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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Find the area bounded by the lines $y = 4 x + 5$ , $y = 5 - x$ , and $4 y = x + 5$ .

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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

$\begin{array}{l} G i v e n t h a t y = 4 x + 5 \dots (i) \\ y = 5 - x \dots (i i) \\ a n d 4 y = x + 5 \dots (i i i) \\ \begin{matrix} x & 0 \\ - 5 / 4 & y \\ 5 & 0 \end{matrix} \begin{matrix} x & 0 \\ 5 & y \\ 5 & 0 \end{matrix} \begin{matrix} x & 0 \\ - 5 & y \\ 5 / 4 & 0 \end{matrix} \\ S o l v i n g e q n (i) a n d (i i) w e g e t \\ 4 x + 5 = 5 - x \\ ∴ x = 0 a n d y = 5 \\ C o o r d i n a t e s o f A = (0, 5) \\ S o l v i n g e q n (i i) a n d (i i i) w e g e t \\ y = 5 - x \\ ∴ 4 y = x + 5 \\ 5 y = 10 \\ ∴ y = 2 a n d x = 3 \\ ∴ C o o r d i n a t e s o f B = (3, 2) \\ S o l v i n g e q n (i) a n d (i i i) w e g e t \\ y = 4 x + 5 a n d 4 y = x + 5 \\ \Rightarrow 4 (4 x + 5) = x + 5 \\ \Rightarrow 16 x + 20 = x + 5 \Rightarrow 15 x = - 15 \\ ∴ x = - 1 a n d y = 1 \\ ∴ C o o r d i n a t e s o f C = (- 1, 1) \\ ∴ A r e a o f r e q u i r e d r e g i o n s \\ = \int_{- 1}^{0} y_{A C} d x + \int_{0}^{3} y_{A B} d x - \int_{- 1}^{3} y_{C B} d x \\ = \int_{- 1}^{0} (4 x + 5) d x + \int_{0}^{3} (5 - x) d x - \int_{- 1}^{3} (\frac{x + 5}{4}) d x \\ = {[4 \frac{x^{2}}{2} + 5 x]}_{- 1}^{0} + {[5 x - \frac{x^{2}}{2}]}_{0}^{3} - \frac{1}{4} {[\frac{x^{2}}{2} + 5 x]}_{- 1}^{3} \\ = [(0 + 0) - (2 - 5)] + [(15 - \frac{9}{2}) - (0 - 0)] - \frac{1}{4} [(\frac{9}{2} + 15) - (\frac{1}{2} - 5)] \\ = 3 + \frac{21}{2} - \frac{1}{4} [\frac{39}{2} + \frac{9}{2}] = 3 + \frac{21}{2} - \frac{1}{4} \times 24 \\ = 3 + \frac{21}{2} - 6 = \frac{21}{2} s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = \frac{21}{2} s q . u n i t s . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Compute the area bounded by the lines $x + 2 y = 2$ , $y - x = 1$ , and $2 x + y = 7$ .

4 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t x + 2 y = 2 \dots (i) \\ y - x = 1 \dots (i i) \\ a n d 2 x + y = 7 \dots (i i i) \\ \begin{matrix} x & 0 \\ 2 & y \\ 1 & 0 \end{matrix} \begin{matrix} x & 0 \\ - 1 & y \\ 1 & 0 \end{matrix} \begin{matrix} x & 0 \\ 7 / 2 & y \\ 7 & 0 \end{matrix} \\ S o l v i n g e q n (i i) a n d (i i i) w e g e t \\ y = 1 + x \\ ∴ 2 x + 1 + x = 7 \\ 3 x = 6 \Rightarrow x = 2 \\ ∴ y = 1 + 2 = 3 \\ C o o r d i n a t e s o f B = (2, 3) \\ S o l v i n g e q n (i) a n d (i i i) w e g e t \\ x + 2 y = 2 \\ ∴ x = 2 - 2 y \\ 2 x + y = 7 \\ 2 (2 - 2 y) + y = 7 \\ \Rightarrow 4 - 4 y + y = 7 \Rightarrow - 3 y = 3 \\ ∴ y = - 1 a n d x = 4 \\ ∴ C o o r d i n a t e s o f C = (4, - 1) a n d C o o r d i n a t e s o f A = (0, 1) \\ T a k i n g t h e l i m i t s o n y - a x i s, w e g e t \\ \int_{- 1}^{3} x_{B C} d y - \int_{- 1}^{1} x_{A C} d y - \int_{1}^{3} x_{A B} d y \\ = \int_{- 1}^{3} \frac{7 - y}{2} d y - \int_{- 1}^{1} (2 - 2 y) d y - \int_{1}^{3} (y - 1) d y \\ = \frac{1}{2} {[7 y - \frac{y^{2}}{2}]}_{- 1}^{3} - 2 {[y - \frac{y^{2}}{2}]}_{- 1}^{1} - {[\frac{y^{2}}{2} - y]}_{1}^{3} \\ = \frac{1}{2} [(21 - \frac{9}{2}) - (- 7 - \frac{1}{2})] - 2 [(1 - \frac{1}{2}) - (- 1 - \frac{1}{2})] - [(\frac{9}{2} - 3) - (\frac{1}{2} - 1)] \\ = \frac{1}{2} [\frac{33}{2} + \frac{15}{2}] - 2 [\frac{1}{2} + \frac{3}{2}] - [\frac{3}{2} + \frac{1}{2}] \\ = \frac{1}{2} \times 24 - 2 \times 2 - 2 = 12 - 4 - 2 = 6 s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = 6 s q . u n i t s . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Draw a rough sketch of the region ${(x, y) : y^{2} \leq 6 a x and x^{2} + y^{2} \leq 16 a^{2}}$ . Also, find the area of the region sketched using the method of integration.

