Tags Class 12th

Class 12th

Get insights from 12k questions on Class 12th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 12th

Follow Ask Question

12k

Questions

Discussions

Active Users

Followers

New answer posted

a year ago

Class 12th Application of Derivatives

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

0 Follower 9 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

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

0 0

New answer posted

a year ago

Class 12th Application of Derivatives

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

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

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

0 0

New answer posted

a year ago

Class 12th Application of Derivatives

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 10 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

a year ago

Class 12th Application of Derivatives

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

0 Follower 7 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

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

0 0

New answer posted

a year ago

Class 12th Application of Derivatives

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:

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

0 0

New answer posted

a year ago

Class 12th Application of Derivatives

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)

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

a year ago

Class 12th Application of Derivatives

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 6 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

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

0 0

New answer posted

a year ago

Class 12th Application of Derivatives

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 4 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

a year ago

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

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

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

0 0

New answer posted

a year ago

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

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

67k Colleges
1.2k Exams
718k Reviews
1850k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Class 12th

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience