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Continuity and Differentiability Physics NCERT Exemplar Solutions Class 12th Chapter Five

If $x = s i n t$ and $y = s i n p t$ , prove that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0$

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t x = s i n t a n d y = s i n p t \\ 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 . t \\ \Rightarrow \frac{d x}{d t} = c o s t a n d \frac{d y}{d t} = c o s p t . p = p . c o s p t \\ \Rightarrow \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{p . c o s p t}{c o s t} \\ \Rightarrow \frac{d y}{d x} = \frac{p . c o s p t}{c o s t} \\ A g a i n 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 \\ \Rightarrow \frac{d}{d x} (\frac{d y}{d x}) = p . \frac{d}{d x} (\frac{c o s p t}{c o s t}) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{c o s t . \frac{d}{d x} (c o s p t) - c o s p t . \frac{d}{d x} (c o s t)}{c o s^{2} t}] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{c o s t . (- s i n p t) . p \frac{d t}{d x} - c o s p t . (- s i n t) . \frac{d t}{d x}}{c o s^{2} t}] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{2} t}] \frac{d t}{d x} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{2} t}] \frac{1}{c o s t} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . (\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{3} t}) \\ N o w w e h a v e t o p r o v e t h a t \\ (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0 \\ L . H . S . = (1 - x^{2}) [p . (\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{3} t})] - x (\frac{p . c o s p t}{c o s t}) + p^{2} y \\ \Rightarrow = (1 - s i n^{2} t) [p . (\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{3} t})] - \frac{p . s i n t c o s p t}{c o s t} + p^{2} . s i n p t \\ \Rightarrow = c o s^{2} t [\frac{- p^{2} c o s t . s i n p t + p c o s p t . s i n t}{c o s^{3} t}] - \frac{p . s i n t c o s p t}{c o s t} + p^{2} . s i n p t \\ \Rightarrow = \frac{- p^{2} c o s t . s i n p t + p c o s p t . s i n t}{c o s t} - \frac{p . s i n t c o s p t}{c o s t} + p^{2} . s i n p t \\ \Rightarrow = \frac{- p^{2} c o s t . s i n p t + p c o s p t . s i n t - p . s i n t c o s p t + p^{2} . s i n p t c o s t}{c o s t} \\ \Rightarrow = \frac{0}{c o s t} = 0 = R . H . S . \\ H e n c e, p r o v e d . \end{array}$

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Continuity and Differentiability Physics NCERT Exemplar Solutions Class 12th Chapter Five

If $x^{m} y^{n} = {(x + y)}^{m + n}$ , prove that

(i) $\frac{d y}{d x} = \frac{y}{x}$

(ii) $\frac{d^{2} y}{d x^{2}} = 0$ .

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (i) G i v e n t h a t x^{m} . y^{n} = {(x + y)}^{m + n} \\ T a k i n g l o g o n b o t h s i d e s, w e g e t, \\ \Rightarrow l o g x^{m} . y^{n} = l o g {(x + y)}^{m + n} [? l o g x y = l o g x + l o g y] \\ \Rightarrow l o g x^{m} + l o g y^{n} = (m + n) l o g (x + y) \\ \Rightarrow m l o g x + n l o g y = (m + n) l o g (x + y) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow m . \frac{d}{d x} l o g x + n . \frac{d}{d x} l o g y = (m + n) \frac{d}{d x} l o g (x + y) \\ \Rightarrow m . \frac{1}{x} + n . \frac{1}{y} . \frac{d y}{d x} = (m + n) \frac{1}{x + y} (1 + \frac{d y}{d x}) \\ \Rightarrow \frac{m}{x} + \frac{n}{y} . \frac{d y}{d x} = \frac{m + n}{x + y} . (1 + \frac{d y}{d x}) \\ \Rightarrow \frac{m}{x} + \frac{n}{y} . \frac{d y}{d x} = \frac{m + n}{x + y} + \frac{m + n}{x + y} . \frac{d y}{d x} \\ \Rightarrow \frac{n}{y} . \frac{d y}{d x} - \frac{m + n}{x + y} . \frac{d y}{d x} = \frac{m + n}{x + y} - \frac{m}{x} \\ \Rightarrow (\frac{n}{y} - \frac{m + n}{x + y}) . \frac{d y}{d x} = \frac{m + n}{x + y} - \frac{m}{x} \\ \Rightarrow (\frac{n x + n y - m y - n y}{y (x + y)}) . \frac{d y}{d x} = (\frac{m x + n x - m x - m y}{x (x + y)}) \\ \Rightarrow (\frac{n x - m y}{y (x + y)}) . \frac{d y}{d x} = (\frac{n x - m y}{x (x + y)}) \\ \Rightarrow \frac{d y}{d x} = \frac{n x - m y}{x (x + y)} * \frac{y (x + y)}{n x - m y} = \frac{y}{x} \\ \Rightarrow \frac{d y}{d x} = \frac{y}{x} H e n c e, p r o v e d . \end{array}$

