ⇔ a ∈ A, b ∈ (B ∪ C)

Since a ∈ A, b ∈ B ⇔ (a, b) ∈ A × B
or a ∈ A, b ∈ C ⇔ (a, b) ∈ A × C

We have (a, b) ∈ (A × B) ∪ (A × C).
Thus A × (B ∪ C) = (A × B) ∪ (A × C).
(iii) (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A).
Solution:
Let (a, b) ∈ (A × B) ∩ (B × A)
We have (a, b) ∈ (A × B) ⇔ a ∈ A, b ∈ B
and (a, b) ∈ (B × A) ⇔ a ∈ B, b ∈ A

Since a ∈ A, a ∈ B ⇔ a ∈ A ∩ B
and b ∈ A, b ∈ B ⇔ b ∈ B ∩ A

We have (a, b) ∈ (A ∩ B) × (B ∩ A).
Thus (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A).
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(iv) C − (B − A) = (C ∩ A) ∪ (C ∩ B ).
Solution:
Let a ∈ C − (B − A)
Let a ∈ C, a /∈ (B − A) ⇔ a ∈ C, a /∈ B, a ∈ A
Since a ∈ C, a ∈ A ⇔ a ∈ C ∩ A

and a ∈ C, a /∈ B ⇔ a ∈ C ∩ B 0
0
We have a ∈ (C ∩ A) ∪ (C ∩ B ).
0
Thus C − (B − A) = (C ∩ A) ∪ (C ∩ B ).

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(v) (B − A) ∩ C = (B ∩ C) − A = B ∩ (C − A).
Solution:
Let a ∈ (B − A) ∩ C ⇔ a ∈ B, a /∈ A, a ∈ C

Since a ∈ B, a ∈ C ⇔ a ∈ B ∩ C
and a ∈ B, a ∈ C, a /∈ A ⇔ a ∈ (B ∩ C) − A
We have (B − A) ∩ C = (B ∩ C) − A

Let a ∈ (B − A) ∩ C ⇔ a ∈ B, a /∈ A, a ∈ C
Since a ∈ C, a /∈ A ⇔ a ∈ (C − A)

and a ∈ B, a ∈ (C − A) ⇔ a ∈ B ∩ (C − A)
We have (B − A) ∩ C = B ∩ (C − A).

(vi) (B − A) ∪ C = (B ∪ C) − (A − C) by taking suitable A, B, C.
Solution: For this problem let us consider the sets given below.
Let A = {a, b, c, d}, B = {c, d, e, f}, C = {a, b, e, f}
B − A = {e, f}
(B − A) ∪ C = {a, b, e, f}
B ∪ C = {a, b, c, d, e, f}
A − C = {c, d}
(B ∪ C) − (A − C) = {a, b, e, f}
(5) Justify the trueness of the statement:
“An element of a set can never be a subset of itself.”
Solution: By the definition of a power set of a set contains the all the subsets of it. It turns out
that each element in the power set is a set. Since each set is a subset of itself, we can easily
show that every element of the power set is a subset of itself. For example, if A = {1, 2} and
B = {1, {1, 2}, 3, 4}, then A ∈ B.
(6) If n(P(A)) = 1024, n(A ∪ B) = 15 and n(P(B)) = 32, find n(A ∩ B).
Solution:
n(P(A)) = 1024 ⇒ n(A) = 10
n(P(B)) = 32 ⇒ n(B) = 5
n(A ∩ B) = n(A) + n(B) − n(A ∪ B)
= 10 + 5 − 15 = 0
n(A ∩ B) = 0
Both sets are disjoint to each other
(7) If n(A ∩ B) = 3 and n(A ∪ B) = 10, then find n(P(A∆B)).
Solution:
A∆B = (A ∪ B) − (A ∩ B)
n(A∆B) = n(A ∪ B) − n(A ∩ B)
= 10 − 3
n(A∆B) = 7
n(P(A∆B)) = 2 7
= 128
(8) For a set A, A × A contains 16 elements and two of its elements are (1, 3) and (0, 2). Find the
elements of A.
Solution: n(A × A) = 16. n(A) = 4. Let the elements of A be {a, b, c, d}. Clearly the two
elements given can be represented as (a, b) and (c, d). Hence the elements of A are {0, 1, 2, 3}.
(9) Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B,
find A and B, where x, y, z are distinct elements.
Solution: A = {x, y, z} and B = {1, 2}

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(10) If A × A has 16 elements, S = {(a, b) ∈ A × A : a < b} ; (−1, 2) and (0, 1) are two elements of
S, then write the remaining elements of S.
Solution: n(A) = 4. Clearly from the two distinct elements of S, the elements of A =
{−1, 0, 1, 2}. The elements of S = {(−1, 0), (−1, 1), (−1, 2), (0, 1), (0, 2), (1, 2)}.


Exercise - 1.2



1. Discuss the following relations for reflexivity, symmetricity and transitivity:
(i) The relation R defined on the set of all positive integers by “mRn if m divides n”.
Solution: Since each positive integer divides itself, R is reflexive;
Clearly m divides n ; n divides m whenever m 6= n
Hence R is not symmetric.
Let p divides q. Then q = kp for some positive integer k
Let q divides r. Then r = mq for some positive integer m
Since r = mq = m(kp) = (mk)p we have p divides r
Hence R is transitive.
(ii) Let P denote the set of all straight lines in a plane. The relation R defined by “`Rm if ` is
perpendicular to m”.
Solution: No straight line can be perpendicular to itself. Hence P is not reflexive.
Since ` is perpendicular to m ⇒ m is perpendicular to `. Hence R is symmetric.
No three lines will be perpendicular to each other in Cartesian Plane. Hence P is not transitive.
(iii) Let A be the set consisting of all the members of a family. The relation R defined by “aRb if
a is not a sister of b”.
Solution: We know that a is not a sister of a. Hence R is reflexive. But a is not a sister of
b ; b is not a sister of a.So R is not symmetric. a is not a sister of b and b is not a sister of
c ; a is not a sister of c. Hence R is not transitive.
(iv) Let A be the set consisting of all the female members of a family. The relation R defined by
“aRb if a is not a sister of b”.
Solution: We know that a is not a sister of a. Hence R is reflexive. But a is not a sister of
b ⇒ b is not a sister of a because both a and b are females. Hence R is symmetric. Let a is
not a sister of b and b is not a sister of c. As a, b, c are all females, it follows that a cannot be
a sister of c. Hence R is transitive.
(v) On the set of natural numbers the relation R defined by “xRy if x + 2y = 1”.
Solution: If xRx then for some natural number 3x = 1 which is not possible in natural
numbers. Hence R is not reflexive.
If xRy then x + 2y = 1. If yRx then 2x + y = 1. Solving both equations we get x = y which
has no solution in natural numbers. Hence R is not symmetric.
If xRy and yRz then x + 2y = 1 and y + 2z = 1. Solving we get x + 2y = y + 2z. that is,
x + 2z cannot be equal to 1 when x, z are natural numbers. Hence R is not transitive.
2. Let X = {a, b, c, d} and R = {(a, a), (b, b), (a, c)}. Write down the minimum number of ordered
pairs to be included to R to make it
(i) reflexive (ii) symmetric (iii) transitive (iv) equivalence
Solution: Since (c, c), (d, d) are not in R, it is not reflexive. It is enough to include (c, c), (d, d) to
make it reflexive. (a, c) ∈ R but (c, a) /∈ R. It is enough to include (c, a) to make it symmetric.
After including the above mentioned elements, R = {(a, a), (b, b), (c, c), (d, d), (a, c), (c, a)} we
see that R is transitive. Hence the minimum number of ordered pairs to be included to R to make
it equivalence relation is THREE.

3. Let A = {a, b, c} and R = {(a, a), (b, b), (a, c)}. Write down the minimum number of ordered
pairs to be included to R to make it

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(i) reflexive (ii) symmetric (iii) transitive (iv) equivalence
Solution: Since (c, c) is not in R, it is not reflexive. It is enough to include (c, c) to make it
reflexive. (a, c) ∈ R but (c, a) /∈ R. It is enough to include (c, a) to make it symmetric. After
including the above mentioned elements, R = {(a, a), (b, b), (c, c), (a, c), (c, a)} we see that R is
transitive. Hence the minimum number of ordered pairs to be included to R to make it equivalence
relation is TWO.

4. Let P be the set of all triangles in a plane and R be the relation defined on P as aRb if a is similar
to b. Prove that R is an equivalence relation.
Solution: Each triangle is similar to itself. Hence R is reflexive. a is similar to b ⇒ b is similar to
a. Hence R is symmetric. Let a is similar to b and b is similar to c. Then a is similar to c. Hence
R is transitive. Since R is reflexive, symmetric and transitive, it is an equivalence relation.
5. On the set of natural numbers let R be the relation defined by aRb if 2a + 3b = 30. Write down
the relation by listing all the pairs. Check whether it is
(i) reflexive (ii) symmetric (iii) transitive (iv) equivalence
Solution: R = {(3, 8), (6, 6), (9, 4), (12, 2)}. R is not reflexive, not symmetric, transitive (How?).
Hence it is not an equivalence relation.
Note: The first elements viz, 3, 6, 9, 12 and second elements viz 2, 4, 6, 8 are in Arithmetic
Progression. The values of first elements are in the form 2n + 1 (odd!) while the values of second
elements are in the form 2(5 − n)(even!). To make a total even, both terms should be even. Hence
second term is multiplied by even. But it is interesting to note that first term is multiplied by odd
while the second term is multiplied by even.
6. Prove that the relation ”friendship” is not an equivalence relation on the set of all people in
Chennai.

7. On the set of natural numbers let R be the relation defined by aRb if a + b ≤ 6. Write down the
relation by listing all the pairs. Check whether it is
(i) reflexive (ii) symmetric (iii) transitive (iv) equivalence
Solution:
R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)}
The relation is not reflexive since (4, 4), (5, 5) /∈ R. The relation is symmetric. It is not transitive
since (4,1),(1,4) belongs to R but (4, 4) /∈ R.
Note: Interesting to note that even when b 6= c whenever (a, b), (b, c) ∈ R,we find that (a, c) ∈ R
both a and c are same.
8. Let A = {a, b, c}. What is the equivalence relation of smallest cardinality on A? What is the
equivalence relation of largest cardinality on A?
Solution:
The smallest set of equivalence relation on A is = {(a, a), (b, b), (c, c)}. The largest set of relation
on A is = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}.

9. In the set Z of integers, define mRn if m − n is divisible by 7.Prove that R is an equivalence
relation.
Solution:
m − n = 7k where k is any integer. We see that mRm since zero is divisible by 7. Hence R is
reflexive.
mRn ⇒ nRm. Hence R is symmetric.
Let mRn and nRp. That is, m − n = 7k and n − p = 7t for some integers k, t. Now, (m − n) +
(n − p) = m − p = 7(k + t) ⇒ mRp. Hence R is transitive.
Hence R is an equivalence relation.

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Exercise - 1.3



1. Suppose that 120 students are studying in 4 sections of eleventh standard in a school. Let A denote
the set of students and B denote the set of the sections. Define a relation from A to B as “x related
to y if the student x belongs to the section y”. Is this relation a function? What can you say about
the inverse relation? Explain your answer.
Solution:
Since a student cannot belongs to two different sections this relation is a function. Since the section
consist of more than one student, inverse function does not exist.


−x + 4 if − ∞ < x ≤ −3


x + 4 if − 3 < x < −2


2
2. Write the values of f at −4, 1, −2, 7, 0 if f(x) = x − x if − 2 ≤ x < 1
 2
x − x if 1 ≤ x < 7




0 otherwise
Solution:
f(−4) = 8 because −4 ≤ −3
f(1) = 0 because 1 ≤ 7
f(−2) = 6 because −2 ≤ 1
f(7) = 0 because 7 ≥ 7
f(0) = 0 because 0 ≤ 1
 2
x + x − 5 if x ∈ (−∞, 0)

 2

x + 3x − 2 if x ∈ (3, ∞)
3. Write the values of f at −3, 5, 2, −1, 0 if f(x) =
x 2 if x ∈ (0, 2)


x − 3 otherwise
 2
Solution:
f(−3) = 1 because −3 ≤ 0
f(5) = 38 because 5 ≥ 3
2
f(2) = 1 because 2 − 3 = 1
f(−1) = −5 because −1 ≥ 0
2
f(0) = −3 because 0 − 3 = −3
4. State whether the following relations are functions or not. If it is a function check for one–to–
oneness and ontoness. If it is not a function, state why?
(i) If A = {a, b, c} and f = {(a, c), (b, c), (c, b)}; (f : A → A).
Solution:
It is a function and it is not one–to–one and not onto.
(ii) If X = {x, y, z} and f = {(x, y), (x, z), (z, x)}; (f : X → X).
Solution:
It is not a function because x has two images y and z.
5. Let A = {1, 2, 3, 4} and B = {a, b, c, d}. Give a function from A → B for each of the following:
(i) neither one–to–one nor onto. (ii) not one–to–one but onto.

(iii) one-to–one but not onto. (iv) one-to–one and onto.