4 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} W e a r e g i v e n t h a t {(x, y) : y^{2} \leq 6 a x a n d x^{2} + y^{2} \leq 16 a^{2}} \\ E q u a t i o n o f P a r a b o l a i s y^{2} = 6 a x \dots (i) \\ a n d e q u a t i o n o f c i r c l e i s x^{2} + y^{2} = 16 a^{2} \dots (i i) \\ S o l v i n g e q n . (i) a n d (i i) w e g e t \\ x^{2} + 6 a x = 16 a^{2} \\ \Rightarrow x^{2} + 6 a x - 16 a^{2} = 0 \\ \Rightarrow x^{2} + 8 a x - 2 a x - 16 a^{2} = 0 \\ \Rightarrow x (x + 8 a) - 2 a (x + 8 a) = 0 \\ \Rightarrow (x + 8 a) (x - 2 a) = 0 \\ ∴ x = 2 a a n d x = - 8 a . (Rejected a s i t o u t o f r e g i o n) \\ A r e a o f t h e required s h a d e d r e g i o n = 2 [\int_{0}^{2 a} \sqrt[]{6 a x} d x + \int_{2 a}^{4 a} \sqrt[]{16 a^{2} - x^{2}}] \\ = 2 [\sqrt[]{6 a} \int_{0}^{2 a} \sqrt[]{x} d x + \int_{2 a}^{4 a} \sqrt[]{{(4 a)}^{2} - x^{2}}] \\ = 2 \sqrt[]{6 a} . \frac{2}{3} {[x^{3 / 2}]}_{0}^{2 a} + 2 {[\frac{x}{2} \sqrt[]{{(4 a)}^{2} - x^{2}} + \frac{16 a^{2}}{2} s i n^{- 1} \frac{x}{4 a}]}_{2 a}^{4 a} \\ = \frac{4 \sqrt[]{6}}{3} \sqrt[]{a} [{(2 a)}^{3 / 2} - 0] + {[x \sqrt[]{{(4 a)}^{2} - x^{2}} + 16 a^{2} s i n^{- 1} \frac{x}{4 a}]}_{2 a}^{4 a} \\ = \frac{4 \sqrt[]{6}}{3} \sqrt[]{a} . 2 \sqrt[]{2} . a^{3 / 2} + [0 + 16 a^{2} s i n^{- 1} (\frac{4 a}{4 a}) - 2 a \sqrt[]{16 a^{2} - 4 a^{2}} - 16 a^{2} s i n^{- 1} \frac{2 a}{4 a}] \\ = \frac{8 \sqrt[]{12}}{3} a^{2} + [16 a^{2} s i n^{- 1} (1) - 2 a \sqrt[]{12 a^{2}} - 16 a^{2} s i n^{- 1} \frac{1}{2}] \\ = \frac{16 \sqrt[]{3}}{3} a^{2} + [16 a^{2} . \frac{π}{2} - 2 a . 2 \sqrt[]{3} a - 16 a^{2} . \frac{π}{6}] \\ = \frac{16 \sqrt[]{3}}{3} a^{2} + 8 π a^{2} - 4 \sqrt[]{3} a^{2} - \frac{8}{3} π a^{2} \\ = (\frac{16 \sqrt[]{3}}{3} - 4 \sqrt[]{3}) a^{2} + \frac{16}{3} π a^{2} = \frac{4 \sqrt[]{3}}{3} a^{2} + \frac{16}{3} π a^{2} \\ = \frac{4}{3} (\sqrt[]{3} + 4 π) a^{2} s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = \frac{4}{3} (\sqrt[]{3} + 4 π) a^{2} s q . u n i t s . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Find the area of the region bounded by the triangle whose vertices are $(- 1, 1)$ , $(0, 5)$ , and $(3, 2)$ , using integration.

