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3 months ago

Application of Derivatives Maths NCERT Exemplar Solutions Class 12th Chapter Six

The function y = x (x – 3) ² decreases for the values of x given by:

(A) 1 < x < 3

(B) x < 0

(C) x > 0

(D) 0 < x < 3/2

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} T h e g i v e n f u n c t i o n i s y = x {(x - 3)}^{2} \\ \frac{d y}{d x} = x . 2 (x - 3) + {(x - 3)}^{2} . 1 \\ \Rightarrow \frac{d y}{d x} = 2 x (x - 3) + {(x - 3)}^{2} \\ F o r increasing a n d decreasing \frac{d y}{d x} = 0 \\ ∴ 2 x (x - 3) + {(x - 3)}^{2} = 0 \\ \Rightarrow (x - 3) (2 x + x - 3) = 0 \Rightarrow (x - 3) (3 x - 3) = 0 \\ \Rightarrow 3 (x - 3) (x - 1) = 0 \\ ∴ x = 1, x = 3 \\ T h e p o s s i b l e intervals a r e (- \infty, 1), (1, 3), (3, \infty) \\ N o w \frac{d y}{d x} = (x - 3) (x - 1) \\ \Rightarrow F o r (- \infty, 1) = (-) (-) = (+) increasing \\ \Rightarrow F o r (1, 3) = (-) (+) = (-) decreasing \\ \Rightarrow F o r (3, \infty) = (+) (+) = (+) increasing \\ So the function decreas e s i n (1, 3) o r 1 < x < 3 \\ H e n c e, &thin \end{array}$

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New answer posted

3 months ago

Application of Derivatives Maths NCERT Exemplar Solutions Class 12th Chapter Six

Let the function f: R → R be defined by f(x) = 2x + cosx, then f:

(A) Has a minimum at x = π

(B) Has a maximum at x = 0

(C) Is a decreasing function

(D) Is an increasing function

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} G i v e n t h a t f (x) = 2 x + c o s x \\ f^{'} (x) = 2 - s i n x \\ Since f^{'} (x) > 0 \forall x \\ S o f (x) i s a n increasing f u n c t i o n . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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3 months ago

Application of Derivatives Maths NCERT Exemplar Solutions Class 12th Chapter Six

The interval on which the function f(x) = 2x³ + 9x² + 12x – 1 is decreasing is:

(A) [–1, ∞)

(B) [–2, –1]

(C) (–∞, –2]

(D) [–1, 1]

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n f u n c t i o n i s f (x) = 2 x^{3} + 9 x^{2} + 12 x - 1 \\ f^{'} (x) = 6 x^{2} + 18 x + 12 \\ F o r increasing a n d decreasing f^{'} (x) = 0 \\ ∴ 6 x^{2} + 18 x + 12 = 0 \\ \Rightarrow x^{2} + 3 x + 2 = 0 \Rightarrow x^{2} + 2 x + x + 2 = 0 \\ \Rightarrow x (x + 2) + 1 (x + 2) = 0 \Rightarrow (x + 2) (x + 1) = 0 \\ \Rightarrow x = - 2, x = - 1 \\ T h e p o s s i b l e intervals a r e (- \infty, - 2), (- 2, - 1), (- 1, \infty) \\ N o w f^{'} (x) = (x + 2) (x + 1) \\ \Rightarrow f^{'} {(x)}_{(- \infty, - 2)} = (-) (-) = (+) increasing \\ \Rightarrow f^{'} {(x)}_{(- 2, - 1)} = (+) (-) = (-) decreasing \\ \Rightarrow f^{'} {(x)}_{(- 1, - \infty)} = (+) (+) = (+) increasing \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

3 months ago

Application of Derivatives Maths NCERT Exemplar Solutions Class 12th Chapter Six

The two curves x³ – 3xy² + 2 = 0 and 3x²y – y³ – 2 = 0 intersect at an angle of:

(A) π/4

(B) π/3

(C) π/2

(D) π/6

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} T h e g i v e n c u r v e a r e x^{3} - 3 x y^{2} + 2 = 0 (i) \\ a n d 3 x^{2} y - y^{3} - 2 = 0 (i i) \\ D i f f e r e n t i a t i n g e q u a t i o n (i) w . r . t, x, w e g e t \\ 3 x^{2} - 3 (x . 2 y \frac{d y}{d x} + y^{2} . 1) = 0 \\ \Rightarrow x^{2} - 2 x y \frac{d y}{d x} - y^{2} = 0 \Rightarrow 2 x y \frac{d y}{d x} = x^{2} - y^{2} \\ ∴ \frac{d y}{d x} = \frac{x^{2} - y^{2}}{2 x y} \\ S o S l o p e l i e s o f t h e c u r v e m_{1} = \frac{x^{2} - y^{2}}{2 x y} \\ D i f f e r e n t i a t i n g e q u a t i o n (i i) w . r . t, x, w e g e t \\ 3 [x^{2} \frac{d y}{d x} + y . 2 x] - 3 y^{2} . \frac{d y}{d x} = 0 \\ x^{2} \frac{d y}{d x} + 2 x y - y^{2} . \frac{d y}{d x} = 0 \Rightarrow (x^{2} - y^{2}) \frac{d y}{d x} = - 2 x y \\ ∴ \frac{d y}{d x} = \frac{- 2 x y}{x^{2} - y^{2}} \\ S o S l o p e l i e s o f t h e c u r v e m_{2} = \frac{- 2 x y}{x^{2} - y^{2}} \\ N o w m_{1} * m_{2} = \frac{x^{2} - y^{2}}{2 x y} * \frac{- 2 x y}{x^{2} - y^{2}} = - 1 \\ S o t h e a n g l e b e t w e e n t h e c u r v e s i s \frac{π}{2} . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

3 months ago

Application of Derivatives Maths NCERT Exemplar Solutions Class 12th Chapter Six

The slope of the tangent to the curve x = t² + 3t – 8, y = 2t² – 2t – 5 at the point (2, –1) is:

(A) 22/7

(B) 6/7

(C) -6/7

(D) –6

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n c u r v e i s x = t^{2} + 3 t - 8 a n d y = 2 t^{2} - 2 t - 5 \\ D i f f e r e n t i a t i n g b o t h e q u a t i o n s w . r . t, t \\ \frac{d x}{d t} = 2 t + 3 a n d \frac{d y}{d t} = 4 t - 2 \\ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{4 t - 2}{2 t + 3} \\ N o w (2, - 1) l i e s o n t h e c u r v e \\ ∴ 2 = t^{2} + 3 t - 8 \Rightarrow t^{2} + 3 t - 10 = 0 \\ \Rightarrow t^{2} + 5 t - 2 t - 10 = 0 \Rightarrow t (t + 5) - 2 (t + 5) = 0 \\ \Rightarrow (t + 5) (t - 2) = 0 \\ ∴ t = 2, t = - 5 a n d - 1 = 2 t^{2} - 2 t - 5 \\ \Rightarrow 2 t^{2} - 2 t - 4 = 0 \\ \Rightarrow t^{2} - t - 2 = 0 \Rightarrow t^{2} - 2 t + t - 2 = 0 \\ \Rightarrow t (t - 2) + 1 (t - 2) = 0 \Rightarrow (t + 1) (t - 2) = 0 \\ \Rightarrow t = - 1 a n d t = 2 \\ S o t = 2 i s c o m m o n v a l u e \\ ∴ S l o p e = {\frac{d y}{d x}}_{x = 2} = \frac{4 * 2 - 2}{2 * 2 + 3} = \frac{6}{7} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

3 months ago

Application of Derivatives Maths NCERT Exemplar Solutions Class 12th Chapter Six

The tangent to the curve y = e²? at the point (0,1) meets the x-axis at:

(A) (0, 1)

(B) (–1/2, 0)

(C) (2, 0)

(D) (0, 2)

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f t h e c u r v e i s y = e^{2 x} \\ S l o p e o f t h e tangent \frac{d y}{d x} = 2 e^{2 x} \Rightarrow {\frac{d y}{d x}}_{(0, 1)} = 2 e^{0} = 2 \\ ∴ E q u a t i o n o f tangent t o t h e c u r v e a t (0, 1) i s \\ y - 1 = 2 (x - 0) \\ \Rightarrow y - 1 = 2 x \Rightarrow y - 2 x = 1 \\ Since t h e tangent m e e t x - a x i s, w h e r e y = 0 \\ ∴ 0 - 2 x = 1 \Rightarrow x = - \frac{1}{2} \\ So, the Point i s (- \frac{1}{2}, 0) \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

3 months ago

Application of Derivatives Maths NCERT Exemplar Solutions Class 12th Chapter Six

The points at which the tangents to the curve y = x³ – 12x + 18 are parallel to the x-axis are:

(A) (2, –2), (–2, –34)

(B) (2, 34), (–2, 0)

(C) (0, 34), (–2, 0)

(D) (2, 2), (–2, 34)

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} G i v e n t h a t y = x^{3} - 12 x + 18 \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e h a v e \\ \Rightarrow \frac{d y}{d x} = 3 x^{2} - 12 \\ Since t h e tangents a r e p a r a l l e l t o x - a x i s, t h e n \frac{d y}{d x} = 0 \\ ∴ 3 x^{2} - 12 = 0 \Rightarrow x = \pm 2 \\ ∴ y_{x = 2} = {(2)}^{3} - 12 (2) + 18 = 8 - 24 + 18 = 2 \\ y_{x = - 2} = {(- 2)}^{3} - 12 (- 2) + 18 = - 8 + 24 + 18 = 34 \\ ∴ Points a r e (2, 2) a n d (- 2, 34) \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted

3 months ago

Application of Derivatives Maths NCERT Exemplar Solutions Class 12th Chapter Six

The equation of the tangent to the curve y(1 + x²) = 2 – x, where it crosses the x-axis, is:

(A) x + 5y = 2

(B) x – 5y = 2

(C) 5x – y = 2

(D) 5x + y = 2

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y (1 + x^{2}) = 2 - x \dots (i) \\ I f i t c u t s x - a x i s, t h e n y - c o o r d i n a t e i s 0 . \\ ∴ 0 (1 + x^{2}) = 2 - x \Rightarrow x = 2 \\ P u t x = 2 i n e q u a t i o n (i) \\ y (1 + 4) = 2 - 2 \Rightarrow y (5) = 0 \Rightarrow y = 0 \\ Point o f c o n t a c t = (2, 0) \\ D i f f e r e n t i a t i n g e q . (i) w . r . t . x, w e h a v e \\ y * 2 x + (1 + x^{2}) \frac{d y}{d x} = - 1 \\ \Rightarrow 2 x y + (1 + x^{2}) \frac{d y}{d x} = - 1 \Rightarrow (1 + x^{2}) \frac{d y}{d x} = - 1 - 2 x y \\ ∴ \frac{d y}{d x} = \frac{- 1 - 2 x y}{(1 + x^{2})} \Rightarrow {\frac{d y}{d x}}_{(2, 0)} = \frac{- 1}{(1 + 4)} = \frac{- 1}{5} \\ E q u a t i o n o f tangent i s y - 0 = \frac{- 1}{5} (x - 2) \\ \Rightarrow 5 y = - x + 2 \Rightarrow x + 5 y = 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

3 months ago

Application of Derivatives Maths NCERT Exemplar Solutions Class 12th Chapter Six

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

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

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New answer posted

3 months ago

Application of Derivatives Maths NCERT Exemplar Solutions Class 12th Chapter Six

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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