Solution:
(a) neither one–to–one nor onto. f(x) = a where x ∈ A
f = {(1, a), (2, a), (3, a), (4, a)}
(b) not one–to–one but onto. This is not possible since both A and B has same number of
elements. If the function is not one–to–one then it has more than one element has same image.

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Hence atleast two elements in A have one image. Then for remaining two elements we cannot
have three images.
(c) one–to–one but not onto. This is also not possible and similar explanation as above could be
given.
(d) one–to–one and onto. f = {(1, a), (2, b), (3, c), (4, d)}
1
6. Find the domain of .
1 − 2sinx
Solution:
1
The function is defined for all x ∈ R except for 1 − 2sinx = 0. That is, except sinx = . That is,
π n π o 2
except x = 2nπ ± , n ∈ Z. Hence the domain is R − C 2nπ ± , n ∈ Z.
6 6
√
4 − x 2
7. Find the largest possible domain of the real valued function f(x) = √ .
2
x − 9
Solution:
2
2
If x ≥ −3 or x ≤ 3, then x will be less than 9 and hence x − 9 will become negative or zero.
If it is negative then it has no square root in R. If it is zero, f is not defined. So x must lie outside
the interval [−3, 3].That is, x must lie on (−∞, −3) ∪ (3, ∞).
2
2
Also if x ≤ −2 and x ≥ 2, then 4 − x will become negative . Then, 4 − x has no square root in
R. So x must lie on the interval [−2, 2].
Combining these two conditions, the largest possible domain forf is

[−2, 2] ∩ ((−∞, −3) ∪ (3, ∞))

That is, ∅.
1
8. Find the range of the function .
2cosx − 1
Solution:
Clearly,
−1 ≤ cos x ≤ 1
⇒ 2 ≥ 2 cos x ≥ −2
⇒ −2 ≤ 2 cos x ≤ 2
⇒ −3 ≤ 2 cos x − 1 ≤ 1
1 1 1
By taking reciprocals, we get ≤ − and ≥ 1. Hence the range of f is
2 cos x − 1 3 2 cos x − 1

1
−∞, − ∪ [1, ∞).
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9. Show that the relation xy = −2 is a function for a suitable domain. Find the domain and the range
of the function.
Solution:
−2
y =
x
Hence the domain is R − {0}
The range is also R − {0}
10. If f, g : R → R are defined by f(x) = |x| + x and g(x) = |x| − x, find g ◦ f and f ◦ g.
Solution:
When x > 0, we have f(x) = x + x = 2x and g(x) = x − x = 0. Since g(x) is a constant
function, (g ◦ f)(x) = 0 when x > 0.
When x < 0, we have f(x) = x − x = 0 which is a constant function. Here g(constant) = 0. In
both cases (g ◦ f)(x) = 0
When x > 0, we have g(x) = x − x = 0 and f(x) = x + x = 2x. Since g(x) is a constant
function, (f ◦ g)(x) = 0 when x > 0.

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When x < 0, we have f(x) = x − x = 0 which is a constant function. Here (f ◦ g)(x) = 0.In
both cases (f ◦ g)(x) = 0
11. If f, g, h are real valued functions defined on R , then prove that (f + g) ◦ h = (f ◦ h) + (g ◦ h).
What can you say about (f ◦ (g + h))? Justify your answer.
Solution:
Let f : A → B, g : A → B, h : C → A are three real-valued functions.Let a ∈ A. Thenf(a) ∈
B, g(a) ∈ B, f(a) + g(a) ∈ B, in other words(f + g)(a) ∈ B.
We know that (f + g)(a) = f(a) + g(a)for somea ∈ Aandf(a) + g(a) ∈ B
Let b = f(a) + g(a), h(c) = a wherea ∈ A
Now (f + g)(h(c)) = b, f(h(c))
******* INCOMPLETE AND WRONG DEDUCTION
2
f(x) = x , g(x) = x, h(x) = x if f satisfies f(x + y) = f(x) + f(y)
12. Iff : R → R is defined by f(x) = 3x − 5, prove that f is a bijection and find its inverse.
Solution:
To prove f is one–to–one
Let x, y ∈ R and x 6= y.
3x 6= 3y

3x − 5 6= 3y − 5
f(x) 6= f(y)
Hence f is one–to–one.
To prove f is onto
If f is not onto then for some value of f the pre-image of f should not be in R
let y = 3x − 5
y + 5
Then x =
3
For none of the value of y ∈ R we could find x /∈ R
Hence our assumption is wrong and f is onto. Hence f is bijective.
x + 5
While proving f is onto, we derived the inverse of function f. That is, f −1 (x) =
3
13. The weight of the muscles of a man is a function of his body weight x and can be expressed as
W(x) = 0.35x. Determine the domain of this function.
Solution:
The domain is (0, ∞)
2
14. The distance of an object falling is a function of time t and can be expressed as s(t) = −6t .
Graph the function and determine if it is one–to–one.
Solution:
Since time is non-negative, t 1 6= t 2 ⇒ s(t 1 ) 6= s(t 2 ) Hence the function is one–to–one.
15. The total cost of airfare on a given route is comprised of the base cost C and the fuel surcharge S
in rupee. Both C and S are functions of the mileage m; C(m) = 0.4m + 50 and S(m) = 0.03m.
Determine a function for the total cost of a ticket in terms of the mileage and find the airfare for
flying 1600 miles.
Solution: The function is
f(m) = C(m) + S(m)

= 0.4m + 50 + 0.03m
= 0.43m + 50
The airfare for flying 1600 miles = f(1600) = 0.43(1600) + 50 = 738

16. A salesperson whose annual earnings can be represented by the function A(x) = 30, 000+ 0.04x,
where x is the rupee value of the merchandise he sells. His son is also in sales and his earnings

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are represented by the function S(x) = 25, 000 + 0.05x. Find (A + S)(x) and determine the total
family income if they each sell Rs.1, 50, 00, 000 worth of merchandise.
Solution:
(A + S)(x) = A(x) + S(x) = 30, 000 + 0.04x + 25, 000 + 0.05x
= 55, 000 + 0.09x (A + S)(1, 50, 00, 000) = 14, 05, 000

17. The function for exchanging American dollars for Singapore Dollar on a given day is f(x) =
1.23x, where x represents the number of American dollars. On the same day the function for
exchanging Singapore Dollar to Indian Rupee is g(y) = 50.50y, where y represents the number
of Singapore dollars. Write a function which will give the exchange rate of American dollars in
terms of Indian rupee.
Solution:
The function is (g ◦ f)(x) = g(f(x)) = g(1.23x) = 50.50 ∗ 1.23x = 62.115x
18. The owner of a small restaurant can prepare a particular gourmet meal at a cost of Rs.100. He
estimates that if the menu price of the meal is x rupees, then the number of customers who will
order that meal at that price in an evening is given by the function D(x) = 200 − x. Express his
day revenue, total cost and profit on this meal as functions of x.
Solution:
Cost per meal = 100. Price per meal = x. Demand per day D(x) = 200 − x
Day Revenue R(x) = x × (200 − x) = 200x − x 2
Total Cost C(x) = 100 × D(x) = 20000 − x 2
Profit = R(x) − C(x) = 200x − 20000
5x 160
19. The formula for converting from Fahrenheit to Celsius temperatures is y = − . Find the
9 9
inverse of this function and determine whether the inverse is also a function.
Solution:
9y = 5x − 160

9y + 160
x =
5
9x
Hence the inverse of the function is f −1 (x) = + 32. Since the inverse is a linear polynomial
5
and have non-zero denominator, the inverse is also a function

20. A simple cipher takes a number and codes it, using the function f(x) = 3x−4. Find the inverse of
this function, determine whether the inverse is also a function and verify the symmetrical property
about the line y = x (by drawing the lines).
Solution:
y + 4 x + 4
Let y = 3x − 4 Then x = Then f −1 (x) =
3 3


Exercise - 1.4


3
1. For the given curve y = x , try to draw
3
3
3
3
i) y = −x ii) y = x + 1 iii) y = x − iv) y = (x + 1) with the same scale.
1

2. For the given curve y = x 3 try to draw
1 1 1 1

− iv) y = (x + 1)
i)y = −x 3 ii) y = x 3 + 1 iii) y = x 3 √ 3
3
3. Graph the functions f(x) = x and g(x) = 3 x on the same coordinate plane. Find fog and graph
it on the plane as well. Explain your results.
4. From the curve y = sinx, graph the functions (i) y = sin(−x) (ii) y = −sin(−x) (iii) y =
π
sin( + x) which is cos (iv) y = sin which is also cos (refer trigonometry)
2

10

Exercise - 1.5


−x
x
1. If A = {(x, y) : y = e , x ∈ R} and B = {(x, y) : y = e , x ∈ R} then n(A ∩ B) is
a)∞ b)0 c)1 d)2
Answer: c) 1.
2. If A = {(x, y) : y = sinx, x ∈ R} and B = {(x, y) : y = cosx, x ∈ R} then A ∩ B contains
a) no element b) infinitely many elements c) only one element d) cannot be determined.
Answer: b) infinitely many elements
2
2
3. The relation R defined on a set A = {0, −1, 1, 2} by xRy if |x + y | ≤ 2, then which one of the
following is True?
a) R = {(0, 0), (0, −1), (0, 1), (−1, 0), (−1, 1), (1, 2), (1, 0)}
Answer:False. (1, 2) /∈ R
b) R −1 = {(0, 0), (0, −1), (0, 1), (−1, 0), (1, 0)}
Answer: False. (−1, 1) ∈ R −1
c) Domain of R is {0, −1, 1, 2}
Answer: False.
d) Range of R is {0, −1, 1}
Answer: True.

4. If f(x) = |x − 2| + |x + 2|, x ∈ R , then f(x) =
 
−2x x ∈ (−∞, −2] 2x x ∈ (−∞, −2]
 
a) f(x) = 4 x ∈ (−2, 2] b) f(x) = 4 x ∈ (−2, 2]
 
2x x ∈ (2, ∞) −2x x ∈ (2, ∞)
 
 
−2x x ∈ (−∞, −2] −2x x ∈ (−∞, −2]
 
c) f(x) = −4 x ∈ (−2, 2] d) f(x) = 2 x ∈ (−2, 2]
 
2x x ∈ (2, ∞) 2x x ∈ (2, ∞)
 
Answer: 
−2x x ∈ (−∞, −2]

(a) f(x) = 4 x ∈ (−2, 2]

2x x ∈ (2, ∞)

5. Let R be the set of all real numbers. Consider the following subsets of the plane R × R
S = {(x, y) : y = x + 1 and 0 < x < 2};
T = {(x, y) : x − y is an integer }


Then which of the following is true?

(a) T is an equivalence relation but S is not an equivalence relation.
(b) Neither S nor T is an equivalence relation
(c) Both S and T are equivalence relation
(d) S is an equivalence relation but T is not an equivalence relation.
Answer: T is an equivalence relation but S is not an equivalence relation.
0 0
6. Let A and B be subsets of the universal set N, the set of natural numbers. Then A ∪[(A∩B)∪B ]
is
0
(a) A (b) A (c) B (d) N
Answer: (d) N
0 0 0 0
(A ∩ B) ∪ B = (A ∪ B ) ∩ (B ∪ B ) = (A ∪ B ) ∪ N = N
0
A ∪ [N] = N

11

7. The number of students who take both the subjects Mathematics and Chemistry is 70. This
represents 10% of the enrolment in Mathematics and 14% of the enrolment in Chemistry. How
many students take at least one of these two subjects?
(a) 1120 (b) 1130 (c) 1100 (d) insufficient data
Answer:
100
Enrolment in Mathematics = % of 70% = 700
10
100
Enrolment in Chemistry = % of 70% = 500
14
Enrolment in atleast one of these two subjects = 1200 − 70 = 1130
8. If .(A × B) ∩ (A × C)] = 8 and n(B ∩ C) = 2 , then n(A) is
(a) 6 (b) 4 (c) 8 (d) 16

Answer: (A × B) ∩ (A × C) = A × (B ∩ C).
n(A × (B ∩ C) = n(A) × n(B ∩ C) = 8
Since n(B ∩ C) = 2, n(A) = 4
9. If n(A) = 2 and n(B ∪ C) = 3, then n[(A × B) ∪ (A × C)] is
(a) 2 3 (b) 3 2 (c) 6 (d) 5

Answer: (c) 6
[(A × B) ∪ (A × C)] = A × (B ∪ C)
n[A × (B ∪ C)] = n(A) × n(B ∪ C) = 2 × 3 = 6
10. If two sets A and B have 17 elements in common, then the number of elements common to the
set A × B and B × A is
(a) 2 17 (b) 17 2 (c) 34 (d) insufficient data
Answer:(b) 17 2
11. For non–empty sets A and B, if A ⊂ B then (A × B) ∩ (B × A) equal
a) A ∩ B (b) A × A (c) B × B (d) none of these.

