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New answer posted

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Class 12th Application of Derivatives

Kindly consider the following

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = s i n x - c o s x - 2 a x + b, a \geq 1 \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t \\ f^{'} (x) = c o s x + s i n x - 2 a \\ F o r decreasing f u n c t i o n, f^{'} (x) < 0 \\ ∴ c o s x + s i n x - 2 a < 0 \\ \Rightarrow 2 (\frac{}{2} c o s x + \frac{1}{2} s i n x) - 2 a < 0 \\ \Rightarrow \frac{}{2} c o s x + \frac{1}{2} s i n x - a < 0 \\ \Rightarrow (c o s \frac{π}{6} c o s x + s i n \frac{π}{6} s i n x) - a < 0 \\ \Rightarrow c o s (x - \frac{π}{6}) - a < 0 \\ Since c o s x \in [- 1, 1] a n d a \geq 1 \\ ∴ f^{'} (x) < 0 \\ H e n c e, t h e g i v e n f u n c t i o n i s decreasing f u n c t i o n i n R . \end{array}$

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New answer posted

a year ago

Class 12th Application of Derivatives

Kindly consider the following

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

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

Sol.

$\begin{array}{l} G i v e n t h a t f (x) = 2 x + c o t^{- 1} x + l o g (- x) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t \\ f^{'} (x) = 2 - \frac{1}{1 + x^{2}} + \frac{1}{- 1} * \frac{d}{d x} (- x) \\ = 2 - \frac{1}{1 + x^{2}} + \frac{(\frac{1}{2} * (2 x - 1))}{- 1} \\ = 2 - \frac{1}{1 + x^{2}} + \frac{x -}{(- x)} \\ = 2 - \frac{1}{1 + x^{2}} - \frac{- x}{(- x)} \\ = 2 - \frac{1}{1 + x^{2}} - \frac{1}{} \\ F o r increasing f u n c t i o n, f^{'} (x) \geq 0 \\ ∴ 2 - \frac{1}{1 + x^{2}} - \frac{1}{} \geq 0 \\ \Rightarrow \frac{2 (1 + x^{2}) - 1 +}{1 + x^{2}} \geq 0 \Rightarrow 2 + 2 x^{2} - 1 + \geq 0 \\ \Rightarrow 2 x^{2} + 1 + \geq 0 2 x^{2} + 1 \geq - \\ S q u a r i n g b o t h s i d e s, w e g e t 4 x^{4} + 1 + 4 x^{2} \geq 1 + x^{2} \\ \Rightarrow 4 x^{4} + 4 x^{2} - x^{2} \geq 0 \Rightarrow 4 x^{4} + 3 x^{2} \geq 0 \Rightarrow x^{2} (4 x^{2} + 3) \geq 0 \\ w h i c h i s t r u e f o r a n y v a l u e o f x \in R \\ H e n c e, t h e g i v e n f u n c t i o n i s a n increasing f u n c t i o n o v e r R . \end{array}$

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New answer posted

a year ago

Class 12th Application of Derivatives

Show that the line x/a + y/b = 1 touches the curve y = b(-x/a) at the point where the curve intersects the y-axis.

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t e q u a t i o n o f t h e c u r v e i s y = b . e^{- x / a} \\ a n d t h e e q u a t i o n o f t h e l i n e i s \frac{x}{a} + \frac{y}{b} = 1 \\ L e t t h e c o o r d i n a t e s o f t h e point w h e r e t h e c u r v e intersects t h e y - a x i s b e (0, y_{1}) \\ N o w, d i f f e r e n t i a t i n g y = b . e^{- x / a} b o t h s i d e s w . r . t . x, w e g e t \\ \frac{d y}{d x} = b . e^{- x / a} (- \frac{1}{a}) = - \frac{b}{a} . e^{- x / a} \\ So, t h e s l o p e o f t h e tangent, m_{1} = - \frac{b}{a} . e^{- x / a} \\ D i f f e r e n t i a t i n g \frac{x}{a} + \frac{y}{b} = 1 b o t h s i d e s w . r . t . x, w e g e t \\ ∴ \frac{1}{a} + \frac{1}{b} . \frac{d y}{d x} = 0 \\ So, t h e s l o p e o f t h e l i n e, m_{2} = - \frac{b}{a} \\ I f t h e l i n e t o u c h e s t h e c u r v e, t h e n m_{1} = m_{2} \\ \Rightarrow - \frac{b}{a} . e^{- x / a} = - \frac{b}{a} \Rightarrow e^{- x / a} = 1 \\ \Rightarrow \frac{- x}{a} l o g e = l o g 1 (T a k i n g l o g o n b o t h s i d e s) \\ \Rightarrow \frac{- x}{a} = 0 \Rightarrow x = 0 \\ P u t t i n g x = 0 i n e q u a t i o n y = b . e^{- x / a} \\ \Rightarrow y = b . e^{0} = b \\ H e n c e, t h e g i v e n e q u a t i o n o f t h e c u r v e intersects a t (0, b) i . e . o n y - a x i s . \end{array}$

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New answer posted

a year ago

Class 12th Application of Derivatives

At what points on the curve x² + y² – 2x – 4y + 1 = 0 are the tangents parallel to the y-axis?

0 Follower 8 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Six

Find the equation of the normal lines to the curve 3x² – y² = 8 which are parallel to the line x + 3y = 4.

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 12th Application of Derivatives

Prove that the curves y² = 4x and x² + y² – 6x + 1 = 0 touch each other at the point (1, 2).

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t t h e e q u a t i o n o f t h e t w o c u r v e s a r e y^{2} = 4 x \dots (i) \\ a n d x^{2} + y^{2} - 6 x + 1 = 0 \dots (i i) \\ D i f f e r e n t i a t i n g e q (i) w . r . t . x, w e g e t \\ 2 y \frac{d y}{d x} = 4 \Rightarrow \frac{d y}{d x} = \frac{2}{y} \\ S l o p e o f t h e tangent at (1, 2), m_{1} = \frac{2}{2} = 1 \\ D i f f e r e n t i a t i n g e q (i i) w . r . t . x, w e g e t \\ \Rightarrow 2 x + 2 y . \frac{d y}{d x} - 6 = 0 \Rightarrow 2 y . \frac{d y}{d x} = 6 - 2 x \Rightarrow \frac{d y}{d x} = \frac{6 - 2 x}{2 y} \\ ∴ S l o p e o f t h e tangent at t h e s a m e point (1, 2), m_{2} = \frac{6 - 2 * 1}{2 * 2} = \frac{4}{4} = 1 \\ W e s e e t h a t m_{1} = m_{2} = 1 at t h e point (1, 2) . \\ H e n c e, t h e g i v e n c i r c l e s t o u c h e a c h o t h e r a t t h e s a m e point (1, 2) . \end{array}$

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New answer posted

a year ago

Class 12th Application of Derivatives

Find the coordinates of the point on the curve √x + √y = 4 at which the tangent is equally inclined to the axes

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alok kumar singh

Contributor-Level 10

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

Sol:

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New answer posted

a year ago

Class 12th Application of Derivatives

Prove that the curves xy = 4 and x² + y² = 8 touch each other.

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n c i r c l e s a r e 2 x y = 4 \dots (i) \\ a n d x^{2} + y^{2} = 8 \dots (i i) \\ D i f f e r e n t i a t i n g e q (i) w . r . t . x, w e g e t \\ x . \frac{d y}{d x} + y . 1 = 0 \Rightarrow \frac{d y}{d x} = - \frac{y}{x} \Rightarrow m_{1} = \frac{1}{y} (m_{1} = s l o p e o f t h e tangent to the curve) \\ D i f f e r e n t i a t i n g e q (i i) w . r . t . x, w e g e t \\ ∴ 2 x + 2 y . \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{x}{y} \Rightarrow m_{2} = - \frac{x}{y} (m_{2} = s l o p e o f t h e o t h e r tangent to the curve) \\ T o f i n d t h e point o f c o n t a c t o f t h e t w o c i r c l e s \\ t h e n m_{1} = m_{2} \\ \Rightarrow - \frac{y}{x} = - \frac{x}{y} \Rightarrow x^{2} = y^{2} \\ P u t t i n g t h e v a l u e o f y^{2} i n e q . (i i) \\ x^{2} + x^{2} = 8 \Rightarrow 2 x^{2} = 8 \Rightarrow x^{2} = 4 \\ ∴ x = \pm 2 \\ ? x^{2} = y^{2} \Rightarrow y = \pm 2 \\ ∴ T h e point o f c o n t a c t o f t h e t w o c i r c l e s a r e (2, 2) a n d (- 2, 2) . \end{array}$

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New answer posted

a year ago

Class 12th Application of Derivatives

Find the condition that the curves 2x = $y^{2}$ and 2xy = k intersect orthogonally.

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

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

Sol.

$\begin{array}{l} T h e t w o c i r c l e s intersect o r t h o g o n a l l y i f t h e a n g l e b e t w e e n t h e tangents d r a w n t o t h e t w o c i r c l e s \\ a t t h e point o f t h e i r intersection i s 9 0^{0} . \\ E q u a t i o n o f t h e t w o c i r c l e s a r e g i v e n a s \\ 2 x = y^{2} \dots (i) \\ a n d 2 x y = k \dots (i i) \\ D i f f e r e n t i a t i n g e q (i) a n d e q (i i) w . r . t . x, w e g e t \\ 2.1 = 2 y . \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{1}{y} \Rightarrow m_{1} = \frac{1}{y} (m_{1} = s l o p e o f t h e tangent) \\ \Rightarrow 2 x y = k \\ \Rightarrow 2 [x . \frac{d y}{d x} + y . 1] = 0 \\ ∴ \frac{d y}{d x} = - \frac{y}{x} \Rightarrow m_{2} = - \frac{y}{x} (m_{2} = s l o p e o f t h e o t h e r tangent) \\ I f t h e t w o tangents a r e p e r p e n d i c u l a r t o e a c h o t h e r, \\ t h e n m_{1} * m_{2} = - 1 \\ \Rightarrow \frac{1}{y} * - \frac{y}{x} = - 1 \Rightarrow \frac{1}{x} = 1 \Rightarrow x = 1 \\ N o w s o l v i n g 2 x = y^{2} [F r o m (i)] \\ a n d &thinsp \end{array}$

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New answer posted

a year ago

Class 12th NCERT

A particle is dropped from a height H. The de-Broglie wavelength of the particle as a function of height is proportional to (a) H (b) H^1/2 (c) H⁰ (d) H^-1/2

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Pallavi Pathak

Contributor-Level 10

Explanation- velocity of a freely falling body is v= $\sqrt[]{2 g h}$

And $? = \frac{h}{m v} = \frac{h}{m \sqrt[]{2 g h}}$

$? = h$ ^-1

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