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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Let $f (x) = {\begin{matrix} x^{} + 2 & if x \leq - 1, \\ c x^{2} & if x > - 1 \end{matrix}$ , find ' $c$ ' if $l i m_{x \to - 1} f (x)$ exists

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n f (x) = {\begin{matrix} x + 2 & i f x \leq - 1, \\ c x^{2} & i f x > - 1 \end{matrix} \\ L H L f (x) = \underset{x \to - 1^{-}}{l i m} (x + 2) = \underset{h \to 0}{l i m} (- 1 - h + 2) \\ = \underset{h \to 0}{l i m} (1 - h) = 1 \\ R H L f (x) = \underset{x \to - 1^{+}}{l i m} c x^{2} = \underset{h \to 0}{l i m} c {(- 1 + h)}^{2} = c \\ Since t h e limits e x i t . \\ ∴ L H L = R H L \\ ∴ c = 1 \end{array}$

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Let $f (x) = {\begin{matrix} \frac{k c o s x}{π - 2 x} & when x \neq \frac{π}{2}, \\ 3 & when x = \frac{π}{2} \end{matrix}$ , and if $l i m_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2})$ , find the value of $k$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n f (x) = {\begin{matrix} \frac{k c o s x}{π - 2 x} & w h e n x \neq \frac{π}{2}, \\ 3 & w h e n x = \frac{π}{2} \end{matrix} \\ L H L f (x) = \underset{x \to {\frac{π}{2}}^{-}}{l i m} \frac{k c o s x}{π - 2 x} = \underset{h \to 0}{l i m} \frac{k c o s (\frac{π}{2} - h)}{π - 2 (\frac{π}{2} - h)} \\ = \underset{h \to 0}{l i m} \frac{k s i n h}{π - π + 2 h} = \underset{h \to 0}{l i m} \frac{k s i n h}{2 h} = \frac{k}{2} . 1 = \frac{k}{2} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \\ R H L f (x) = \underset{x \to {\frac{π}{2}}^{+}}{l i m} \frac{k c o s x}{π - 2 x} = \underset{h \to 0}{l i m} \frac{k c o s (\frac{π}{2} + h)}{π - 2 (\frac{π}{2} + h)} \\ = \underset{h \to 0}{l i m} \frac{- k s i n h}{π - π - 2 h} = \underset{h \to 0}{l i m} \frac{- k s i n h}{- 2 h} = \frac{k}{2} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \\ W e a r e g i v e n t h a t \underset{x \to \frac{π}{2}}{l i m} f (x) = 3 \\ S o, \frac{k}{2} = 3 \Rightarrow k = 6 \end{array}$

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Show that $l i m_{x \to 4} \frac{? x - 4 ?}{x - 4}$ does not exist.

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n \underset{x \to 4}{l i m} \frac{| x - 4 |}{x - 4} \\ L H L = \underset{x \to 4^{-}}{l i m} \frac{- (x - 4)}{x - 4} = - 1 [? | x - 4 | = - (x - 4) i f x < 4] \\ R H L = \underset{x \to 4^{+}}{l i m} \frac{(x - 4)}{x - 4} = 1 [? | x - 4 | = (x - 4) i f x > 4] \\ L H L \neq R H L \\ H e n c e, t h e limit d o e s n o t e x i s t . \end{array}$