Answer: (b) A × A
Explanation: Let (x, y) ∈ (A × B) ∩ (B × A).
(x, y) ∈ (A × B) ⇔ x ∈ A, y ∈ B
(x, y) ∈ (B × A) ⇔ x ∈ B, y ∈ A
A ⊂ B ⇔ A ∩ B = A
x ∈ A, x ∈ B ⇔ x ∈ (A ∩ B), x ∈ A
y ∈ A, y ∈ B ⇔ y ∈ (A ∩ B), y ∈ A
(x, y) ∈ A × A
(A × B) ∩ (B × A) = A × A
12. The number of relations on a set containing 3 elements is

(a) 9 (b) 81 (c) 512 (d) 1024
Answer:The number of relations = 2 3 2 = 512.
13. Let R be the universal relation on a set X with more than one element. Then R is
(a) not reflexive (b) not symmetric (c) transitive (d) none of the above

Answer: (c) transitive.
14. Let X = {1, 2, 3, 4} and R = {(1, 1), (1, 2), (1, 3), (2, 2), (3, 3), (2, 1), (3, 1), (1, 4), (4, 1)}. Then
R is (a) reflexive (b) symmetric (c) transitive (d) equivalence
Answer: R is symmetric. R is not reflexive since (4, 4) /∈ R.R is not transitive since
(3, 1), (1, 2) ∈ R but (3, 2) ∈ R. Hence R is not transitive. It follows that the R is not an
equivalence relation either.

12

1
15. The range of the function is
1 − 2 sin x
1 1
(a)(−∞, −1) ∪ ( , ∞) (b)(−1, )
3 3
1 1
(c)[−1, ] (d)(−∞, −1] ∪ [ , ∞)
3 3
1
Answer:(d)(−∞, −1] ∪ [ , ∞)
3
16. The range of the function f(x) = |bxc − x|, x ∈ R is
(a)[0, 1] (b)[0, ∞)
(c)[0, 1) (d)(0, 1)
Answer: (a) [0,1]
2
17. The rule f(x) = x is a bijection if the domain and the co–domain are given by

(a) R, R (b) R, (0, ∞) (c) (0, ∞), R (d) [0, ∞), [0, ∞)
Answer: (d) [0, ∞), [0, ∞)
18. The number of constant functions from a set containing m elements to a set containing n elements
is
(a) mn (b) m (c) n (d) m + n
Answer: (c) n
19. The function f : [0, 2π] → [−1, 1] defined by f(x) = sin x is
(a) one-to–one (b) onto (c) bijection (d) cannot be defined

Answer:(b) onto
2
20. If the function f : [−3, 3] → S defined by f(x) = x is onto, then S is
(a) [−9, 9] (b) R (c) [−3, 3] (d) [0, 9]
Answer:(a) [−9, 9]
21. Let X = {1, 2, 3, 4}, Y = {a, b, c, d} and f = {(1, a), (4, b), (2, c), (3, d), (2, d)}. Then fis
(a) an one–to–one (b) an onto function
function
(c) a function which (d) not a function
is not one–to–one
Answer:(d) not a function. because (2, c), (2, d) ⇒ 2 has more than two images.

x x < 1

22. The inverse of f(x) = x 2 1 ≤ x ≤ 4 is
 √
8 x x > 4

a) f −1 (x) = b) f −1 (x) =
 
x x < 1 −x x < 1
√ √


x 1 ≤ x ≤ 16 x 1 ≤ x ≤ 16
x 2 x 2


 x > 16  x > 16
64 64
c) f −1 (x) = d) f −1 (x) =
 
x 2 x < 1 2x x < 1
√ √


x 1 ≤ x ≤ 16 x 1 ≤ x ≤ 16
x 2 x 2


 x > 16  x > 16
16 8
23. Let f : R → R be defined by f(x) = 1 − |x|. Then the range of f is
(a) R (b) (1, ∞) (c) (−1, ∞) (d) (−∞, 1]
Answer: (d) (−∞, 1]
24. The function f : R → R is defined by f(x) = sinx + cosx is

13

(a) an odd function (b) neither an odd function nor an even func-
tion
(c) an even function (d) both odd function and even function.
Answer:(b) neither an odd function nor an even function
Explanation: f(−x) = − sin x + cos x
Neither f(−x) = f(x) nor f(−x) = −f(x)
2
4
(x + cosx)(1 + x )
25. The function f : R → R is defined by f(x) = + e −|x| is
3
(x − sinx)(2x − x )
(a)an odd function (b)neither an odd function nor an even function
(c)an even function (d)both odd function and even function.
Answer:(c) an even function
Explanation
4
2
(−x) + cos − x)(1 + (−x) )
f(−x) = + e −|−x|
3
((−x) − sin(−x))(2(−x) − (−x) )
4
2
(x + cosx)(1 + x )
f(−x) = + e −|x|
−(x − sinx)(−1)(2x − x )
3
Hence f(−x) = f(x).
Exercise - 2.6


√
1. Classify each element of { 7, −1 , 0, 3.14, 4, 22 } as a member of
4 7
(i) N (ii) Q (iii) R − Q (iv) Z.
N Q R − Q Z
Solution: √
,
4 0, −1 22 7 0, 3.14
√ 4 7
2. Prove that 3 is an irrational number.
√ √ m
Suppose that 3 is a rational number. Let 3 = , where m and n are positive integers with no
n
2
2
2
common factors greater than 1. Then, we have m = 3n , which implies that m is a multiple of
2
2
2
3 and hence m is a multiple of 3. Let m = 3k. Then, we have 3n = 9k which gives n = 3k 2
Thus, n is also multiple of 3. It follows, that m and n have a common factor 3. Thus, we arrived
√
at a contradiction. Hence, 3 is an irrational number.
3. Are there two distinct irrational numbers such that their difference is a rational number? Justify.
Solution: We can find suitable irrational numbers such that their difference is a rational number.
√
√
For example 2 + 1 is an irrational number, then 2 − 1 is also an irrational number. But the
difference is 2 which is a rational number.
4. Find two irrational numbers such that their sum is a rational number. Can you find two irrational
numbers whose product is a rational number.
Solution: As discussed in the previous example, the sum of two irrational numbers can be rational
√
√
number. Example : 3 + 2, 3 − 2. The product of these two numbers is rational number. Note
that if x and y are irrational numbers than atleast one of x + y, x − y is irrational number.
1
5. Find a positive number smaller than . Justify.
2 1000
1 1 1 1
Solution: is a positive number smaller than . Justification: Let > . Multiply
2 1001 2 1000 2 1001 2 1000
both sides by 2 1001 . We have 1 > 2 which is a contradiction. Hence the result.



Exercise - 2.7

14




1. Solve for x:
3 1

(i) |3 − x| < 7. (ii) |4x − 5| ≥ −2. (iii) 3 − x ≤ . (iv) |x| − 10 < −3


4 4


1 1 3

(v) x − < x − .

4 2 4
Solution:
(i) |3 − x| < 7.
−7 < (3 − x) < 7
⇒ 7 > (x − 3) > −7
⇒ 10 > x > −4
⇒ −4 < x < 10


(ii) |4x − 5| ≥ −2.
Since LHS is always positive and RHS has negative integer, the values of x is R.

3 1

(iii) 3 − x ≤ .


4 4

1 3 1
− ≤ 3 − x ≤
4 4 4
1 3 1
≥ x − 3 ≥ −
4 4 4
13 3 11
≥ x ≥
4 4 4
11 13
≤ x ≤
3 3
(iv) |x| − 10 < −3.
−7 < x < 7

1 1 3

(v) x − < x − .

4 2 4


1 1 3 1 1 3
x − < − x + OR x − < x −
4 2 4 4 2 4
3 1 1
x < 1 OR x < −
2 2 2
2
x < OR x < −1
3
The solution set is {−∞, −1}
1
2. Solve < 6 and express the solution using the interval notation.
|2x − 1|
1
Solution: Given |2x − 1| >
6
1 1
(2x − 1) > OR (2x − 1) < −
6 6
7 5
2x > OR 2x <
6 6
7 5
x > OR x <
12 12

15


5 7
x ∈ −∞, ∪ , ∞
12 12
3. Solve −3|x| + 5 ≤ −2 and graph the solution set in a number line.
Solution:

−3|x| ≤ −7
7
|x| ≥
3
7 7
x ≥ OR x ≤ −
3 3

7 7
x ∈ −∞, − OR x ∈ , ∞
3 3
4. Solve 2|x + 1| − 6 ≤ 7 and graph the solution set in a number line.
Solution:
13
|x + 1| ≤
2
13 13
(x + 1) ≤ (or) (x + 1) ≥ −
2 2
11 15
x ≤ (or) x ≥ −
2 2


15 11
x ∈ − ,
2 2
1
5. Solve |10x − 2| < 1.
5
Solution:
|10x − 2| < 5
(10x − 2) < 5 (or) (10x − 2) > −5
7 3
x < (or) x > −
10 10

3 7
x ∈ − ,
10 10
6. Solve |5x − 12| < −2.
Solution:

(5x − 12) < −2 (or) (5x − 12) > 2
14
x < 2 (or) x >
5

14
x ∈ (−∞, 2) ∪ , ∞ ⇒ x ∈ ∅
5
Hence No solution
1 1
7. Solve < .
|x| − 3 2
Solution:
|x| − 3 > 2 ⇒ |x| > 5

x > 5(or)x < −5

x ∈ (−∞, −5) ∪ (5, ∞)

16

Exercise - 2.8
















1. Represent the following inequalities in the interval notation:
(i) x ≥ −1 and x < 4 x ∈ [−1, 4)

(ii) x ≤ 5 and x ≥ −3 x ∈ [−3, 5]

(iii) x < −1 or x < 3 x ∈ (−∞, 3)

(iv) −2x > 0 or 3x − 4 < 11 x ∈ (−∞, 5)
2. Solve 23x < 100 when
(i) x is a natural number {1, 2, 3, 4}
(ii) x is an integer. {. . . , −3, −2, −1, 0, 1, 2, 3, 4}
3. Solve −2x ≥ 9 when
9
(i) x is a real number (−∞, − ]
2
(ii) x is an integer (. . . , −7, −6, −5]

(iii) x is a natural number. {}
4. Solve:
3(x − 2) 5(2 − x)
(i) ≤
5 3
(×15) 9(x − 2) ≤ 25(2 − x)
34x ≤ 68

x ≤ 2

5 − x x
(ii) < − 4.
3 2
5 − x x − 8
<
3 2
2(5 − x) < 3(x − 8)

5x > 34
34
x >
5
5. To secure A grade one must obtain an average of 90 marks or more in 5 subjects each of maximum
100 marks. If one scored 84, 87, 95, 91 in first four subjects, what is the minimum mark one scored
in the fifth subject to get A grade in the course?
Solution:Let the minimum mark to be scored in the fifth subject be x. Minimum Total to be scored
to obtain average of 90 is 450. Total of the remaining subjects is 357. Hence the mark to be scored
in the fifth subject should be x > 93.
6. A manufacturer has 600 litres of a 12 percent solution of acid. How many litres of a 30 percent
acid solution must be added to it so that the acid content in the resulting mixture will be more than
15 percent but less than 18 percent?

17

Solution:
Quantity of existing solution = 600 litres

Amount of acid in the existing solution = 600 × 12%

= 72 litres

Let the quantity of 30%to be added = x
Amount of acid in the solution will be = 30% × x = 0.3x

Given that resulting mixture should be > 15% of total

15
72 + 0.3x > 600 + x
100
72 + 0.3x > 0.15x + 90
0.15x > 18

x > 120
Also given mixture should be less than 18% of total
18
72 + 0.3x < 600 + x
100
72 + 0.3x > 0.18x + 108

0.12x > 36

x > 300
Hence quantity of solution to be added should be between 120 litres and 300 litres.
7. Find all pairs of consecutive odd natural numbers both of which are larger than 10 and their sum
is less than 40.
Solution: Let the pair be 2x − 1 and 2x + 1.
(2x − 1) > 10 ⇒ x > 5

(2x − 1) + (2x + 1) = 4x < 40 ⇒ x < 10

Hence 5 < x < 10. The numbers are (11, 13), (13, 15), (15, 17), and (17, 19)
8. A model rocket is launched from the ground. The height h reached by the rocket after t seconds
2
from lift off is given by h(t) = −5t +100t, 0 ≤ t ≤ 20. At what time the rocket is 495 feet above
the ground?
Solution:
h(t) ≥ 495
2
−5t + 100t ≥ 495
2
t − 20t − 99 ≤ 0
(t − 11)(t − 9) ≤ 0
Hence when t is between 9 and 11 the rocket is 495 feet above the ground.
9. A and B are shopping for a new car with 6 percent sales tax. There is an extra cost of rupees
5000/− as Registration cost. They want to buy a good quality car and plan to spend at least Rs. 5
lakhs but not more rupees 6 lakhs. What is the range of the price of the car?
Solution:Let the price of the car be x.

18

The additional cost of sales tax and registration cost amounts to 1.6x + 5000.
Given 1.6x + 5000 > 5, 00, 000.
Hence we have 1.6x > 4, 95, 000.

That is, x > |3, 09, 375.

But Maximum they could spend is |6, 00, 000.

Hence we have 1.6x < 5, 95, 000.

