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Ncert Solutions Maths class 12th Application of Derivatives

**If y = x? – 10 and x changes from 2 to 1.99, what is the change in *y

(A) 0.32

(B) 0.032

(C) 5.68

(D)** 5.968

4 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} G i v e n t h a t y = x^{4} - 10 \\ \frac{d y}{d x} = 4 x^{3} \\ Δ x = 2.00 - 1.99 = 0.01 \\ ∴ Δ y = \frac{d y}{d x} . Δ x = 4 x^{3} . Δ x \\ = 4 \times {(2)}^{3} \times 0.01 = 32 \times 0.01 = 0.32 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Ncert Solutions Maths class 12th Application of Derivatives

**If the curves ay + x² = 7 and x³ = y cut orthogonally at (1, 1), then the value of a is:

(A) 1

(B) 0

(C) –6

(D)** 6

7 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f t h e g i v e n c u r v e s a r e a y + x^{2} = 7 \dots (i) \\ a n d x^{3} = y \dots (i i) \\ D i f f e r e n t i a t i n g e q . (i) w . r . t . x, w e h a v e \\ a . \frac{d y}{d x} + 2 x = 0 \\ \Rightarrow \frac{d y}{d x} = - \frac{2 x}{a} \\ ∴ m_{1} = - \frac{2 x}{a} (m_{1} = \frac{d y}{d x}) \\ N o w, d i f f e r e n t i a t i n g e q . (i i) w . r . t . t, w e h a v e \\ 3 x^{2} = \frac{d y}{d x} \Rightarrow m_{2} = 3 x^{2} (m_{2} = \frac{d y}{d x}) \\ T h e t w o c u r v e s a r e s a i d t o b e o r t h o g o n a l i f t h e a n g l e b e t w e e n t h e tangents a t t h e point o f \\ intersection i s 9 0^{0} . \\ ∴ m_{1} \times m_{2} = - 1 \\ \Rightarrow - \frac{2 x}{a} \times 3 x^{2} = - 1 \Rightarrow - \frac{6 x^{3}}{a} = - 1 \Rightarrow 6 x^{3} = a \\ (1, 1) i s t h e point o f intersection o f t w o c u r v e s . \\ ∴ 6 {(1)}^{3} = a \Rightarrow a = 6 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Ncert Solutions Maths class 12th Application of Derivatives

**If the curves ay + x² = 7 and x³ = y cut orthogonally at (1, 1), then the value of a is:

(A) 1

(B) 0

(C) –6

(D)** 6

5 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f t h e g i v e n c u r v e s a r e a y + x^{2} = 7 \dots (i) \\ a n d x^{3} = y \dots (i i) \\ D i f f e r e n t i a t i n g e q . (i) w . r . t . x, w e h a v e \\ a . \frac{d y}{d x} + 2 x = 0 \\ \Rightarrow \frac{d y}{d x} = - \frac{2 x}{a} \\ ∴ m_{1} = - \frac{2 x}{a} (m_{1} = \frac{d y}{d x}) \\ N o w, d i f f e r e n t i a t i n g e q . (i i) w . r . t . t, w e h a v e \\ 3 x^{2} = \frac{d y}{d x} \Rightarrow m_{2} = 3 x^{2} (m_{2} = \frac{d y}{d x}) \\ T h e t w o c u r v e s a r e s a i d t o b e o r t h o g o n a l i f t h e a n g l e b e t w e e n t h e tangents a t t h e point o f \\ intersection i s 9 0^{0} . \\ ∴ m_{1} \times m_{2} = - 1 \\ \Rightarrow - \frac{2 x}{a} \times 3 x^{2} = - 1 \Rightarrow - \frac{6 x^{3}}{a} = - 1 \Rightarrow 6 x^{3} = a \\ (1, 1) i s t h e point o f intersection o f t w o c u r v e s . \\ ∴ 6 {(1)}^{3} = a \Rightarrow a = 6 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Ncert Solutions Maths class 12th Application of Derivatives

**The equation of the normal to the curve *3x² – y² = 8, which is parallel to the line **x + 3y = 8*, is:

(A) 3x – y = 8

(B) 3x + y + 8 = 0

(C) x + 3y ± 8 = 0

(D)** x + 3y = 0

4 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n o f t h e c u r v e 3 x^{2} - y^{2} = 8 \\ D i f f e r e n t i a t i n g b o t h w . r . t . x, w e g e t \\ 6 x - 2 y . \frac{d y}{d x} = 0 \\ \Rightarrow \frac{d y}{d x} = \frac{3 x}{y} \\ \frac{3 x}{y} i s t h e s l o p e o f t h e tangent . \\ ∴ s l o p e o f t h e n o r m a l = \frac{- 1}{\frac{d y}{d x}} = \frac{- y}{3 x} \\ N o w x + 3 y = 8 i s p a r a l l e l t o t h e n o r m a l \\ D i f f e r e n t i a t i n g b o t h w . r . t . t, w e h a v e \\ 1 + 3 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{1}{3} \\ ∴ \frac{- y}{3 x} = - \frac{1}{3} \Rightarrow y = x \\ P u t t i n g y = x i n e q . (i) w e g e t \\ 3 x^{2} - x^{2} = 8 \Rightarrow 2 x^{2} = 8 \Rightarrow x^{2} = 4 \\ ∴ x = \pm 2 a n d y = \pm 2 \\ S o t h e points a r e (2, 2) a n d (- 2, - 2) . \\ E q u a t i o n o f n o r m a l t o t h e g i v e n c u r v e a t (2, 2) i s \\ y - 2 = - \frac{1}{3} (x - 2) \\ \Rightarrow 3 y - 6 = - x + 2 \Rightarrow x + 3 y - 8 = 0 \\ E q u a t i o n o f n o r m a l a t (- 2, - 2) i s \\ &thin \end{array}$