3 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} T h e c o o r d i n a t e s o f t h e v e r t i c e s o f Δ A B C a r e g i v e n b y A (- 1, 1), B (0, 5) C (3, 2) . \\ E q u a t i o n o f A B i s y - 1 = \frac{5 - 1}{0 + 1} (x + 1) \\ \Rightarrow y - 1 = 4 x + 4 \Rightarrow y = 4 x + 4 + 1 \\ \Rightarrow y = 4 x + 5 \dots (i) \\ E q u a t i o n o f B C i s y - 5 = \frac{2 - 5}{3 - 0} (x - 0) \\ \Rightarrow y - 5 = - x \\ \Rightarrow y = 5 - x \dots (i i) \\ E q u a t i o n o f C A i s y - 1 = \frac{2 - 1}{3 + 1} (x + 1) \\ \Rightarrow y - 1 = \frac{1}{4} x + \frac{1}{4} \Rightarrow y = \frac{1}{4} x + \frac{1}{4} + 1 \\ \Rightarrow y = \frac{1}{4} x + \frac{5}{4} = \frac{1}{4} (5 + x) \dots (i i i) \\ A rea o f Δ A B C = \int_{- 1}^{0} (4 x + 5) d x + \int_{0}^{3} (5 - x) d x - \int_{- 1}^{3} \frac{1}{4} (5 + x) d x \\ = \frac{4}{2} {[x^{2}]}_{- 1}^{0} + 5 {[x]}_{- 1}^{0} + 5 {[x]}_{0}^{3} - \frac{1}{2} {[x^{2}]}_{0}^{3} - \frac{1}{4} {[5 x + \frac{x^{2}}{2}]}_{- 1}^{3} \\ = 2 (0 - 1) + 5 (0 + 1) + 5 (3 - 0) - \frac{1}{2} (9 - 0) - \frac{1}{4} [(15 + \frac{9}{2}) - (- 5 + \frac{1}{2})] \\ = - 2 + 5 + 15 - \frac{9}{2} - \frac{1}{4} (\frac{39}{2} + \frac{9}{2}) \\ = 18 - \frac{9}{2} - \frac{1}{4} \times \frac{48}{2} = 18 - \frac{9}{2} - 6 = 12 - \frac{9}{2} = \frac{15}{2} s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = \frac{15}{2} s q . u n i t s . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Find the area bounded by the curve $y = s i n x$ between $x = 0$ and $x = 2 π$

5 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Find the area of the region bounded by the curves $y^{2} = 2 x$ and $x^{2} + y^{2} = 4 x$ .

7 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} E q u a t i o n s o f t h e c u r v e a r e g i v e n b y \\ x^{2} + y^{2} = 4 x \dots (i) \\ a n d y^{2} = 2 x \dots (i i) \\ \Rightarrow x^{2} - 4 x + y^{2} = 0 \\ \Rightarrow x^{2} - 4 x + 4 - 4 + y^{2} = 0 \\ {(x - 2)}^{2} + y^{2} = 4 \\ C l e a r l y i t i s t h e e q u a t i o n o f a c i r c l e h a v i n g i t s c e n t r e (2, 0) a n d r a d i u s 2 . \\ S o l v i n g x^{2} + y^{2} = 4 x a n d y^{2} = 2 x \\ x^{2} + 2 x = 4 x \\ \Rightarrow x^{2} + 2 x - 4 x = 0 \\ \Rightarrow x^{2} - 2 x = 0 \\ \Rightarrow x (x - 2) = 0 \\ ∴ x = 0, 2 \\ Area of the r equired r e g i o n = 2 [\int_{0}^{2} \sqrt[]{4 - {(x - 2)}^{2}} d x - \int_{0}^{2} \sqrt[]{2 x} d x] \\ [? P a r a b o l a a n d c i r c l e b o t h a r e s y m m e t r i c a l a b o u t x - a x i s] \\ = 2 {[\frac{x - 2}{2} \sqrt[]{4 - {(x - 2)}^{2}} + \frac{4}{2} s i n^{- 1} \frac{x - 2}{2}]}_{0}^{2} - 2 . \sqrt[]{2} . \frac{2}{3} {[x^{3 / 2}]}_{0}^{2} \\ = 2 [(0 + 0) - (0 + 2 s i n^{- 1} (- 1))] - \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Find the area bounded by the curves $y = \sqrt[]{x}$ and $x = 2 y + 3$ in the first quadrant and the x-axis.

