NCERT Solutions Class 11th Mathematics. FREE Flip-BOOK by Study Innovations. 614 Pages

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Convert the following in the polar form:
(i) , (ii)
Answer :
(i) Here,

Let r cos θ = –1 and r sin θ = 1

On squaring and adding, we obtain

r2 (cos2 θ + sin2 θ) = 1 + 1

⇒ r2 (cos2 θ + sin2 θ) = 2 [cos2 θ + sin2 θ = 1]
⇒ r2 = 2

∴z = r cos θ + i r sin θ
This is the required polar form.
(ii) Here,

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Let r cos θ = –1 and r sin θ = 1

On squaring and adding, we obtain

r2 (cos2 θ + sin2 θ) = 1 + 1
⇒r2 (cos2 θ + sin2 θ) = 2

⇒ r2 = 2 [cos2 θ + sin2 θ = 1]

∴z = r cos θ + i r sin θ

This is the required polar form.

Q6 :

Solve the equation
Answer :

The given quadratic equation is
This equation can also be written as
On comparing this equation with ax2 + bx + c = 0, we obtain
a = 9, b = –12, and c = 20
Therefore, the discriminant of the given equation is
D = b2 – 4ac = (–12)2 – 4 × 9 × 20 = 144 – 720 = –576

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x

x
are …–3, –2, –1, 0, 1, 2, 3, 4.

x
…–3, –2, –1, 0, 1, 2, 3, 4.
Hence, in this case, the solution set is {…–3, –2, –1, 0, 1, 2, 3, 4}.

The given inequality is –1 x

x

are …, –5, –4, –3
x
…, –5, –4, –3.
Hence, in this case, the solution set is {…, –5, –4, –3}.

x– 3 < 7

are …, –4, –3, –2, –1, 0, 1 x x
x
…, –4, –3, –2, –1, 0, 1.
Hence, in this case, the solution set is {…, –4, –3, –2, –1, 0, 1}.
x

x ∈(–∞

x

(i) The integers greater than – are –1, 0, 1, 2, …
x

–1, 0, 1, 2 …
Hence, in this case, the solution set is {–1, 0, 1, 2, …}.

xis a real number, the solutions of the given inequality are all the real numbers, which are greater than –2
this case, the solution set is (– 2 ∞

xx

⇒x x

⇒x x

⇒x xxx

⇒x

x

∠

x– 7 > x–
⇒ x– 7 + 7 > x– 1 +
⇒x x
⇒ x– x x+ 6 – x
⇒ – x

x,which are less than –3, are the solutions of the given inequality.
Hence, the solution set of the given inequality is (–∞, –3)

x â”°Â x â”°Â
∠
⇒ x â”°Â x

⇒x â”°Â x

⇒ x â”°Â x

⇒ x - x â”°Â x x

⇒ x â”°Â

x

â”°Â

x â”°Â x

⇒ x â”°Â x

⇒ x x â”°Â xx

⇒ x â”°Â

⇒ x â”°Â
⇒ x â”°Â
⇒ x â”°Â

x

∠

x
Hence, the solution set of the given inequality is (–∞

x,which are less than –6, are the solutions of the given inequality.
Hence, the solution set of the given inequality is (–∞, –6)

x
Hence, the solution set of the given inequality is (–∞

x
Hence, the solution set of the given inequality is (–∞, 120].

x
Hence, the solution set of the given inequality is (4, ∞)

â”°Â
x
Hence, the solution set of the given inequality is (–∞

x
∞

x
Hence, the solution set of the given inequality is (–∞

xx
⇒x x
⇒x

â”°Â



x

x x

x
x
⇒ x< 10 –
⇒ x< 8 … (i
∴x x
⇒x
⇒ x> 11 –
⇒x

xx

x

x
x– 2) c

â”°Â â”°Â
x
x

x x≤

xx
⇒x ≤

⇒ x ≤91 –
⇒ x≤

∴ x≥x
⇒ x≥x
⇒ x ≥8 … (2

≤x≤

xy
xy

â”°Â

xy
xy

2(0) + 0 â”°Â≦ 6 or 0 â”°Â≦ 6, which is false

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x y â”°Â xy
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x xy
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y â”°Â

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y y

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x â”°Â y x
y â”°Â xy
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y

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x y â”°Â

xy x y

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xy

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x â”°Â

xy yx

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xy x

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x y â”°Â
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y â”°Â

xy x y xy
y x
xy
x

y

â”°Â

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x y â”°Â xy xy xy xy
x y â”°Â y â”°Â xy
xy
x â”°Â

â”°Â â”°Â â”°Â

x yâ”°Â xy xy xy xy
x y â”°Â xy xy xy

x y â”°Â

xy
xy

≤ 3, ≥ 12, ≥ 0, ≥ 1

x y ┰¤ 3 ... (1)

x y â”°Â≦ 12 ... (2)

y â”°Â≦ 1 ... (3)

xy xy y

xy xy

xy xy

yy

x â”°Â≦ 0, represents the region on the right hand side of y y

y