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Ncert Solutions Maths class 12th Application of Derivatives

**The curve y = x^1/? has at (0, 0):

(A) A vertical tangent (parallel to y-axis)

(B) A horizontal tangent (parallel to x-axis)

(C) An oblique tangent

(D)** No tangent

6 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} E q u a t i o n o f c u r v e i s y = x^{1 / 5} \\ D i f f e r e n t i a t i n g w . r . t . x, w e g e t \frac{d y}{d x} = \frac{1}{5} x^{- 4 / 5} \\ (a t x = 0) \frac{d y}{d x} = \frac{1}{5} {(0)}^{- 4 / 5} = \frac{1}{5} \times \frac{1}{0} = \infty \\ \frac{d y}{d x} = \infty \\ ∴ T h e tangent is parallel to y-axis. \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Ncert Solutions Maths class 12th Application of Derivatives

A ladder, 5 meters long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when the lower end of the ladder is 2 meters from the wall is: (A)  1/10 radian/sec (B)  1/20 radian/sec (C) 20 radian/sec (D) 10 radian/sec

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Ncert Solutions Maths class 12th Application of Derivatives

The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases when the side is 10 cm is:

(A) 10 cm²/s

(B) √3 cm²/s

(C) 10 cm²/s

(D) 10/3 cm²/s

7 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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Ncert Solutions Maths class 12th Application of Derivatives

Prove that f(x) = sinx + cosx has a maximum value at x = , $\frac{π}{6}$ .

6 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

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Ncert Solutions Maths class 12th Application of Derivatives

Prove that f(x) = sinx + cosx has a maximum value at x = , <math> <mrow> <mo> </mo> <mfrac> <mrow> <mi>π</mi> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> </mrow> </math> .

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Ncert Solutions Maths class 12th Application of Derivatives

At what point is the slope of the curve y = –x³ + 3x² + 9x – 27 maximum? Also, find the maximum slope.

11 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol.

$\begin{array}{l} G i v e n t h a t y = - x^{3} + 3 x^{2} + 9 x - 27 \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t \\ \frac{d y}{d x} = - 3 x^{2} + 6 x + 9 \\ L e t s l o p e o f t h e c u r v e \frac{d y}{d x} = Z \\ ∴ z = - 3 x^{2} + 6 x + 9 \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t \frac{d z}{d x} = - 6 x + 6 \\ F o r l o c a l maxima a n d l o c a l minima, \frac{d z}{d x} = 0 \\ ∴ - 6 x + 6 = 0 \Rightarrow x = 1 \\ \Rightarrow \frac{d^{2} z}{d x^{2}} = - 6 < 0 M a x i m a \\ P u t x = 1 i n e q u a t i o n o f t h e c u r v e y = {(- 1)}^{3} + 3 {(1)}^{2} + 9 (1) - 27 \\ = - 1 + 3 + 9 - 27 = - 16 \\ M a x i m u m s l o p e = - 3 {(1)}^{2} + 6 (1) + 9 = 12 \\ H e n c e, (1, - 16) i s t h e point a t w h i c h t h e s l o p e o f t h e g i v e n c u r v e i s maximum a n d \\ M a x i m u m s l o p e = 12 \end{array}$

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Ncert Solutions Maths class 12th

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**If y = x? – 10 and x changes from 2 to 1.99, what is the change in *y

(A) 0.32

(B) 0.032

(C) 5.68

(D)** 5.968

**If the curves ay + x² = 7 and x³ = y cut orthogonally at (1, 1), then the value of a is:

(A) 1

(B) 0

(C) –6

(D)** 6

**If the curves ay + x² = 7 and x³ = y cut orthogonally at (1, 1), then the value of a is:

(A) 1

(B) 0

(C) –6

(D)** 6

**The equation of the normal to the curve *3x² – y² = 8, which is parallel to the line **x + 3y = 8*, is:

(A) 3x – y = 8

(B) 3x + y + 8 = 0

(C) x + 3y ± 8 = 0

(D)** x + 3y = 0

**The curve y = x^1/? has at (0, 0):

(A) A vertical tangent (parallel to y-axis)

(B) A horizontal tangent (parallel to x-axis)

(C) An oblique tangent

(D)** No tangent

The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases when the side is 10 cm is:

(A) 10 cm²/s

(B) √3 cm²/s

(C) 10 cm²/s

(D) 10/3 cm²/s

Prove that f(x) = sinx + cosx has a maximum value at x = , $\frac{π}{6}$ .

At what point is the slope of the curve y = –x³ + 3x² + 9x – 27 maximum? Also, find the maximum slope.