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Class 12th College of Engineering, Pune

For lateral entry in COEP, PICT or any colleges of Pune, is MHT CET or any entrance exam required or admission is based on 12th percentage?

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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

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

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

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

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

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

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

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New question posted

a year ago

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

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New answer posted

a year ago

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

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted

a year ago

Class 12th Application of Derivatives

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

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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New question posted

a year ago

Class 12th Application of Derivatives

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

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New answer posted

a year ago

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.

0 Follower 10 Views

alok kumar singh

Contributor-Level 10

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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New answer posted

a year ago

Class 12th Application of Derivatives

Show that f(x) = tan?¹(sinx + cosx) is an increasing function in (0, $\frac{π}{4}$ ).

0 Follower 22 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = t a n^{- 1} (s i n x + c o s x) i n (0, \frac{π}{4}) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t \\ f^{'} (x) = \frac{1}{1 + {(s i n x + c o s x)}^{2}} . \frac{d}{d x} (s i n x + c o s x) \\ \Rightarrow f^{'} (x) = \frac{1 * (c o s x - s i n x)}{1 + {(s i n x + c o s x)}^{2}} \\ \Rightarrow f^{'} (x) = \frac{(c o s x - s i n x)}{1 + s i n^{2} x + c o s^{2} x + 2 s i n x c o s x} \\ \Rightarrow f^{'} (x) = \frac{(c o s x - s i n x)}{1 + 1 + 2 s i n x c o s x} \Rightarrow f^{'} (x) = \frac{c o s x - s i n x}{2 + 2 s i n x c o s x} \\ F o r a n increasing f u n c t i o n, f^{'} (x) \geq 0 \\ ∴ \frac{c o s x - s i n x}{2 + 2 s i n x c o s x} \geq 0 \\ \Rightarrow c o s x - s i n x \geq 0 [? (2 + s i n 2 x) \geq 0 i n (0, \frac{π}{4})] \\ \Rightarrow c o s \geq s i n x, w h i c h i s t r u e f o r (0, \frac{π}{4}) \\ H e n c e, t h e g i v e n f u n c t i o n f (x) i s a n increasing f u n c t i o n i n (0, \frac{π}{4}) . \end{array}$

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