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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

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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Asked by Payal Gupta

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Payal Gupta | Contributor-Level 10

a year ago

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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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

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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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

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

6 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

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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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

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

3 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

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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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

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

4 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

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

(A) $\sqrt[]{2}$ sq. units

(B) $\sqrt[]{2}$ +1sq. units

(C) $\sqrt[]{2} - 1$ sq. units

(D) $2 \sqrt[]{2} - 1$ sq. units

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Draw a rough sketch of the given curve $y = 1 + ? x + 1 ?$ , $x = - 3$ , $x = 3$ , $y = 0$  and find the area of the region bounded by them using integration.

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Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n e q u a t i o n s a r e y = 1 + | x + 1 |, x = - 3 a n d x = 3, y = 0 \\ T a k i n g y = 1 + | x + 1 | \\ \Rightarrow y = 1 + x + 1 \\ \Rightarrow y = x + 2 a n d y = 1 - x - 1 \Rightarrow y = - x \\ O n s o l v i n g w e g e t x = - 1 \\ ∴ A r e a o f r e q u i r e d r e g i o n s \\ = \int_{- 3}^{- 1} - x d x + \int_{- 1}^{3} (x + 2) d x \\ = - {[\frac{x^{2}}{2}]}_{- 3}^{- 1} + {[\frac{x^{2}}{2} + 2 x]}_{- 1}^{3} = - [\frac{1}{2} - \frac{9}{2}] + [(\frac{9}{2} + 6) - (\frac{1}{2} - 2)] \\ = - (- 4) + [\frac{21}{2} + \frac{3}{2}] = 4 + 12 = 16 s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = 16 s q . u n i t s . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Find the area bounded by the curve $y = 2 c o s x$ and the x-axis from $x = 0$ to $x = 2 π$ .

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n e q u a t i o n o f t h e c u r v e i s y = 2 c o s x \\ ∴ A r e a o f s h a d e d r e g i o n \\ = \int_{0}^{2 π} 2 c o s x d x = \int_{0}^{\frac{π}{2}} 2 c o s x d x + \int_{\frac{π}{2}}^{\frac{3 π}{2}} | 2 c o s x | d x + \int_{\frac{3 π}{2}}^{2 π} 2 c o s x d x \\ = 2 {[s i n x]}_{0}^{\frac{π}{2}} + | {[s i n x]}_{\frac{π}{2}}^{\frac{3 π}{2}} | + 2 {[s i n x]}_{\frac{3 π}{2}}^{2 π} \\ = 2 [s i n \frac{π}{2} - s i n 0] + | 2 (s i n \frac{3 π}{2} - s i n \frac{π}{2}) | + 2 [s i n 2 π - s i n \frac{3 π}{2}] \\ = 2 (1) + | 2 (- 1 - 1) | + 2 (0 + 1) = 2 + 4 + 2 = 8 s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = 8 s q . u n i t s . \end{array}$

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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

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

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

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

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

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

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

(A) $\sqrt[]{2}$ sq. units

(B) $\sqrt[]{2}$ +1sq. units

(C) $\sqrt[]{2} - 1$ sq. units

(D) $2 \sqrt[]{2} - 1$ sq. units

Find the area bounded by the curve $y = 2 c o s x$ and the x-axis from $x = 0$ to $x = 2 π$ .