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Maths NCERT Exemplar Solutions Class 12th Chapter Eight

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Maths NCERT Exemplar Solutions Class 12th Chapter Eight Exemplar Solution

The area of the region bounded by the curve $y = x + 1$ and the lines $x = 2$ and $x = 3$ is:

(A) $\frac{7}{2}$ sq. units

(B) $\frac{9}{2}$ sq. units

(C) $\frac{11}{2}$ sq. units

(D) $\frac{13}{2}$ sq. units

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Maths NCERT Exemplar Solutions Class 12th Chapter Eight Exemplar Solution

The area of the region bounded by the circle $x^{2} + y^{2} = 1$ is:

(A) $2 π$ sq. units

(B) $π$ sq. units

(C) $3 π$ sq. units

(D) $4 π$ sq. units

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n e q u a t i o n o f c i r c l e i s x^{2} + y^{2} = 1 \\ \Rightarrow y = \sqrt[]{1 - x^{2}} \\ ∴ Since t h e c i r c l e i s s y m m e t r i c a l a b o u t t h e a x e s . \\ ∴ R equired a r e a = 4 * \int_{0}^{1} \sqrt[]{1 - x^{2}} d x \\ = 4 {[\frac{x}{2} \sqrt[]{1 - x^{2}} + \frac{1}{2} s i n^{- 1} x]}_{0}^{1} \\ = 4 [0 + \frac{1}{2} s i n^{- 1} (1) - 0 - 0] = 4 * \frac{1}{2} * \frac{π}{2} = π s q . u n i t s \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Eight Exemplar Solution

The area of the region bounded by the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ is:

(A) $20 π$ sq. units

(B) $20 π^{2}$ sq. units

(C) $16 π^{2}$ sq. units

(D) $25 π$ sq. units

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n e q u a t i o n o f e l l i p s e i s \frac{x^{2}}{25} + \frac{y^{2}}{16} = 1 \\ \Rightarrow \frac{y^{2}}{16} = 1 - \frac{x^{2}}{25} \Rightarrow y^{2} = \frac{16}{25} (25 - x^{2}) \\ ∴ y = \frac{4}{5} \sqrt[]{25 - x^{2}} \\ ∴ Since t h e e l l i p s e i s s y m m e t r i c a l a b o u t t h e a x e s . \\ ∴ R equired a r e a = 4 \int_{0}^{5} \frac{4}{5} \sqrt[]{25 - x^{2}} d x = 4 * \frac{4}{5} \int_{0}^{5} \sqrt[]{{(5)}^{2} - x^{2}} d x \\ = \frac{16}{5} {[\frac{x}{2} \sqrt[]{{(5)}^{2} - x^{2}} + \frac{25}{2} s i n^{- 1} \frac{x}{5}]}_{0}^{5} \\ = \frac{16}{5} [0 + \frac{25}{2} s i n^{- 1} (\frac{5}{5}) - 0 - 0] = \frac{16}{5} [\frac{25}{2} . s i n^{- 1} (1)] \\ = \frac{16}{5} [\frac{25}{2} . \frac{π}{2}] = 20 π s q . u n i t s \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Eight Exemplar Solution

The area of the region bounded by the curve $y = s i n x$ between the ordinates $x = 0$ , $x = \frac{π}{2}$ , and the x-axis is:

(A) 2 sq. units

(B) 4 sq. units

(C) 3 sq. units

(D) 1 sq. unit

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n e q u a t i o n o f c u r v e i s y = s i n x b e t w e e n x = 0 a n d x = \frac{π}{2} \\ A r e a o f r equired r e g i o n = \int_{0}^{\frac{π}{2}} s i n x d x \\ = {[- c o s x]}_{0}^{\frac{π}{2}} = - [c o s \frac{π}{2} - c o s 0] = - [0 - 1] = 1 s q . u n i t s \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Eight Exemplar Solution

The area of the region bounded by the parabola $y^{2} = x$ and the straight line $2 y = x$ is:

(A) $\frac{4}{3}$ sq. units

(B) 1 sq. unit

(C) $\frac{2}{3}$ sq. units

(D) $\frac{1}{3}$ sq. units

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n e q u a t i o n o f p a r a b o l a i s y^{2} = x \dots (i) \\ a n d e q u a t i o n o f s t r a i g h t l i n e i s 2 y = x \dots (i i) \\ S o l v i n g e q n . (i) a n d (i i) w e g e t \\ {(\frac{x}{2})}^{2} = x \Rightarrow \frac{x^{2}}{4} = x \Rightarrow x^{2} = 4 x \\ \Rightarrow x (x - 4) = 0 ∴ x = 0, 4 \\ Required a r e a = \int_{0}^{4} \sqrt[]{x} d x - \int_{0}^{4} \frac{x}{2} d x \\ = \frac{2}{3} {[x^{3 / 2}]}_{0}^{4} - \frac{1}{2} . \frac{1}{2} {[x^{2}]}_{0}^{4} \\ = \frac{2}{3} [{(4)}^{3 / 2} - 0] - \frac{1}{4} [{(4)}^{2} - 0] \\ = \frac{2}{3} * 8 - \frac{1}{4} * 16 = \frac{16}{3} - 4 = \frac{4}{3} s q . u n i t s \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Eight Exemplar Solution

The area of the region bounded by the parabola $y^{2} = x$ and the straight line $2 y = x$ is:

(A) $\frac{4}{3}$ sq. units

(B) 1 sq. unit

(C) $\frac{2}{3}$ sq. units

(D) $\frac{1}{3}$ sq. units

0 Follower 6 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n ? ? e q u a t i o n ? ? o f ? ? p a r a b o l a ? ? i s ? ? y^{2} = x ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \dots (i) \\ a n d ? ? e q u a t i o n ? ? o f ? ? s t r a i g h t ? ? l i n e ? ? i s ? ? 2 y = x ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \dots (i i) \\ S o l v i n g ? ? e q n . (i) ? ? a n d ? ? (i i) ? ? w e ? ? g e t \\ ? ? ? ? ? {(\frac{x}{2})}^{2} = x ? ? ? ? ? \frac{x^{2}}{4} = x ? ? ? ? ? x^{2} = 4 x \\ ? x (x ? 4) = 0 ? ? ? ? ? ? ? ? ? ? ? ? ? x = 0, 4 \\ Required ? ? a r e a = ?_{0}^{4} \sqrt[]{x} ? d x ? ?_{0}^{4} \frac{x}{2} ? d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} {[x^{3 / 2}]}_{0}^{4} ? \frac{1}{2} . \frac{1}{2} {[x^{2}]}_{0}^{4} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [{(4)}^{3 / 2} ? 0] ? \frac{1}{4} [{(4)}^{2} ? 0] \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} * 8 ? \frac{1}{4} * 16 = \frac{16}{3} ? 4 = \frac{4}{3} ? s q . ? u n i t s \\ H e n c e, ? ? t h e ? ? c o r r e c t ? ? o p t i o n ? ? i s ? ? (a) . \end{array}$

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Eight Exemplar Solution

The area of the region bounded by the curve y = cosx between $x = 0$ and $x = π$ is:

(A) 2 sq. units

(B) 4 sq. units

(C) 3 sq. units

(D) 1 sq. unit

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : y = c o s x, x = 0, x = π \\ Required a r e a = \int_{0}^{\frac{π}{2}} c o s x d x + | \int_{\frac{π}{2}}^{π} c o s x d x | d x \\ = {[s i n x]}_{0}^{\frac{π}{2}} + | {[s i n x]}_{\frac{π}{2}}^{π} | \\ = [s i n \frac{π}{2} - s i n 0] + | [s i n π - s i n \frac{π}{2}] | \\ = (1 - 0) + | 0 - 1 | = 1 + 1 = 2 s q . u n i t s \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Eight Exemplar Solution

Area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle $x^{2} + y^{2} = 32$ is:

(A) $16 π$ sq. units

(B) $4 π$ sq. units

(C) $32 π$ sq. units

(D) 24 sq. units

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n e q u a t i o n o f c i r c l e i s x^{2} + y^{2} = 32 \Rightarrow x^{2} + y^{2} = {(4 \sqrt[]{2})}^{2} \\ a n d t h e l i n e i s y = x a n d t h e x - a x i s . \\ S o l v i n g t h e t w o e q u a t i o n s w e h a v e \\ x^{2} + x^{2} = 32 \Rightarrow 2 x^{2} = 32 \\ \Rightarrow x^{2} = 16 ∴ x = \pm 4 \\ Required a r e a = \int_{0}^{4} x d x + \int_{4}^{4 \sqrt[]{2}} \sqrt[]{{(4 \sqrt[]{2})}^{2} - x^{2}} d x \\ = \frac{1}{2} {[x^{2}]}_{0}^{4} + {[\frac{x}{2} \sqrt[]{{(4 \sqrt[]{2})}^{2} - x^{2}} + \frac{32}{2} s i n^{- 1} \frac{x}{4 \sqrt[]{2}}]}_{4}^{4 \sqrt[]{2}} \\ = \frac{1}{2} [16 - 0] + [0 + 16 s i n^{- 1} (\frac{4 \sqrt[]{2}}{4 \sqrt[]{2}}) - 2 \sqrt[]{32 - 16} - 16 s i n^{- 1} \frac{x}{4 \sqrt[]{2}}] \\ = 8 + [16 s i n^{- 1} (1) - 8 - 16 s i n^{- 1} \frac{1}{\sqrt[]{2}}] \\ = 8 + 16 . \frac{π}{2} - 8 - 16 . \frac{π}{4} = 8 π - 4 π = 4 π s q . u n i t s \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Eight Exemplar Solution

The area of the region bounded by the curve $y = \sqrt[]{16 - x^{2}}$ and the x-axis is:

(A) 8 sq. units

(B) $20 π$ sq. units

(C) $16 π$ sq. units

(D) $256 π$ sq. units

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} H e r e, e q u a t i o n o f c u r v e i s y = \sqrt[]{16 - x^{2}} \\ Required a r e a = 2 [\int_{0}^{4} \sqrt[]{16 - x^{2}} d x] \\ = 2 {[\frac{x}{2} \sqrt[]{16 - x^{2}} + \frac{16}{2} s i n^{- 1} \frac{x}{4}]}_{0}^{4} \\ = 2 [(0 + 8 s i n^{- 1} \frac{4}{4}) - (0 + 0)] \\ = 2 [8 s i n^{- 1} (1)] = 16 . \frac{π}{2} = 8 π s q . u n i t s \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

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Maths NCERT Exemplar Solutions Class 12th Chapter Eight Exemplar Solution

The area of the region bounded by the curve $x^{2} = 4 y$ and the straight line $x = 4 y - 2$ is:

(A) $\frac{3}{8}$ sq. units

(B) $\frac{5}{8}$ sq. units

(C) $\frac{7}{8}$ sq. units

(D) $\frac{9}{8}$ sq. units

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : T h e e q u a t i o n o f p a r a b o l a i s x^{2} = 4 y \dots (i) \\ a n d e q u a t i o n o f s t r a i g h t l i n e x = 4 y - 2 \dots (i i) \\ S o l v i n g e q n (i) a n d (i i) w e g e t \\ y = \frac{x^{2}}{4} \Rightarrow x = 4 (\frac{x^{2}}{4}) - 2 \\ \Rightarrow x = x^{2} - 2 \Rightarrow x^{2} - x - 2 = 0 \\ \Rightarrow x^{2} - 2 x + x - 2 = 0 \Rightarrow x (x - 2) + 1 (x - 2) = 0 \\ \Rightarrow (x - 2) (x + 1) = 0 ∴ x = - 1, x = 2 \\ Required a r e a = \int_{- 1}^{2} \frac{x + 2}{4} d x - \int_{- 1}^{2} \frac{x^{2}}{4} d x \\ = \frac{1}{4} {[\frac{x^{2}}{2} + 2 x]}_{- 1}^{2} - \frac{1}{4} . \frac{1}{3} {[x^{3}]}_{- 1}^{2} \\ = \frac{1}{4} [(\frac{4}{2} + 2) - (\frac{1}{2} - 2)] - \frac{1}{12} [8 + 1] \\ = \frac{1}{4} [6 + \frac{3}{2}] - \frac{1}{12} [9] = \frac{1}{4} * \frac{15}{2} - \frac{3}{4} = \frac{15}{8} - \frac{3}{4} = \frac{9}{8} s q . u n i t s \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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