Solving we get the maximum limit as |3, 71, 875.
Hence the range of cost of the car is between |3, 09, 375&|3, 71, 875.
10. A Plumber can be paid according to the following schemes: In the first scheme he will be paid
rupees 500 plus rupees 70 per hour, and in the second scheme he will paid rupees 120 per hour.
If it takes x hours to complete the job, then for what value of x does the first scheme give better
wages?
Solution:
Given the payment on first scheme is 500 + 70x

The payment on the second scheme is 120x.

To get better wages on first scheme, 500 + 70x > 120x

50x < 500
x < 10
The first scheme give better wages only when the job completed less than 10 hours.
11. A and B are working on similar jobs but their annual salaries differ by more than Rs 6000. If B
earns rupees 27000 per month, then what are the possibilities of A’s salary per month?
Solution:
Let the salary of A be |x. Given |27000 − x| > 500.

(27000 − x) > 500 (27000 − x) < −500

26500 > x (or) 27500 < x
x < 26500 x > 27500


Hence Salary of A is either less than |21000 or more than |33000.


12. Forensic Scientists use h = 61.4 + 2.3F to predict the height h in centimetres for a female whose
thigh bone (femur) measures F cm. If the height of the female lies between 160 to 170 cm, then
find the range of values for the length of the thigh bone?
Solution:Given 160 < h < 170
61.4 + 2.3F > 160 61.4 + 2.3F < 170

2.3F > 98.6 (and) 2.3F < 108.6

F > 42.87 F < 47.22

Hence the range of values for the length of the thigh bone is between 42.87cm and 47.22cm.





Exercise - 2.9

19





1. Construct a quadratic equation with roots 7 and −3.
Solution:Sum of the roots is 4 and product of the roots is −21.
2
Hence the required quadratic equation is x − 4x − 21 = 0.
√
2. A quadratic polynomial has one of its zeros 1 + 5 and it satisfies p(1) = 2. Find the quadratic
polynomial.
Solution: Since the quadratic polynomial has one of its zero in an irrational number, it has roots
√
√
1 + 5, 1 − 5 also will be the roots. Sum of the roots is 2 and product is −4. Hence the
2
quadratic polynomial is x −2x−4 = 0. Now p(1) = 2. Hence the quadratic polynomial becomes

2
2
− (x − 2x − 4) = 0
5 √
2
3. If α and β are the roots of the quadratic equation x + 2x + 3 = 0, form a quadratic polynomial
1 1
with zeroes , .
α β
Solution: √
Sum of the roots = α + β = − 2
Product of the roots = αβ = 3
√
1 1 α + β 2
SUM + = = −
α β αβ 3
1 1
PRODUCT =
αβ √ 3
2
The quadratic polynomial is 3x + 2x + 1 = 0
2
4. If one root of k(x − 1) = 5x − 7 is double the other root, show that k = 2 or −25.
Solution:Let the roots be α and 2α.
2
Simplifying the equation we getkx − (5 + 2k)x + k + 7
(5 + 2k)
Sum of the roots = 3α =
k
(k + 7)
2
Product of the roots = 2(α) =
k
2
5 + 2k (k + 7)
2 =
3k k
2
2(5 + 2k) = 9k(k + 7)
2
2
2(25 + 4k + 20k) = 9k + 63k
2
2
50 + 8k + 40k = 9k + 63k
2
k + 23k − 50 = 0
(k − 2)(k + 25) = 0

The values of k are = 2 or − 25
2
5. If the difference of the roots of the equation 2x − (a + 1)x + a − 1 = 0 is equal to their product,
then prove that a = 2.
Solution:
a + 1
Sum of the roots =
2
a − 1
Difference of the roots = Product of the roots =
2
sum of roots + Difference of roots a
One of the root = =
2 2

20

Substituting this value in the given equation, we get
a a
2
2( ) − (a + 1) + a − 1 = 0
2 2
a 2 (a)(a + 1)
− + a − 1 = 0
2 2
2
a − a(a + 1) + 2a − 2 = 0
2
2
a − a − a + 2a − 2 = 0
a = 2
2
6. Find the condition that one of the roots of ax + bx + c may be (i) negative of the other, (ii)
thrice the other, (iii) reciprocal of the other.
Solution:Let one of the roots be α
(i) The roots are α and −α. The sum of the roots is 0. Hence b = 0.
b
(ii) The roots are α and 3α. Sum of the roots = 4α = - .
c a
2
Product of the roots are 3(α) =
a
2
b c
Hence, 3 − =
4a a
2
3b = 16ac
1 c
(iii) The roots are α and . Product of the roots = 1 = .
α a
Hence c = a.
2
2
7. If the equations x − ax + b = 0 and x − ex + f = 0 have one root in common and if the second
equation has equal roots, then prove that ae = 2(b + f).
e
2
Solution: The equation x − ex + f = 0 has equal roots. The roots are .
2
e
2
The Product of the roots = = f. Since this root satisfy the first equation,
2
e e e
2
we have − a + b = 0. That is, f − a + b = 0. Hence ae = 2(b + f).
2 2 2
2
2
2
8. Discuss the nature of roots of (i) −x + 3x + 1 = 0, (ii) 4x − x − 2 = 0, (iii) 9x + 5x = 0.
Solution:
2
Equation b − 4ac Nature of the roots
2
(i) −x + 3x + 1 = 0 9 + 4 = 13 Real and distinct
2
(ii) 4x − x − 2 = 0 1 + 32 = 33 Real and distinct
2
(iii) 9x + 5x = 0 25 + 0 = 25 Real and distinct
9. Without sketching the graphs, find whether the graphs of the following functions will intersect the
x-axis and if so in how many points.
2
2
2
(i) y = x + x + 2, (ii) y = x − 3x − 7, (iii) y = x + 6x + 9.
Solution:
2
(i) x + x + 2 = 0 gives no real solution. Hence the function will not intersect the x axis.
2
(ii)x −3x−7 = 0 gives 2 distinct values. Hence the function will intersect the x axis in 2 points.
2
(iii)x + 6x + 9 = 0 gives 2 equal values. Hence the function will touch the x axis at 1 point.

21

2
10. Write f(x) = x + 5x + 4 in completed square form.
2
Solution:f(x) = x + 5x + 4
2
f(x) = (x + 5x) + 4

25 25
2
= x + 5x + − + 4
4 4

25 9
2
= x + 5x + −
4 4
2 2
5 3
= x + −
2 2
2
11. Find all values of m so that the quadratic function g(x) = (m − 2)x + 8x + m + 4 is negative
for all real values of x.
2
Solution:g(x) = (m − 2)x + 8x + m + 4 < 0
2
The quadratic function is negative for all real values of x if coefficient of x is negative. Hence
the values of m should be less than 2.

Exercise - 2.10


2
2
1. Solve for x (i) 2x + x − 15 ≤ 0. (ii) −x + 3x − 2 ≥ 0.
Solution:

5
2
(i) 2x + x − 15 ≤ 0. On factorizing the polynomial we get 2 (x + 3) x − ≤ 0
2

5
2
Interval Sign of (x + 3) Sign of x − Sign of 2x + x − 15
2
(−∞, −3) − − +


5
−3, + − −
2

5
, ∞ + + +
2


5
So the inequality is satisfied in −3, .
2
2
2
(ii) −x + 3x − 2 ≥ 0. can be re-written as x − 3x + 2 ≤ 0. On factorizing the polynomial we
get (x − 1)(x − 2) ≤ 0

2
Interval Sign of (x − 1) Sign of (x − 2) Sign of x − 3x + 2 ≤ 0

(−∞, 1) − − +


(1, 2) + − −

(2, ∞) + + +



So the inequality is satisfied in [1, 2].

22

2
2. Solve the inequality x + ax + a − 1 > 0.

Solution: We use completion of squares method to factorize the polynomial
2
f(x) = x + ax + a − 1
2 2
a a
2
= x + ax + − + a − 1
4 4

a a
h i 2 2
= x + − − a + 1
2 4
a a
h i 2 h i 2
= x + − − 1
2 2
= (x + a − 1)(x + 1)
The inequality holds true only when,either x + a > 1, x > −1 (or) x + a < 1, x < −1.
Exercise - 2.11


2
1. Find the zeros of the polynomial function f(x) = 4x − 25.
5
2
Solution:f(a) = 4a − 25 = 0 ⇒ a = ± . Hence the zeros of the polynomial function f(x) =
2
5 5
− and + .
2 2
2
3
2. If x = −2 is one root of x − x − 17x = 22, then find the other roots of equation.
2
2
3
Solution:x − x − 17x − 22 = (x + 2)(ax + bx + c).
3
Comparing the coefficients of x , we have a = 1.Comparing constant terms, we have c = −11.
2
2
Comparing coefficients of x we have 2a+b = −1 ⇒ b = −3. Hence the equation is x −3x−11.
√
!
3 ± 53
Solving for the roots we have, .
2
4
3. Find the real roots of x = 16.
2
4
Solution:Since x − 16 = (x + 4)(x + 2)(x − 2), the real roots are −2, 2.
2
2
4. Solve (2x + 1) − (3x + 2) = 0.
2
2
Solution: Since (3x + 2) > (2x + 1), we can rewrite as (3x + 2) − (2x + 1) = 0. Hence we
have [(3x + 2) + (2x + 1)] × [(3x + 2) − (2x + 1)] = (5x + 3)(x + 1). Hence values of x are
3
− , −1.
5
Exercise - 2.12
4
1. Factorize: x + 1. (Hint: Try completing the square.)
Solution:
4
2 2
x + 1 = (x ) + 1
2
2
2 2
= (x ) + 2x − 2x + 1
2
2 2
= [(x ) + 2x + 1] − 2x 2
√
2
2
= (x + 1) − ( 2x) 2
√ √
2
2
= (x + 1 + 2x)(x + 1 − 2x)
√ √
2
2
= (x + 2x + 1)(x − 2x + 1)
3
2
2
2. If x + x + 1 is a factor of the polynomial 3x + 8x + 8x + a, then find the value of a.
2
2
3
Solution:Let 3x + 8x + 8x + a = (x + x + 1)(px + q)

23

2
3
By comparing the coefficients of x we have p = 3. By comparing the coefficients of x we have
2
2
3
p + q = 8 ⇒ q = 5. Hence 3x + 8x + 8x + a = (x + x + 1)(3x + 5). By comparing the
coefficients of constant term we have a = 5.








Exercise - 2.13



3
x (x − 1)
1. Find all values of x for which > 0.
(x − 2)
Solution: The critical points are 0,1,2.

3
x (x − 1)
Interval Sign of x Sign of (x − 1) Sign of (x − 2) Sign of > 0
(x − 2)
(−∞, 0) − − − −


(0, 1) + − − +

(1, 2) + + − −


(2, ∞) + + + +



So the inequality is satisfied in (0, 1) ∪ (2, ∞).
2x − 3
2. Find all values of x that satisfies the inequality < 0.
(x − 2)(x − 4)
3
Solution:The critical points are , 2, 4.
2
Interval Sign of 2x − 3 Sign of (x − 2) Sign of (x − 4) Sign of 2x − 3(x − 2)(x − 4) < 0

3
(−∞, ) − − − −
2
3
( , 2) + − − +
2
(2, 4) + + − −


(4, ∞) + + + +


3
So the inequality is satisfied in (−∞, ) ∪ (2, 4).
2
2
x − 4
3. Solve ≤ 0.
2
x − 2x − 15
Solution: The critical points are −3, −2, 2, 5

24


2
x − 4
Interval Sign of (x + 2) Sign of (x − 2) Sign of (x − 5) Sign of x + 3 Sign of ≤ 0.
2
x − 2x − 15
(−∞, −3) − − − − +


(−3, −2) − − − − +

(−2, 2) + − − + +


(2, 5) + + − + −

(5, ∞) + + + + +



So the inequality is satisfied in [2, 5].
x − 2 5
4. Solve ≥ .
x + 4 x + 3
(x + 3)(x − 2)
Solution:The inequality can be rewritten as ≥ 0. The critical points are -4,-3,2.
5(x + 4)


(x + 3)(x − 2)
Interval Sign of (x + 3) Sign of (x − 2) Sign of (x + 4) Sign of ≥ 0
5(x + 4)

(−∞, −4) − − − −

(−4, −3) − − + +


(−3, 2) + − + −

(2, ∞) + + + +



So the inequality is satisfied in [−4, 3] ∪ [2, ∞].