$\begin{array}{l} (i i) G i v e n t h a t \frac{d y}{d x} = \frac{y}{x} \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\frac{y}{x}) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{x . \frac{d y}{d x} - y . 1}{x^{2}} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{x . \frac{y}{x} - y}{x^{2}} [? \frac{d y}{d x} = \frac{y}{x}] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{y - y}{x^{2}} = \frac{0}{x^{2}} = 0 \\ H e n c e, \frac{d^{2} y}{d x^{2}} = 0 H e n c e, p r o v e d . \end{array}$

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New answer posted

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Continuity and Differentiability Physics NCERT Exemplar Solutions Class 12th Chapter Five

Find the values of $p$ and $q$ so that $f (x) = {\begin{matrix} x^{2} + 3 x + p, & x \leq 1 \\ q x + 2, & x > 1 \end{matrix}$ is differentiable at $x = 1$ .

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar $\begin{array}{l} G i v e n t h a t f (x) = {\begin{matrix} x^{2} + 3 x + p, & x \leq 1 \\ q x + 2, & x > 1 \end{matrix} a t x = 1 . \\ L . H . L . f' (c) = \underset{x \to 1^{-}}{l i m} \frac{f (x) - f (c)}{x - c} \\ f' (1) = \underset{x \to 1^{-}}{l i m} \frac{f (x) - f (1)}{x - 1} \\ = \underset{x \to 1^{-}}{l i m} \frac{(x^{2} + 3 x + p) - (1 + 3 + p)}{x - 1} \\ = \underset{h \to 0}{l i m} \frac{[{(1 - h)}^{2} + 3 (1 - h) + p] - (1 + 3 + p)}{1 - h - 1} \\ = \underset{h \to 0}{l i m} \frac{[1 + h^{2} - 2 h + 3 - 3 h + p] - (4 + p)}{- h} \\ = \underset{h \to 0}{l i m} \frac{[h^{2} - 5 h + 4 + p] - [4 + p]}{- h} \\ = \underset{h \to 0}{l i m} \frac{h^{2} - 5 h + 4 + p - 4 - p}{- h} \\ = \underset{h \to 0}{l i m} \frac{h^{2} - 5 h}{- h} = \underset{h \to 0}{l i m} \frac{h [h - 5]}{- h} = 5 \\ R . H . L . f' (1) = \underset{x \to 1^{+}}{l i m} \frac{f (x) - f (1)}{x - 1} \\ = \underset{x \to 1^{+}}{l i m} \frac{(q x + 2) - (1 + 3 + p)}{x - 1} \\ = \underset{h \to 0}{l i m} \frac{[q (1 + h) + 2] - [4 + p]}{1 + h - 1} \\ = \underset{h \to 0}{l i m} \frac{q + q h + 2 - 4 - p}{h} = \underset{h \to 0}{l i m} \frac{q + q h - 2 - p}{h} \\ F o r e x i s t i n g t h e limit \\ q - 2 - p = 0 \Rightarrow q - p = 0 \dots (1) \\ \underset{h \to 0}{\Rightarrow l i m} \frac{q h - 0}{h} = q \\ I f L . H . L . f' (1) = R . H . L . f' (1) t h e n q = 5 . \\ N o w p u t t \end{array}$

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Continuity and Differentiability Class 12th

139. Kindly consider the following

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alok kumar singh

Contributor-Level 10

139. Kindly go through the solution

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Continuity and Differentiability Class 12th

137. Kindly consider the following

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alok kumar singh

Contributor-Level 10

137. Yes, Let us take $f (x) = | x - 1 | + | x - 2 | .$

So, x = 1, x= 2 divides the real line into three disjoint intervals $(- \infty, 1], [1, 2]$ and $[2, \infty) .$

For $x \in (- \infty, 1] .$

$f (x) = - (x - 1) + [- (x - 2)] = - x + 1 - x + 2 = 3 - 2 x .$

For $x \in [1, 2] .$

$f (x) = (x - 1) - (x - 2) = 1 .$

For $x \in [2, \infty)$

$f (x) = x - 1 + x - 2 = 2 x - 3 .$

Hence, these polynomial fun are all continous and desirable. for all real values of x or, except x = 1 and x = 2.