3 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t y = \sqrt[]{x}, x = 2 y + 3, f i r s t q u a d r a n t a n d x - a x i s . \\ S o l v i n g y = \sqrt[]{x} a n d x = 2 y + 3, w e g e t \\ y = \sqrt[]{2 y + 3} \Rightarrow y^{2} = 2 y + 3 \\ \Rightarrow y^{2} - 2 y - 3 = 0 \Rightarrow y^{2} - 3 y + y - 3 = 0 \\ \Rightarrow y (y - 3) + 1 (y - 3) = 0 \Rightarrow (y + 1) (y - 3) = 0 \\ ∴ y = - 1, 3 \\ A r e a o f t h e s h a d e d r e g i o n = \int_{0}^{3} (2 y + 3) d y - \int_{0}^{3} y^{2} d y \\ = {[2 \frac{y^{2}}{2} + 3 y]}_{0}^{3} - \frac{1}{3} {[y^{3}]}_{0}^{3} \\ = [(9 + 9) - (0 + 0)] - \frac{1}{3} [27 - 0] \\ = 18 - 9 = 9 s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = 9 s q . u n i t s . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Find the area enclosed by the curve $y = - x^{2}$ and the straight line $x + y + 2 = 0$ .

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 a short answer type question as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Find the area of the region bounded by  $y = \sqrt[]{x}$ and $y = x$ .

4 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} W e a r e g i v e n t h e e q u a t i o n s o f c u r v e y = \sqrt[]{x} a n d l i n e y = x . \\ S o l v i n g y = \sqrt[]{x} \Rightarrow y^{2} = x a n d y = x, w e g e t \\ x^{2} = x \Rightarrow x^{2} - x = 0 \\ \Rightarrow x (x - 1) = 0 ∴ x = 0, 1 \\ A r e a o f t h e required s h a d e d r e g i o n = \int_{0}^{1} \sqrt[]{x} d x - \int_{0}^{1} x d x \\ = \frac{2}{3} {[x^{3 / 2}]}_{0}^{1} - \frac{1}{2} {[x^{2}]}_{0}^{1} \\ = \frac{2}{3} [{(1)}^{3 / 2} - 0] - \frac{1}{2} [{(1)}^{2} - 0] \\ = \frac{2}{3} - \frac{1}{2} = \frac{4 - 3}{6} = \frac{1}{6} s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = \frac{1}{6} s q . u n i t s . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Determine the area under the curve $y = \sqrt[]{a^{2} - x^{2}}$ included between the lines $x = 0$ and .

5 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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Find the area bounded by the lines $y = 4 x + 5$ , $y = 5 - x$ , and $4 y = x + 5$ .

Compute the area bounded by the lines $x + 2 y = 2$ , $y - x = 1$ , and $2 x + y = 7$ .

Draw a rough sketch of the region ${(x, y) : y^{2} \leq 6 a x and x^{2} + y^{2} \leq 16 a^{2}}$ . Also, find the area of the region sketched using the method of integration.

Find the area of the region bounded by the triangle whose vertices are $(- 1, 1)$ , $(0, 5)$ , and $(3, 2)$ , using integration.

Find the area bounded by the curve $y = s i n x$ between $x = 0$ and $x = 2 π$

Find the area of the region bounded by the curves $y^{2} = 2 x$ and $x^{2} + y^{2} = 4 x$ .

Find the area bounded by the curves $y = \sqrt[]{x}$ and $x = 2 y + 3$ in the first quadrant and the x-axis.

Find the area enclosed by the curve $y = - x^{2}$ and the straight line $x + y + 2 = 0$ .

Find the area of the region bounded by  $y = \sqrt[]{x}$ and $y = x$ .

Determine the area under the curve $y = \sqrt[]{a^{2} - x^{2}}$ included between the lines $x = 0$ and .