Exercise - 2.14


Resolve the following rational expressions into partial fractions.
1 3x + 1 x x 1
1. 2. 3. 4. 5.
2
4
2
x − a 2 (x − 2)(x + 1) (x + 1)(x − 1)(x + 2) (x − 1) 3 x − 1
2
1
2
3
(x − 1) 2 x + x + 1 x + 2x + 1 x + 12 6x − x + 1
6. 7. 8. 9. 10.
3
2
3
2
2
2
x + x x − 5x + 6 x + 5x + 6 (x + 1) (x − 2) x + x + x + 1
3
2
2x + 5x − 11 7 + x x − 1
11. 12. 13. .
2
2
2
x + 2x − 3 (1 + x)(1 + x ) x + x + 1
1
1.
2
x − a 2
1 A B
Let = +
2
x − a 2 x − a x + a
Then 1 = A(x + a) + B(x − a)

25

1 1
When x = a, A = When x = −a, B = −
2a 2a
1 1 1
Hence = −
2
x − a 2 (2a)(x − a) (2a)(x + a)
3x + 1
2.
(x − 2)(x + 1)
3x + 1 A B
Let = +
(x − 2)(x + 1) x − 2 x + 1
Then 3x + 1 = A(x + 1) + B(x − 2)
7 2
When x = 2, A = When x = −1, B =
3 3
3x + 1 7 2
Hence = +
(x − 2)(x + 1) 3(x − 2) 3(x + 1)
x
3.
2
(x + 1)(x − 1)(x + 2)
x Ax + B C D
Let = + +
2
2
(x + 1)(x − 1)(x + 2) x + 1 x − 1 x + 2
2
2
Then x = (Ax + B)(x − 1)(x + 2) + (C)(x + 1)(x + 2) + (D)(x + 1)(x − 1)
1 2
When x = 1, C= When x = −2, D=
6 15
When x = 0, 2B= 2C − D

2 2
= −
6 15
1
B =
10
When x = −1, 2A= −1 + 2B − 2C + 4D

1 1 8 3
= −1 + − + = −
5 3 15 5
x 1 − 3x 1 2
Hence = + +
2
2
(x + 1)(x − 1)(x + 2) 10(x + 1) 6(x − 1) 15(x + 2)
x
4.
(x − 1) 3
x A B C
Let = + +
(x − 1) 3 x − 1 (x − 1) 2 (x − 1) 3
2
Then x = A(x − 1) + B(x − 1) + C
When x = 1, C = 1

When x = 0, A - B = −1


When x = −1, 4A -2B = −2 ⇒ 2A - B = −1

Solving we get A = 0,B = 1,C = 1.
x 1 1
Hence = +
(x − 1) 3 (x − 1) 2 (x − 1) 3

26

1
5.
4
x − 1
1 A B Cx + D
Let = + +
2
4
x − 1 x − 1 x + 1 x + 1
2
2
Then 1 = A(x + 1)(x + 1) + B(x − 1)(x + 1) + (Cx + D)(x + 1)(x − 1)
1
When x = 1, A =
4
1
When x = −1, B = −
4
1
When x = 0, D = −1 - A + B = −
2
1 -15A - 5B - 3D
When x = 2, C = −
2
= 0
1 1 1 1 1 1
Hence = − − = −
2
2
2
4
x − 1 4(x − 1) 4(x + 1) 2(x + 1) 2(x − 1) 2(x + 1)
(x − 1) 2
6.
3
x + x
(x − 1) 2 A Bx + C
Let = +
2
3
x + x x (x + 1)
2
2
Then (x − 1) = A(x + 1) + (Bx + C)(x)
When x = 0 A = 1
2
Comparing x coefficients, we have B = 0
Comparing x coefficients, we have C = −2

(x − 1) 2 1 2
Hence = −
3
x + x x (x + 1)
2
2
x + x + 1
7.
x − 5x + 6
2
2
x + x + 1 B C
Let = A + +
2
x − 5x + 6 x − 3 x − 2
2
Then x + x + 1 = A(x − 3)(x − 2) + B(x − 2) + C(x − 3)
When x = 2, C = −7

When x = 3, B = 13

When x = 4, A = 1
2
x + x + 1 13 7
Hence = 1 + −
x − 5x + 6 x − 3 x − 2
2

27

3
x + 2x + 1
8.
x + 5x + 6
2
3
x + 2x + 1 C D
Let = Ax + B + +
2
x + 5x + 6 x + 3 x + 2
2
3
Then x + 2x + 1 = (Ax + B)(x + 5x + 6) + C(x + 2) + D(x + 3)
When x = −2, D = −11
When x = −3, C = 32


1 − 3D − 2C
When x = 0, B = = −5
6
3
By comparing the coefficients of x we have A = 1.
3
x + 2x + 1 32 11
Hence = x − 5 + −
2
x + 5x + 6 x + 3 x + 2
1
x + 12
9.
2
(x + 1) (x − 2)
x + 12 A B C
Let = + +
2
(x + 1) (x − 2) x + 1 (x + 1) 2 x − 2
Then x + 12 = A(x + 1)(x − 2) + B(x − 2) + C(x + 1) 2
14
When x = 2 C =
9
11
When x = −1 B = −
3
14
2
Comparing x coefficients, we have A = −
9
x + 12 14 14 11
Hence = − −
2
(x + 1) (x − 2) 9(x − 2) 9(x + 1) 3(x + 1) 2
2
6x − x + 1
10.
2
3
x + x + x + 1
2
6x − x + 1 Ax + B C
Let = +
3
2
2
x + x + x + 1 x + 1 x + 1
2
2
Then 6x − x + 1 = (Ax + B)(x + 1) + (C)(x + 1)
When x = −1 C = 4
When x = 0 B = −3
2
Comparing x coefficients, we have A = 2
2
6x − x + 1 2x - 3 4
Hence = +
2
x + x + x + 1 x + 1 x + 1
3
2
2
2x + 5x − 11
11.
2
x + 2x − 3
2
2x + 5x − 11 B C
Let = A + +
2
x + 2x − 3 x − 1 x + 3
2
Then 2x + 5x − 11 = A(x − 1)(x + 3) + B(x + 3) + C(x − 1)

28

When x = 1 B = −1


When x = −3 C = 2

Comparing coefficients of x 2 we have A = 2
2
2x + 5x − 11 1 2
Hence = 2 − +
x + 2x − 3 x − 1 x + 3
2
7 + x
12.
2
(1 + x)(1 + x )
7 + x A Bx + C
Let = +
2
(1 + x)(1 + x ) 1 + x 1 + x 2
2
Then 7 + x = A(1 + x ) + (Bx + C)(1 + x)
When x = −1 A = 3
Comparing coefficients of x 2 we have A + B = 0

Comparing coefficients of x we have B + C = 1
Solving the above equations, we have B = −3 and C = 4
7 + x 3 4 − 3x
Hence = +
2
(1 + x)(1 + x ) 1 + x 1 + x 2
3
x − 1
13.
x + x + 1
2
2
3
x − 1 (x − 1)(x + x + 1)
Since = ,
2
x + x + 1 x + x + 1
2
We have = x − 1





Exercise - 2.15



Determine the region in the plane determined by the inequalities:

(1) x ≤ 3y, x ≥ y.
(2) y ≥ 2x, −2x + 3y ≤ 6.
(3) 3x + 5y ≥ 45, x ≥ 0, y ≥ 0.
(4) 2x + 3y ≤ 35, y ≥ 2, x ≥ 5.
(5) 2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0.
(6) x − 2y ≥ 0, 2x − y ≤ −2, x ≥ 0, y ≥ 0.
(7) 2x + y ≥ 8, x + 2y ≥ 8, x + y ≤ 6.








Exercise - 2.16



1. Simplify:

29

2 2

3
(a) (125) 3 = (5 ) 3 = 5 2 = 25
−3 −3

4
(b) (16) 4 , = (2 ) 4 = 2 (−3) = 1
8
−2 −2

3
(c) (−1000) 3 , = (−(10) ) 3 = −10 (−2) = 1
100
!
1
3
(d) (3) (−6) , = (3) (−2) = 1
9
−2

(27) 3 −1
(e) . = (27) 3 = (3) (−1) = 1
−1 3

(27) 3
−1
3
2. Evaluate (256) −1/2 4 .
Solution:
3
−1 −1/2 4 −1
3 −1 3
3
(256) −1/2 4 = 2 (8) = 2 (−4) 4 = 2 = 8
3. If (x 1/2 + x −1/2 2 1/2 − x −1/2 ) for x > 1.
) = 9/2, then find the value of (x
p
1/2
−1/2
) − 4.
Solution: We know that (x − x ) = (x 1/2 + x −1/2 2
q q
Hence we have 9 − 4 = 1 = 1
2 4 2
2n 2 −n
3 9 3
4. Simplify and hence find the value of n: = 27.
3 3n
Solution:
2n 2 −n
3 9 3
= 27.
3 3n
2n
4
3
3n
3 (3 )(3 −n ) = 3 (3 )
n + 4 = 3n + 3
1
Hence n =
2
5. Find the radius of the spherical tank whose volume is 32π/3 units.
4 32π
3
3
Solution:Since Volume (V ) = ( )πr = we have r = 8. Hence radius r = 2 units.
3 3
6. Solve (1 − x) 1/4 + (15 + x) 1/4 = 2.
1
4
Solution: Let u = (1 − x) 4 and hence x = 1 − u .The given equation can be rewritten as
1
4 ( )
u + (16 − u ) 4 = 2
1
4 ( )
(16 − u ) 4 = 2 − u
Raising to fourth power
16 − u 4 = (2 − u) 4
2
16 − u 4 = 16 − 32u + 24u + 32u = 0
2
3
4
2u + 8u − 24u + 32u = 0
2
2u(u − 2)(u − 2u + 8) = 0

30

1
( )
2
u − 2u + 8 has no solution in R. From other factors, u = 0, u = 2. (1 − x) 4 = 0 ⇒ x = 1.
1
( )
(1 − x) 4 = 2 ⇒ x = −15. Hence the values of x are 1, −15.
√
7. Solve (x + 1) 1/3 = x − 3.
2
Solution:Squaring both sides, we have (x + 1) 3 = x − 3. Raising to the power of 3 we have
3
2
3
2
2
(x + 1) = (x − 3) . The expression now becomes, x + 2x + 1 = x − 9x + 27x − 27.
2
3
Hence we have x − 10x + 25x − 28 = 0 a 0 = 28, a n = 1The dividers of a 0 :
1, 2, 4, 7, 14, 28, The dividers of a n : 1
1, 2, 4, 7, 14, 28
Therefore, check the following rational numbers : ±
1
7
is a root of the expression, so factor out x − 7
1
3
2
x − 10x + 25x − 28
Compute to get the rest of the equation :
x − 7
2
2
x − 3x + 4 = (x − 7) (x − 3x + 4) √
7 + 6
8. Simplify by rationalising the denominator. √ .
3 − 2
Solution:Multiplying both numerator and denominator by the conjugate of the denominator, we
√
√
√
√
√
(7 + 6)(3 + 2) 21 + 3 6 + 7 2 + 2 3
have =
9 − 2 7
1 1 1 1 1
9. Simplify √ − √ √ + √ √ − √ √ + √ .
3 − 8 8 − 7 7 − 6 6 − 5 5 − 2
Solution: Note that the given expression is
1 1 1 1 1
√ √ − √ √ + √ √ − √ √ + √ √ . Multiplying both numerator and
9 − 8 8 − 7 7 − 6 6 − 5 5 − 4 √ √
denominator by the conjugate of the denominator for each fraction, we have, (3 + 8) − ( 8 +
√
√
√
√
√
√
7) + ( 7 + 6) − ( 6 + 5) + ( 5 + 2) = 3 + 2 = 5
√ √ x + 1
2
10. If x = 2 + 3 find .
x − 2
2
Solution: √ √ √ √
2
2
x + 1 = ( 2 + 3) + 1 = 5 + 2 6 + 1 = 6 + 2 6
√ √ √ √
2
2
x − 2 = ( 2 + 3) − 2 = 5 + 2 6 − 2 = 3 + 2 6
√ √ √ √
6 − 2 6 (6 − 2 6)(3 − 2 6) 18 − 6 6 − 24
√ = √ √ =
3 + 2 6 (3 + 2 6)(3 − 2 6) −24
√
−6 − 6 6
=
−15
√
6(1 + 6)
= −
15
√
2(1 + 6)
= −
√ 5
11. Find the square root of 7 + 2 10.
p √ √
Solution: Let 7 + 2 10 = a + b 10 where a, b are rationals.