ie, $\forall x \in R - {1, 2} .$

For differentiavity at x = 1,

LHD = $= \underset{x \to 1^{-}}{l i m} \frac{f (x) - f (1)}{x - 1} = \underset{x \to 1^{-}}{l i m} \frac{3 - 2 x - 1}{x - 1} = \underset{x \to 1^{-}}{l i m} \frac{2 - 2 x}{x - 1} .$

$= \underset{x \to 1^{-}}{l i m} \frac{- 2 (x - 1)}{x - 1}$

$= \underset{x \to 1^{-}}{l i m} - 2$

= -2

RHD = $= \underset{x \to 1^{+}}{l i m} \frac{f (x) - f (1)}{x - 1} = \underset{x \to 1^{+}}{l i m} \frac{1 - 1}{x - 1} = \underset{x \to 1^{+}}{l i m} \frac{0}{x - 1} = 0 .$

as L.HD ≠ R.HD

f is not differentiable at x =1.

For continuity at x = 1.

L.HL= $= \underset{x \to 1^{-}}{l i m} f (x) = \underset{x \to 1^{-}}{l i m} = 1 .$

RHL = $\underset{x \to 1^{+}}{l i m} f (x) = \underset{x \to 1^{+}}{l i m} 1 = 1$ \ LHL = RHS

f is continuous at x = 1

For continuity & differentiability at x = 2

$= \underset{x \to 2^{-}}{l i m} f (x) = \underset{x \to 2^{-}}{l i m} 1 = 1 .$

$= \underset{x \to 2^{+}}{l i m} f (x) = \underset{x \to 2^{+}}{l i m} (2 x - 3) = 4 - 3 = 1 .$

? LHL = RHL

f is continuous at x = 2

$\begin{array}{l} = \underset{x \to 2^{-}}{l i m} \frac{f (x) - f (2)}{x - 2} = \underset{x \to 2^{-}}{l i m} \frac{1 - 1}{x - 2} = \underset{x \to 2^{-}}{l i m} = \frac{0}{x - 2} \end{array}$

$= \underset{x \to 2^{+}}{l i m} \frac{f (x) - f (2)}{x - 2} = \underset{x \to 2^{+}}{l i m} \frac{2 x - 3 - 1}{x - 2}$

$= \underset{x \to 2^{+}}{l i m} \frac{2 (x - 2)}{x - 2}$

$= \underset{x \to 2^{+}}{l i m} 2$

= 2

? LHD ≠ RHD

f is not differentiable at x = 2.

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New answer posted

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Continuity and Differentiability Class 12th

136. Kindly consider the following

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alok kumar singh

Contributor-Level 10

136. Given, $s i n (A + B) = s i n A c o s B + c o s A s i n B$

Differentiating w r t. 'x' we get,

$\frac{d}{d x} s i n (A + B) = \frac{d}{d x} (s i n A c o s B + c o s A s i n B)$

$\to c o s (A + B) - \frac{d}{d x} (A + B) = s i n A \frac{d}{d x} c o s B + c o s B \frac{d}{d x} s i n A + c o s A \frac{d}{d x} s i n B + s i n B \frac{d}{d x} c o s A$

$\to c o s (A + B) (\frac{d A}{d x} + \frac{d B}{d x}) = - s i n A s i n B \frac{d B}{d x} + c o s B c o s A \frac{d A}{d x} + c o s A c o s B \frac{d B}{d x} - s i n A s i n B \frac{d A}{d x}$

$\to c o s (A + B) (\frac{d A}{d x} + \frac{d B}{d t}) = c o s A c o s B (\frac{d A}{d x} + \frac{d B}{d x}) - s i n A s i n B (\frac{d A}{d x} + \frac{d B}{d x})$

$= (c o s A c o s B - s i n A s i n B) (\frac{d A}{d x} + \frac{d B}{d x}) .$

$\to c o s (A + B) = c o s A c o s B - s i n A s i n B .$

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New answer posted

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Continuity and Differentiability Class 12th

135. Kindly consider the following

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alok kumar singh

Contributor-Level 10

135. Given, $f (x) = | x |^{3} = {\begin{matrix} x^{3} & i f x \geq 0 \\ - x^{3} & i f x < 0 \end{matrix}$

For $x \geq 0, f (x) = | x |^{3} = x^{3}$

and $\begin{matrix} f^{'} (x) = 3 x^{2} & f^{''} (x) = 6 x \end{matrix}$

For $x < 0, f (x) = | x |^{3} = {(- x)}^{3} = - x^{3} .$

so, $\begin{matrix} f^{'} (x) = - 3 x^{2} & f^{''} (x) = - 6 x \end{matrix}$

Hence, $f^{''} (x) = {\begin{matrix} 6 x, & i f x \geq 0 \\ - 6 x, & i f x < 0 \end{matrix}$

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New answer posted

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Continuity and Differentiability Class 12th