31
√ √
Squaring on both sides, 7 + 2 10 = (a + b 10) 2
√
2
2
= a + 10b + 2ab 10

2
Comparing both sides, 7 = a + 10b 2
1
and 1 = ab ⇒ a =
b
1
Hence from first equation 7 = + 10b 2
b 2
2
7b = 1 + 10b 4
√
7 ± 49 − 40
Solving forb 2 =
20
4 1
= or −
5 10
4
Since b is non negative b =
5
5
and a =
4
5 4 √
Hence the square root is = + 10
4 5

1 √
= (25 + 16 10)
20




Exercise - 2.17



x
1. Let b > 0 and b 6= 1. Express y = b in logarithmic form. Also state the domain and range of the
logarithmic function.
Solution: The logarithmic form is log y = x. The domain is (0, ∞) and the range is (−∞, ∞)
b
2. Compute log 27 − log 9 .
9
27
Solution:
3
log 27 − log 9 = log 2 3 − log 3 3 2
9 27 3 3
x
x
Using the expression log y a =
a
y
3 2
= −
2 3
5
=
6
3. Solve log x + log x + log x = 11.
8 4 2
1 1 1
Solution:Given + + = 11.
log 8 log 4 log 2
x
x
x
1 1 1
+ + = 11. Hence we have
3 log 2 2 log 2 log 2
x x x
1 1 1
+ + 1 = 11.
log 2 3 2
x

32


1 11
= 11.
log 2 6
x
1
6
= 6 log x = 6 ⇒ x = 2 = 64.
log 2 2
x
4. Solve log 2 8x = 2 log 8 .
2
4
3
Solution:Since log 4 4x = 4x log 4 = 4x and 2 log 8 = 2 log 2 3 = 2 = 8
2
2
4
4
We have 4x = 8 ⇒ x = 2
a + b 1
2
2
5. If a + b = 7ab, show that log = (log a + log b).
3 2
Solution:
2
a + b a + b
2 log = log
3 3
2
2
a + b + 2ab
= log
9
= log ab
= log a + log b
a 2 b 2 c 2
6. Prove log + log + log = 0.
bc ca ab
Solution:
2 2 2
a 2 b 2 c 2 a b c
log + log + log = log
2 2 2
bc ca ab a b c
= 0
16 25 81
7. Prove that log 2 + 16 log + 12 log + 7 log = 1.
15 24 80
Solution:
16 12 7
16 25 81 16 25 81
log 2 + 16 log + 12 log + 7 log = log 2 + log × ×
15 24 80 15 16 24 12 80 7
64 24 28
2 × 5 × 3
= log 2 + log
16
36
16
12
28
3 × 5 × 3 × 2 × 2 × 5 7
64 24 28
2 × 5 × 3
= log 2 + log
28
23
3 × 5 × 2 64
= log 2 + log 5
= log 10
= 1
8. Prove log 2 a log 2 b log 2 c = 18.
a b c
Solution:
2
2
2
(2 log 2 a) (2 log 2 b) (2 log 2 c) = (log 2 a ) (2 log 2 b ) (log 2 c )
a b c a b c
n(n + 1)
3
2
n
9. Prove log a + log a + log a + · · · + log a = log a.
2
Solution:

33

n
2
3
log a + log a + log a + · · · + log a = log a + 2 log a + 3 log a + · · · + n log a
= (1 + 2 + 3 + · · · + n) log a

(n)(n + 1)
= log a
2
log x log y log z
10. If = = , then prove that xyz = 1.
y − z z − x x − y

log x log y log z
Solution: Let = = = log k.
y − z z − x x − y
log x = k(y − z) log y = k(z − x) log z = k(x − y)


log xyz = log x + log y + log z

= k(y − z) + k(z − x) + k(x − y)

= 0

xyz = 1


11. Solve log x − 3 log 1 x = 6.
2
2
Solution: The given expression can be rewritten as
1 3
− = 6
log 2 1
x log x
1 3 2
+ = 6
log 2 log 2
x
x
4
= 6
log 2
x
4 log x = 6
2
3
log x =
2
2
3
√
x = 22 = 2 2

2
12. Solve log (x − 6x + 65) = 2.
5−x
Solution:Rewriting in exponential form we have
2
(5 − x) 2 = x − 6x + 65
2
2
25 + x − 10x = x − 6x + 65
4x = −40

x = −10
Exercise 3.1:

1. Identify the quadrant in which an angle of each given measure lies

34

◦
(i) 25 ◦ − first quadrant (ii) 825 ◦ − second quadrant (iii) −55 − fourth quadrant
◦
◦
(iv) 328 − fourth quadrant (v) −230 − second quadrant
◦
2. For each given angle, find a coterminal angle with measure of θ such that 0 ≤ θ < 360 ◦
◦
◦
◦
(i) 395 ◦ ≡ 35 (ii) 525 ◦ ≡ 165 (iii) 1150 ≡ 70 ◦
◦
◦
◦
(iv) −270 ≡ 90 (v) −450 ≡ 270 ◦
√
2
2
2
3. If a cos θ − b sin θ = c, show that a sin θ + b cos θ = ± a + b − c .
Solution:
p
We know that a sin θ + b cos θ = ± (a sin θ + b cos θ) 2
√
2
2
2
2
= ± a sin θ + b cos θ + 2ab sin θ cos θ
2
2
2
2
2
Now (a cos θ − b sin θ) = a cos θ + b sin θ − 2ab sin θ cos θ
2
2
2
2
2
That is, c = a cos θ + b sin θ − 2ab sin θ cos θ
2
2
2
2
Then 2ab sin θ cos θ = a cos θ + b sin θ − c 2
p
2
2
2
2
2
2
2
2
Hence a sin θ + b cos θ = ± a sin θ + b cos θ + a cos θ + b sin θ − c 2
√
2
2
= ± a + b − c 2
2
4 − 3 (m − 1) 2
6
2
6
4. If sin θ + cos θ = m, show that cos θ + sin θ = , where m ≤ 2.
4
Solution:
2 3
6
(sin θ + cos θ) = [(sin θ + cos θ) ]
2
(sin θ + cos θ) = 1 − 2 sin θ cos θ
2
−2 sin θ cos θ = [m − 1]
6
3
6
2
2
cos θ + sin θ = (cos θ) + (sin θ) 3
4
2
2
2
4
2
= (cos θ + sin θ)(cos θ − cos θ sin θ + sin θ)
2
2
2
2
2
= [(cos θ + sin θ) − 3 cos θ sin θ]
2
= [1 − 3(cos θ sin θ) ]
" 2 #
2
3(m − 1)
= 1 −
4
4
4
cos α sin α
5. If + = 1, prove that
2
2
cos β sin β
4
4
cos β sin β
4
2
2
4
(i) sin α + sin β = 2 sin α sin β (ii) + = 1.
2
2
cos α sin α
Solution:

35

4
4
cos α sin α
1 = +
2
2
cos β sin β
2
4
2
2
4
2
cos β sin β = cos α sin β + sin α cos β
2
2
4
2
2
2
2
(1 − sin β) sin β = (1 − sin α) sin β + sin α(1 − sin β)
2
4
2
2
2
4
2
(1 − sin β) sin β = (1 + sin α − 2 sin α) sin β + sin α(1 − sin β)
2
4
4
2
4
2
4
2
2
2
sin β − sin β = sin β + sin α sin β − 2 sin α sin β + sin α − sin α sin β
2
4
4
2
2 sin α sin β = sin α + sin β
4
4
cos β sin β
(ii) + = 1.
2
cos α sin α
2
4
2
4
2
4
4
cos β sin β cos β sin α + sin β cos α
+ =
2
2
2
2
cos α sin α cos α sin α
2 sin α 1 − cos α + sin α
6. If y = , then prove that = y.
1 + cos α + sin α 1 + sin α
Solution:
α
1 − cos α + sin α 2 sin 2 α 2 + 2 sin cos α 2 sin α
2
2
2
= =
α 2
1 + sin α (sin α + cos ) sin α + cos α
2 2 2 2
α
2 sin α 4 sin cos α 2 sin α
y = = 2 2 = 2
α
1 + cos α + sin α 2 cos 2 α + 2 sin cos α sin α + cos α
2 2 2 2 2
P ∞ 2n P ∞ 2n P ∞ 2n 2n π
7. If x = cos θ, y = sin θ and z = cos θ sin θ, 0 < θ < , then show that
n=0 n=0 n=0 2
1
3
2
xyz = x + y + z. [Hint: Use the formula 1 + x + x + x + . . . = , where |x| < 1].
1 − x
Solution:
1
3
2
Using the formula 1 + x + x + x + . . . = we get,
1 − x
1 1 1
x = , y = and z = .
2
2
2
1 − cos θ 1 − sin θ 1 − sin θ cos θ
2
1 1 1 1 1 1 1 1 1
+ + = + +
yz xz xy y z x z x y
= (1 − sin θ) (1 − sin θ cos θ)
+ (1 − cos θ) (1 − sin θ cos θ) + (1 − cos θ) (1 − sin θ) ********************************************************************
2
= 1 − sin θ − sin θ cos θ + sin θ cos θ
2
+1 − cos θ − sin θ cos θ + cos θ sin θ + 1 − cos θ − sin θ + sin θ cos θ
*******************************************************************
********************************************************************
3
2
2
8. If tan θ = 1 − k , show that sec θ + tan θ cosec θ = 2 − k 2 3/2 . Also, find the values of k for
which this result holds.
2
Solution:Given tan θ = 1 − k 2

36

3
2
2
2
1 sin θ 1 1 sin θ cos θ + sin θ 1
3
sec θ + tan θcosec θ = + × = + = =
3
3
3
3
cos θ cos θ sin θ cos θ cos θ cos θ cos θ
3
3 3
2
2
3
2
= sec θ = (sec θ) 2 = (1 + tan θ) 2 = (2 − k )2
9. If sec θ + tan θ = p, obtain the values of sec θ, tan θ and sin θ in terms of p.
Solution:
1
sec θ − tan θ =
p
2
p + 1
sec θ =
2p
2
p − 1
tan θ =
2p
2
tan θ p − 1
sin θ = =
2
sec θ p + 1
2
10. If cot θ (1 + sin θ) = 4m and cot θ (1 − sin θ) = 4n, then prove that m − n 2 2 = mn.
Solution:
4m = cot θ + cos θ
4n = cot θ + cos θ
4(m + n) = 2 cot θ
4(m − n) = 2 cos θ
2(m + n) = cot θ
2(m − n) = cot θ
2
2
4(m − n ) = cot θ cos θ
2
2
2
2
2
cos θ cos θ cos θ(1 − sin θ) cos θ
2
2
2
2
2
16(m − n ) = cot θ cos θ = = = − cos θ
2
2
2
sin θ sin θ sin θ
2
2
= cot θ − cos θ
= 16mn
3
2 2
2
3
2
11. If cosec θ − sin θ = a and sec θ − cos θ = b , then prove that a b (a + b ) = 1.
Solution:
2
2
1 1 − sin θ cos θ
3
a = cosecθ − sin θ = − sin θ = =
sin θ sin θ sin θ
2 4

2
cos θ 3 cos 3 θ
2
a = = (1)
sin θ 2
sin 3 θ
2
2
1 1 − cos θ sin θ
3
b = secθ − cos θ = − cos θ = =
cos θ cos θ cos θ
2 4

2
sin θ 3 sin 3 θ
2
b = = (2)
cos θ 2
cos 3 θ
4 4

cos 3 θ sin 3 θ
2 2
a b = (1) × (2)
2
2

3 3 θ
2 θ cos
2
sin
= cos 3 θ sin 3 θ (3)

37

4 4 4 2 4 2

cos 3 θ sin 3 θ cos 3 θ cos 3 θ + sin 3 θ sin 3 θ 1
2
2
a + b = + = =
2
2
2
2
2
2

sin 3 θ cos 3 θ sin 3 θ cos 3 θ sin 3 θ cos 3 θ
2 2
2
2
Hence a b (a + b ) = 1.
12. Eliminate θ from the equations a sec θ − c tan θ = b and b sec θ + d tan θ = c.
Solution:Solving both equations for sec θ and tan θ we have
bd + c 2 ac − b 2
sec θ = and tan θ = .
ad + bc ad + bc
2 2 2 2
bd + c ac − b
2
2
Since sec θ − tan θ = 1 we have − = 1.
ad + bc ad + bc
2 2
2 2
2
Simplifying, we have, (bd + c ) = (ac − b ) + (ad + bc) .
Exercise 3.2:
1. Express each of the following angles in radian measure:
◦
◦
(i) 30 (ii) 135 ◦ (iii) −205 ◦ (iv) 150 ◦ (v) 330 .
Solution:
◦
◦
◦
◦
◦
(i) 30 ≡ π (ii) 135 ≡ 3π (iii) −205 ≡ − 41π (iv) 150 ≡ 5π (v) 330 ≡ 11π
6 4 36 6 6
2. Find the degree measure corresponding to the following radian measures
π π 2π 7π 10π
(i) (ii) (iii) (iv) (v) .
3 9 5 3 9
Solution:
π π 2π 7π 10π
◦
◦
◦
◦
(i) ≡ 60 (ii) ≡ 20 (iii) ≡ 72 (iv) ≡ 420 (v) ≡ 200 ◦
3 9 5 3 9
3. What must be the radius of a circular running path, around which an athlete must run 5 times in
order to describe 1 km?
Circumference Circumference
Solution: Radius = ≈ .
2π 6.28
200
Since one round = 200 metres, radius = = 31.85 metres.
6.28
4. In a circle of diameter 40 cm, a chord is of length 20 cm. Find the length of the minor arc of the
chord.
Solution:
Diameter = 40cm. Length of the chord = 20cm. Radius = 20cm. Since radius =
length of chord = 20cm. Hence the formed triangle in the circle is equilateral
◦
triangle with each angle = 60 . We know that ` = rθ
◦
` = 20 × 60 × π
180 ◦
Figure 3.1
` = 20π .Thus length of the minor arc of the chord is 20π
3
3
5. Find the degree measure of the angle subtended at the centre of circle of radius 100 cm by an arc
of length 22cm.
Solution:We know that ` = rθ
` 22 0
◦
Hence θ = = = 0.22radians = 12 36 .
r 100
◦
6. What is the length of the arc intercepted by a central angle of measure 41 in a circle of radius
10 ft?

38

Solution: Since length of the arc = rθ = 41 × 0.017453 = 71.56feet
◦
◦
7. If in two circles, arcs of the same length subtend angles 60 and 75 at the centre, find the ratio of
their radii.
◦
◦
Solution:Let the length of the circle be `. Angle of the circle 1 = 60 . Angle of the circle 2 = 75 .
Let the radius be r 1 and r 2 .
◦
` = r 1 θ = r 1 × 60 × π = πr 1
180 3
◦
` = r 2 θ = r 2 × 75 × π = 5πr 2
180 12

Since length of the arcs are same, πr 1 : 5πr 2 , the ratio of the their radii is given by
3 12
r 1 : r 2 = 5 : 4
Figure 3.2

8. The perimeter of a certain sector of a circle is equal to the length of the arc of a semi-circle having
the same radius. Express the angle of the sector in degrees, minutes and seconds.

Solution:Let the radius of a circle be R.

The perimeter of the semicircle will consist of the semicircular arc and the diameter= πR + 2R.
The perimeter of the sector is arc of the sector + 2R. Equating the two we get πR + 2R = arc of
the sector + 2R, or length arc of the sector = πR = R × θ in radians, or angle subtended by the arc
◦
at the centre is π radians or 180 .
9. An airplane propeller rotates 1000 times per minute. Find the number of degrees that a point on
the edge of the propeller will rotate in 1 second.
Solution:


1000 rotations 1000 rotations 50 rotations 360 degrees 50 rotations
= = . Hence we have, × .
1 minute 60 seconds 3 seconds 1 rotations 3 seconds
360 × 50
Simplifying we get, degrees/second = 6000 degrees/second.
3
10. A train is moving on a circular track of 1500 m radius at the rate of 66 km/hr. What angle will it
turn in 20 seconds?
Solution:Speed of the train = 18.333 m per second.In 1 second, distanceis covered = 18.333m.

. In 10 seconds, distance is covered = 18.333 ∗ 10 = 183.333m.

Angle turned by the train moving on the circular curve in 10 seconds

= arc it made by moving on circular curve/radius of the circular curve

= 183.333/1500 = 0.1222
Hence, the train turnedby 0.1222 Radian (angular measurement)in 10 seconds which is equal to 7
degrees (approximately).
11. A circular metallic plate of radius 8 cm and thickness 6 mm is melted and moulded into a pie (a
sector of the circle with thickness) of radius 16 cm and thickness 4 mm. Find the angle of the
sector.

2
Solution: Volume of the metallic plate = πr h =π64(0.6) = π
Exercise 3.3:

39

1. Find the values of √
3 1 1
◦
◦
◦
(i) sin(480 ) = − (ii) sin(−1110 ) = − (iii) cos(300 ) =
2 2 2
1 1 √
◦
◦
(iv) tan(1050 ) = √ (v) cot(660 ) = −√ (vi) tan 19π = 3
3
√ 3 3
3

(vii) sin − 11π =
√ 3 2
!
5 2 6
2. , is a point on the terminal side of an angle θ in standard position. Determine the
7 7
trigonometric function values of angle θ.
v
√ !
u 2
2
u 5 2 6
p
2
2
r = x + y = t + = 1
7 7
√
!

y 2 6 x 5
sin θ = = ; cos θ = =
r 7 r 7
3. Find the values of other five trigonometric functions for the following:
1
(i) cos θ = − , θ lies in the III quadrant.
2
q √ √
2
Solution:sin θ = ± 1 − − 1 2 = − 3 tan θ = 3; cosecθ = − √ ; sec θ = −2; cot θ = √ 1
2 2 3 3
2
(ii) cos θ = , θ lies in the I quadrant.
3
√
r
4 5
Solution:sin θ = ± 1 − =
9 3
√
q
2

(i) cos θ = − 1 θ lies in the III quadrant. sin θ = ± 1 − − 1 = − 3
2 2 2
√
3 2 √ 1
(3)(i) sin θ = − ; cosec θ = −√ ; sec θ = −2; tan θ = 3; cot θ = √
2
√ 3 √ 3
5 3 3 5 2
(ii) sin θ = ; cosec θ = −√ ; sec θ = ; tan θ = ; cot θ = √
3
√ 5 2 2 √ 5
5 3 3 2 5
(iii) cos θ = ; cosec θ = − ; sec θ = √ ; tan θ = −√ ; cot θ = −
3 2 5 5 √ 2
√ 1 1 2 5
(iv) sec θ = − 5; cos θ = −√ ; cot θ = − ; sin θ = √ ; cosec θ = ;
5 2 5 2
5 12 13 12 5
(v) cos θ = ; sin θ = − ; cosec θ = − ; tan θ = − ; cot θ = − ;
13 13 12 5 12
◦
◦
cot(180 + θ) sin(90 − θ) cos(−θ)
2
4. Prove that = cos θ tan θ.
◦
◦
sin(270 + θ) tan(−θ)cosec(360 + θ)
◦
◦
cot(180 + θ) sin(90 − θ) cos(−θ) cot θ cos θ cos θ
=
◦
◦
sin(270 + θ) tan(−θ)cosec(360 + θ) − cos θ − tan θcosecθ
3
cos θ cos θ sin θ
=
sin θ sin θ
2
= cos θ cot θ
◦
3
2
◦
5. Find all the angles between 0 and 360 which satisfy the equation sin θ = .
√ 4
3 π 2π 4π 5π
Solution: sin θ = ± . Hence the angles are , , ,
2 3 3 3 3

40

6. Show that sin 2 π + sin 2 π + sin 2 7π + sin 2 4π = 2.
9
9
18
Solution: sin 4π = cos π − 4π 18
9 2 9
= sin 2 π + sin 2 π + sin 2 7π + cos 2 π − 4π
18 9 18 2 9
= sin 2 π + sin 2 π + sin 2 7π + cos 2 π
18 9 18 18

= 1 + sin 2 π + sin 2 7π
9 18
= 1 + sin 2 π + cos 2 π − 7π
9 2 18
= 1 + sin 2 π + cos 2 π
9 9
= 1 + 1

= 2
Exercise 3.4:
15 12 π π
1. If sin x = and cos y = , 0 < x < , 0 < y < d ,
17 13 2 2
Solution:Let us first find the values of sin(y) and cos(x).
s
2
12 5
sin(y) = 1 − = .
13 13
s
2
15 8
cos(x) = 1 − = .
17 17
180 40 220
(i) sin(x + y) = sin(x) cos(y) + sin(y)cos(x) = + . =
221 221 221
96 75 171
(ii) cos(x − y) = cos(x) cos(y) + sin(x)sin(y) = + =
221 221 221

15 5
+
tan(x) + tan(y) 8 12 220
(iii) tan(x + y) = = =
1 − tan(x) tan(y) 75 21
1 −
96
3 9 π π
2. If sin A = and cos B = , 0 < A < , 0 < B < ,
5 41 2 2
4 40
Solution: cos(A) = and sin(B) =
5 41
27 160 187
(i) sin(A + B) = sin(A) cos(B) + cos(A) sin(B) = + =
205 205 205
36 120 156
(ii) cos(A − B) = cos(A) cos(B) + sin(A) sin(B) = + =
205 205 205
4 3π 24 3π
3. Find cos(x − y), given that cos x = − with π < x < and sin y = − with π < y < .
5 2 25 2
3 7
Solution:sin x = − and cos y = −
5 25
28 72 110 22
cos(x − y) = cos x cos y + sin x sin y = + = = .
125 125 125 25

41

8 π 24 3π
4. Find sin(x − y), given that sin x = with 0 < x < and cos y = − with π < y < .
17 2 25 2
15 7
Solution:cos x = and sin y = −
17 25
192 85 107
sin(x − y) = sin x cos y − cos x sin y = − + =
425 425 425
7π
5. Find the value of (i) cos 105 ◦ (ii) sin 105 ◦ (iii) tan .
12
Solution:
√ √
1 3 1 − 3
◦
(i) cos 105 = cos(60 + 45) = cos(60)cos(45) − sin(60) sin(45) = √ − √ = √
2 2 2 2 √ 2 2
√
3 1 3 + 1
(ii) sin 105 ◦ = sin(60 + 45) = sin(60)cos(45) + cos(60) sin(45) = √ + √ = √
2 2 2 2 2 2
π π

tan + tan
7π π π 4 3
(iii) tan = tan + = π π
12 4 3 1 − tan tan
√ 4 3
1 + 3 √
= √ = −(2 + 3)
1 − 3
√
3 cos x − sin x
◦
6. Prove that (i) cos(30 + x) = (ii) cos(π + θ) = − cos θ
2
(iii) sin(π + θ) = − sin θ.
Solution:
√ √
3 sin x 3 cos x − sin x
◦
◦
◦
cos(30 + x) = cos(30 ) cos x − sin(30 ) sin x = cos x − =
2 2 2
cos(π + θ) = cos π cos θ − sin π sin θ = − cos θ
sin(π + θ) = sin π cos θ + cos π sin θ = − sin θ
◦
◦
7. Find a quadratic equation whose roots are sin 15 and cos 15 .
√
√
3 − 1 3 + 1 r 3
◦
◦
Solution:We know that sin 15 = √ and cos 15 = √ . Now sum of the roots is =
2 2 2 2 2
1 √
2
and product of the roots = √ . The quadratic equation becomes 4x − 2 6x + 1 = 0.
2
8. Expand cos(A + B + C). Hence prove that
cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin C sin A cos B,
π
if A + B + C = .
2
Solution:

42

cos(A + B + C) = cos[A + (B + C)]

= cos(A) cos(B + C) − sin(A) sin(B + C)

cos(A) cos(B + C) = cos(A)[cos(B) cos(C) − sin(B) sin(C)]
= cos(A) cos(B) cos(C) − cos(A) sin(B) sin(C)

sin(A) sin(B + C) = sin(A)[sin(B) cos(C) + sin(C) cos(B)]

= sin(A) sin(B) cos(C) + sin(A) sin(C) cos(B)

cos(A + B + C) = cos(A) cos(B) cos(C) − cos(A) sin(B) sin(C)

− sin(A) sin(B) cos(C) − sin(A) sin(C) cos(B)

π

Since cos = 0, cos(A + B + C) = 0. Hence we have
2
cos(A) cos(B) cos(C) = cos(A) sin(B) sin(C) + sin(A) sin(B) cos(C) + sin(A) sin(C) cos(B)

9. Prove that
√
◦
◦
(i) sin(45 + θ) − sin(45 − θ) = 2 sin θ.
◦
◦
(ii) sin(30 + θ) + cos(60 + θ) = cos θ.
Solution:
◦
◦
◦
◦
◦
◦
sin(45 + θ) − sin(45 − θ) = sin(45 ) cos θ + sin θ cos(45 ) − sin(45 ) cos θ + sin θ cos(45 )
2 sin θ √
= √ = 2 sin θ
2
◦
◦
◦
◦
◦
◦
sin(30 + θ) + cos(60 + θ) = sin(30 ) cos θ + sin θ cos(30 ) + cos(60 ) cos θ − sin(60 ) sin θ
◦
◦
◦
◦
= sin(30 ) cos θ + sin θ cos(30 ) + sin(30 ) cos θ − cos(30 ) sin θ
◦
= 2 sin(30 ) cos θ
= cos θ
10. If a cos(x + y) = b cos(x − y), show that (a + b) tan x = (a − b) cot y.
Solution:
a cos(x + y) = b cos(x − y)
a cos(x) cos(y) − a sin(x) sin(y) = b cos(x) cos(y) + b sin(x) sin(y)
(a − b) cos(x) cos(y) = (a + b) sin(x) sin(y)
(a − b) = (a + b) tan(x) tan(y)

(a − b) cot(y) = (a + b) tan(x)
◦
◦
◦
11. Prove that sin 105 + cos 105 = cos 45 .
Solution:

43

◦
◦
sin(105) + cos(105) = sin(45) cos(60) + sin(60) cos(45) + cos(45) cos(60) − sin(60) sin(45)
√ √
1 3 1 3
= √ + √ + √ − √
2 2 2 2 2 2 2 2
1
= √
2
= cos 45 ◦
◦
◦
◦
◦
12. Prove that sin 75 − sin 15 = cos 105 + cos 15 .
Solution:
◦
◦
◦
◦
◦
◦
◦
sin 75 − sin 15 = sin(90 − 15 ) − {− cos(90 + 15 )} = cos 15 + cos 105 ◦
◦
◦
13. Show that tan 75 + cot 75 = 4.
Solution:
◦
◦
tan(75 ) + cot(75 ) = tan(45 + 30) + cot(75)
(tan(45) + tan(30)) cos(75)
= +
1 − tan(45) tan(30) sin(75)
√ √
!
1 6 − 2
1 + √ 4
3
= 1 + √ √ !
1 − √ 2 + 6
3 4
√ √ √
3 + 1 6 − 2
= √ + √ √
3 − 1 6 + √ 2
√
= (2 + 3) + (2 − 3)
= 4
14. Prove that cos(A + B) cos C − cos(B + C) cos A = sin B sin(C − A).
Solution:

cos(A + B) cos(C) − cos(B + C) cos A = cos(C) cos(A) cos(B) − cos(C) sin(A) sin(B)
− cos(A)cos(B)cos(C) + cos(A) sin(B) sin(C)

= sin(B) (sin(C) cos(A) − sin(A) cos(C))

= sin(B) sin(C − A)
15. Prove that sin(n + 1)θ sin(n − 1)θ + cos(n + 1)θ cos(n − 1)θ = cos 2θ, n ∈ Z.