134. Kindly consider the following

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alok kumar singh

Contributor-Level 10

134. Given, $x = a (c o s t + t s i n t)$ and $y = a (s i n t - t c o s t) .$

Differentiating w r t. 't' we get,

$\frac{d x}{d t} = a \frac{d}{d t} (c o s t + t s i n t) .$

$= a (- s i n t + t \frac{d}{d t} s i n t + s i n t \frac{d t}{d t}) .$

$= a (- s i n t + t c o s t + s i n t) = a t c o s t$

$\frac{d y}{d t} = \frac{a d}{d t} (c s i n t - t c o s t)$

$= a (c o s t - t \frac{d}{d t} c o s t - c o t \frac{d t}{d t})$

$= a (c o s t + t s i n t - c o s t) = a t s i n t$

$\frac{d y}{d x} = \frac{d y / d t}{d x / a t} = \frac{a t s i n t}{a t c o s t} = t a n t$

So, $\frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (t a n t) = \frac{d}{d t} (t a n t) \cdot \frac{d t}{d x} .$

$= s e c^{2} t * \frac{d t}{d x} .$

$= s e c^{2} t * \frac{1}{(d x / d t)}$

$= s e c^{2} t * \frac{1}{a t c o s t}$

$= \frac{s e c^{3} t}{a t}$

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New answer posted

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Continuity and Differentiability Class 12th

133. Kindly consider the following

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alok kumar singh

Contributor-Level 10

133. Given, $c o s y = x c o s (a + y) .$

$x = \frac{c o s y}{c o s (a + y)}$

Differentiating w r t 'y' we get,

$\frac{d x}{d y} = \frac{d}{d y} (\frac{c o s y}{c o s (a + y)}) .$

$= \frac{c o s (a + y) \frac{d}{d y} c o s y - c o s y \frac{d}{d y} c o s (a + y)}{c o s^{2} (a + y) .}$

$= \frac{c o s (a + y) (- s i n y) - c o s y (- s i n (a + y))}{c o s^{2} (a + y) .}$

$= \frac{- c o s (a + y) s i n y + s i n (a + y) c o s y}{c o s^{2} (a + y)}$

$= \frac{s i n (a + y) c o s y - c o s (a + y) s i n y .}{c o s^{2} (a + y)}$

$\frac{d x}{d y} = \frac{s i n (a + y - y)}{c o s^{2} (a + y)} {\begin{array}{r} ? s i n (A - B) = s i n A c o s B & - c o s A s i n B \end{array}}$

So, $\frac{d y}{d x} = \frac{c o s^{2} (a + y)}{s i n a}$

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New answer posted

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Continuity and Differentiability Class 12th

132. Kindly consider the following

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alok kumar singh

Contributor-Level 10

132. Given, ${(x - a)}^{2} + {(y - b)}^{2} = c^{2} .$

Differentiating w r t 'x' we get

$\frac{d}{d x} {(x - a)}^{2} + \frac{d}{d x} {(y - b)}^{2} = \frac{d}{d x} c^{2}$

$\Rightarrow 2 (x - a) + 2 (y - b) \frac{d y}{d x} = 0$

$\Rightarrow \frac{d y}{d x} = - \frac{2 (x - a)}{2 (y - b)} = - \frac{(x - a)}{(y - b)}$

Again, $\frac{d^{2} y}{d x^{2}} = - {\frac{(y - b) \frac{d}{d x} (x - a) - (x - a) \frac{d}{d x} (y - b)}{{(y - b)}^{2}}}$

$= - {\frac{(y - b) - (x - a) \frac{d y}{d x}}{{(y - b)}^{2}}}$

$= - {\frac{(y - b) + (x - a) \frac{(x - a)}{(y - b)}}{{(y - b)}^{2}}}$

$= - {\frac{{(y - b)}^{2} + {(x - a)}^{2}}{{(y - b)}^{3}}}$

$= - \frac{c^{2}}{{(y - b)}^{3}} {? {(x - a)}^{2} + {(y - b)}^{2} = c^{2}}$

Then, L.H.S = $\frac{{1 + {(\frac{d y}{d x})}^{2}}^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} = \frac{{1 + \frac{{(x - a)}^{2}}{{(y - b)}^{2}}}^{3 / 2}}{\frac{- c^{2}}{{(y - b)}^{2}}}$

$= \frac{{{(y - b)}^{2} + {(x - a)}^{2}}^{3 / 2}}{{(y - b)}^{3}} * \frac{{(y - b)}^{3}}{- c^{2}}$

$= \frac{c^{2 * 3 / 2}}{- c^{2}} = \frac{c^{3}}{- c^{2}} = - c$ Where c is a constant and is independent of a and b.

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