Solution:

sin(n + 1)θ sin(n − 1)θ + cos(n + 1)θ cos(n − 1)θ = cos ((n + 1) − (n − 1)) θ = cos 2θ

2π 4π
16. If x cos θ = y cos θ + = z cos θ + , [0.4cm] find the value of xy + yz + zx.
3 3
Solution:
1


Let x cos θ = y cos θ + 2π = z cos θ + 4π =
3 3 k

44


1 1 1 2π 4π
+ + = k cos θ + k cos θ + + k cos θ +
x y z 3 3
π
= k cos θ + 2k cos (θ + π) cos
3
= k cos θ + k cos (θ + π)
π π

= 2k cos θ + cos
2 2
= 0

1 1 1 yz + zx + xy
+ + =
x y z xyz
Hence xy+yz+zx = 0. ************************************************************************************************
***********************************************************************************

Note that the three vectors in polar form: (1, θ), (1, θ + 120) and (1, θ + 240)
divide the complete angle into three equal parts and from symmetry arguments,
their sum must be equal to 0. Taking this to Cartesian coordinates, this means
that the sum of their x components must be equal to 0. Thus, we infer that,cos θ +
◦
◦
cos(θ + 120 ) + cos(θ + 240 ) = 0.
◦
◦
Let a = cos θ, b = cos (θ + 120 ). Then cos (θ + 240 ) = −(a + b).
a a
Hence y = x and z = − x.
b a + b
2
a a a
2
xy + yz + zx = − − x .
b b(a + b) a + b
2
Calculating coefficient of x separately, we have
2
a a 2 a a(a + b) − a − ab
− − = = 0 Hence xy + yz + zx = 0.
b b(a + b) a + b b(a + b)
***********************************************************************************************
17. Prove that
2
2
(i) sin(A + B) sin(A − B) = sin A − sin B
Solution:
1
sin(A + B) sin(A − B) = (cos(A + B − A + B) − cos(A + B + A − B))
2
1
= (cos(2B) − cos(2A))
2
1
2
2
= 1 − 2 sin (B) − 1 + 2 sin (A)
2
2
2
= sin (A) − sin (B)
2
2
2
2
(ii) cos(A + B) cos(A − B) = cos A − sin B = cos B − sin A
Solution:

45

cos(A + B) cos(A − B) = (cos A cos B − sin A sin B) (cos A cos B + sin A sin B)

2
2
2
2
= cos A cos B − sin A sin B
2
2
2
2

= cos A 1 − sin B − (1 − cos A) sin B
2
2
= cos A − sin B
2
2
= (1 − sin A) − (1 − cos B)
2
2
= cos B − sin A
2
2
(iii) sin (A + B) − sin (A − B) = sin 2A sin 2B
Solution:
1
2
2
sin (A + B) − sin (A − B) = {(1 − cos[2(A + B)]) − (1 − cos[2(A − B)])}
2
1
= (cos 2(A − B) − cos 2(A + B))
2
= sin 2A sin 2B
2
2
(iv) cos 8θ cos 2θ = cos 5θ − sin 3θ
Solution:
1
cos 8θ cos 2θ = cos 10θ + cos 6θ
2
1
2
= (2 cos 5θ − 1 + cos 6θ)
2
1
2
2
= (2 cos 5θ − 2 sin 3θ)
2
2
2
= cos 5θ − sin 3θ
2
2
2
18. Show that cos A + cos B − 2 cos A cos B cos(A + B) = sin (A + B).
Solution:
2
2
2
2
cos A + cos B − 2 cos A cos B cos(A + B) = cos A + cos B − (cos(A + B) + cos(A − B)) cos(A + B)
2
2
2
= cos A + cos B − cos (A + B) − cos(A + B) cos(A − B)
1
2
2
2
= cos A + cos B − cos (A + B) − (cos 2A + cos 2B)
2
1 + cos 2A 1 + cos 2B 1
2
= + − cos (A + B) − (cos 2A + cos 2B)
2 2 2
2
= 1 − cos (A + B)
2
= sin (A + B)
3
19. If cos(α − β) + cos(β − γ) + cos(γ − α) = − , then prove that
2
cos α + cos β + cos γ = sin α + sin β + sin γ = 0.
Solution:

46

2 cos(α − β) + 2 cos(β − γ) + 2 cos(γ − α) + 3 = 0

2 cos α cos β + 2 sin α sin β + 2 cos β cos γ + 2 sin β sin γ + 2 cos γ cos α + 2 sin γ sin α+
2
2
2
2
2
2
sin α + cos α + sin β + cos β + sin γ + cos γ = 0
2
(sin α + sin β + sin γ) + (cos α + cos β + cos γ) 2 = 0
Hence cos α + cos β + cos γ = sin α + sin β + sin γ = 0.
20. Show that
1 + tan A 1 − tan A
◦
◦
(i) tan(45 + A) = (ii) tan(45 − A) = .
1 − tan A 1 + tan A
Solution:
◦
tan(45 ) + tan A
◦
tan(45 + A) =
◦
1 − tan(45 ) tan A
1 + tan A
=
1 − tan A
◦
tan(45 ) − tan A
◦
tan(45 − A) =
◦
1 + tan(45 ) tan A
1 − tan A
=
1 + tan A
cot A cot B − 1
21. Prove that cot(A + B) = .
cot A + cot B
Solution:
1
cot(A + B) =
tan(A + B)
1 − tan A tan B
=
tan A + tan B
Multiplying both Numerator and Denominator by cot A cot B

cot A cot B − 1
We get
cot B + cot A
n 1
22. If tan x = and tan y = , find tan(x + y).
n + 1 2n + 1

Solution:
tan(x) + tan(y)
tan(x + y) =
1 − tan(x) tan(y)
n 1
+
n + 1 2n + 1
= n
1 −
(n + 1)(2n + 1)
n(2n + 1) + n + 1
=
(n + 1)(2n + 1) − n
2
2n + 2n + 1
=
2
2n + 2n + 1
= 1

π 3π

23. Prove that tan + θ tan + θ = −1.
4 4

47

Solution:
π 3π (1 + tan θ) (−1 + tan θ)

tan + θ tan + θ =
4 4 (1 − tan θ) (1 + tan θ)

π 3π

Hence tan + θ tan + θ = −1.
4 4

1 3π 5 π
24. Find the values of tan(α + β), given that cot α = , α ∈ π, and sec β = − , β ∈ , π .
2 2 3 2
Solution:
s

25 4
Given tan α = 2 and tan β = − 1 = − . Hence,
9 3
4
2 − 2
tan(α + β) = 3 =
8 11
1 +
3
k − 1
25. If θ + φ = α and tan θ = k tan φ, then prove that sin(θ − φ) = sin α.
k + 1

Solution:

Given that k tan φ = tan θ
tan θ
Hence k =
tan φ
sin θ sin φ
−
k − 1 tan θ − tan φ cos θ cos φ
Using Componendo & Dividendo rule = =
k + 1 tan θ + tan φ sin θ sin φ
+
cos θ cos φ
sin θ cos φ − cos θ sin φ
cos θ cos φ
=
sin θ cos φ + cos θ sin φ
cos θ cos φ
sin (θ − φ)
=
sin (θ + φ)
sin (θ − φ)
=
sin α
Exercise 3.5:

1. Find the value of cos 2A, A lies in the first quadrant, when
15 4 16
(i) cos A = (ii) sin A = (iii) tan A = .
17 5 63

Solution:
2
15 161
2
(i) cos 2A = 2 cos A − 1 = 2 − 1 =
17 289
2
4 7
2
(ii)cos 2A = 1 − 2 sin A = 1 − 2 = −
5 25

48

256
2
1 − tan A 1 − 3713
(iii)cos 2A = = 3969 =
2
1 + tan A 256 4225
1 +
3969
2. If θ is an acute angle, then find

π θ 1
(i) sin − , when sin θ = .
4 2 25
Solution:

π θ π θ θ π 1 θ θ
sin − = sin cos − sin cos = √ cos − sin
4 2 4 2 2 4 2 2 2
s
2
1 θ θ
= √ cos − sin
2 2 2
1 p
= √ (1 − sin θ)
2
!
r
1 24
= √
2 25
√
2 3
=
5

π θ 8
(ii) cos + , when sin θ = .
4 2 9
Solution:

π θ π θ π θ 1 θ θ
cos + = cos cos − sin sin = √ cos − sin
4 2 4 2 4 2 2 2 2
s
2
1 θ θ
= √ cos − sin
2 2 2
1 p
= √ (1 − sin θ)
2
!
r
1 1
= √
2 9
1
= √
3 2

1 1 1 1
3
3. If cos θ = a + , show that cos 3θ = a + .
2 a 2 a 3

49


1 1
cos θ = a +
2 a
3
cos 3θ = 4 cos θ − 3 cos θ
2
= cos θ (4 cos θ − 3)
!
2
1 1 1
= a + a + − 3
2 a a
!
2
1 1 1
2
= a + a + − 1
2 a a

1 1
3
= a +
2 a 3
3
5
4. Prove that cos 5θ = 16 cos θ − 20 cos θ + 5 cos θ.
Solution:
cos 5θ = cos(3θ + 2θ)
= (cos 3θ cos 2θ − sin 3θ sin 2θ)
3
2
cos 3θ cos 2θ = (4 cos θ − 3 cos θ) (2 cos θ − 1)
5
3
= (8 cos θ − 10 cos θ + 3 cos θ)
3

sin 3θ sin 2θ = 3 sin θ − 4 sin θ (2 sin θ cos θ)
4
2
= 6 sin θ cos θ − 8 sin θ cos θ
3
4
2
sin 3θ sin 2θ = (6 cos θ − 6 cos θ − 8 (1 + cos θ − 2 cos θ) cos θ)
3
3
5
= (6 cos θ − 6 cos θ − 8 cos θ − 8 cos θ + 16 cos θ)
3
5
3
3
5
cos 5θ = (8 cos θ − 10 cos θ + 3 cos θ) − (6 cos θ − 6 cos θ − 8 cos θ − 8 cos θ + 16 cos θ)
3
5
= 16 cos θ − 20 cos θ + 5 cos θ
2
1 − tan α
5. Prove that sin 4α = 4 tan α 2 .
2
(1 + tan α)
Solution:
sin 4α = 2 sin 2α cos 2α
2
2 tan α 1 − tan α
2 sin 2α cos 2α = 2 ×
2
2
1 + tan α 1 + tan α
2
tan α (1 − tan α)
= 4 2
2
(1 + tan α)
◦
6. If A + B = 45 , show that (1 + tan A) (1 + tan B) = 2.
Solution:
tan A + tan B
tan(A + B) = = 1
1 − tan A tan B

50

tan A + tan B + tan A tan B = 1

1 + tan A + tan B + tan A tan B = 2

(1 + tan A) + tan B(1 + tan A) = 2
(1 + tan A)(1 + tan B) = 2


◦
◦
◦
◦
7. Prove that (1 + tan 1 )(1 + tan 2 )(1 + tan 3 ) . . . (1 + tan 44 ) is a multiple of 4.
Solution:
◦
If A + B = 45 ,then (1 + tan A) (1 + tan B) = 2.Here we can rearrange the terms as,
◦
◦
◦
◦
◦
(1 + tan 1 )(1 + tan 44 )(1 + tan 2 ) . . . (1 + tan 22 )(1 + tan 23 ) = 2(22 pairs) =
44( is a multiple of 4)
π π

8. Prove that tan + θ − tan − θ = 2 tan 2θ.
4 4
Solution:
π π 1 + tan θ 1 − tan θ

tan + θ − tan − θ = −
4 4 1 − tan θ 1 + tan θ
2 2
(1 + tan θ) − (1 − tan θ)
=
2
1 − tan θ
4 tan θ
=
2
1 − tan θ
= 2 tan 2θ
◦ √ √ √ √
1
9. Show that cot 7 = 2 + 3 + 4 + 6.
2
Solution:

15 15
cos 2 cos 2

15 2 2 1 + cos 15
cot = = =
2 15 15 15 sin 15
sin 2 sin cos
2 2 2
√ √ √ √
1 3 1 1 3 + 1 6 + 2
cos 15 = cos(45 − 30) = cos 45 cos 30 + sin 45 sin 30 = √ + √ = √ =
2 2 2 2 2 2 4
√ √ √ √
1 3 1 1 3 − 1 6 − 2
sin 15 = sin(45 − 30) = sin 45 cos 30 − cos 45 sin 30 = √ − √ = √ =
2 2 2 2 2 2 4
√ √
!
6 + 2
1 + √ √
4
15 1 + cos 15 4 + 6 + 2
cot = = √ √ ! = √ √
2 sin 15 6 − 2 6 − 2
4
√ √ √ √ √ √ √
√ √ √ √ √
15 4 + 6 + 2 6 + 2 4 6 + 6 + 2 3 + 4 2 + 2 3 + 2
cot = = = 6 + 4 + 3 + 2
2 4 4
n
n
10. Prove that (1 + sec 2θ)(1 + sec 4θ) . . . (1 + sec 2 θ) = tan 2 θ cot θ.