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$\underset{x \to π}{l i m} \frac{1 - s i n \frac{x}{2}}{c o s \frac{x}{2} (c o s \frac{x}{4} - s i n \frac{x}{4})}$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to π}{l i m} \frac{1 - s i n \frac{x}{2}}{c o s \frac{x}{2} (c o s \frac{x}{4} - s i n \frac{x}{4})} \\ = \underset{x \to π}{l i m} \frac{c o s^{2} \frac{x}{4} + s i n^{2} \frac{x}{4} - 2 s i n \frac{x}{4} . c o s \frac{x}{4}}{(c o s^{2} \frac{x}{4} - s i n^{2} \frac{x}{4}) (c o s \frac{x}{4} - s i n \frac{x}{4})} [\begin{array}{l} ? c o s 2 x = c o s^{2} x - s i n^{2} x \\ s i n^{2} x + c o s^{2} x = 1 \end{array}] \\ = \underset{x \to π}{l i m} \frac{{(c o s \frac{x}{4} - s i n \frac{x}{4})}^{2}}{(c o s \frac{x}{4} - s i n \frac{x}{4}) (c o s \frac{x}{4} + s i n \frac{x}{4}) (c o s \frac{x}{4} - s i n \frac{x}{4})} \\ = \underset{x \to π}{l i m} \frac{1}{(c o s \frac{x}{4} + s i n \frac{x}{4})} \\ T a k i n g limit w e h a v e \\ = \frac{1}{c o s \frac{π}{4} + s i n \frac{π}{4}} = \frac{1}{\frac{1}{\sqrt[]{2}} + \frac{1}{\sqrt[]{2}}} = \frac{1}{\frac{2}{\sqrt[]{2}}} = \frac{1}{\sqrt[]{2}} . \end{array}$

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$\underset{x \to \frac{π}{4}}{l i m} \frac{t a n^{3} x - t a n x}{c o s (x + \frac{π}{4})}$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{π}{4}}{l i m} \frac{t a n^{3} x - t a n x}{c o s (x + \frac{π}{4})} \\ = \underset{x \to \frac{π}{4}}{l i m} \frac{t a n x (t a n^{2} x - 1)}{c o s (x + \frac{π}{4})} = \underset{x \to \frac{π}{4}}{l i m} t a n x . \underset{x \to \frac{π}{4}}{l i m} [\frac{- (1 - t a n^{2} x)}{c o s (x + \frac{π}{4})}] \\ = - 1 * \underset{x \to \frac{π}{4}}{l i m} \frac{(1 - t a n x) (1 + t a n x)}{c o s (x + \frac{π}{4})} \\ = \underset{x \to \frac{π}{4}}{l i m} - (1 + t a n x) * \underset{x \to \frac{π}{4}}{l i m} [\frac{1 - t a n x}{c o s (x + \frac{π}{4})}] \\ = - (1 + 1) * \underset{x \to \frac{π}{4}}{l i m} \frac{(c o s x - s i n x)}{c o s x . c o s (x + \frac{π}{4})} = - 2 * \underset{x \to \frac{π}{4}}{l i m} \frac{\sqrt[]{2} (\frac{1}{\sqrt[]{2}} c o s x - \frac{1}{\sqrt[]{2}} s i n x)}{c o s x . c o s (x + \frac{π}{4})} \\ = - 2 \sqrt[]{2} * \underset{x \to \frac{π}{4}}{l i m} \frac{(c o s \frac{π}{4} . c o s x - s i n \frac{π}{4} . s i n x)}{c o s x . c o s (x + \frac{π}{4})} \\ = - 2 \sqrt[]{2} * \underset{x \to \frac{π}{4}}{l i m} \frac{c o s (x + \frac{π}{4})}{c o s x . c o s (x + \frac{π}{4})} = - 2 \sqrt[]{2} * \underset{x \to \frac{π}{4}}{l i m} \frac{1}{c o s x} \\ T a k i n g limit w e h a v e \\ = \frac{- 2 \sqrt[]{2}}{c o s \frac{π}{4}} = \frac{- 2 \sqrt[]{2}}{\frac{1}{\sqrt[]{2}}} = - 2 * 2 = - 4 . \end{array}$

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$\underset{x \to 0}{l i m} \frac{s i n (α + β) x - s i n (α - β) x + s i n 2 α x}{c o s^{2} β x - c o s^{2} α x} . x$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{s i n (α + β) x - s i n (α - β) x + s i n 2 α x}{c o s^{2} β x - c o s^{2} α x} . x \\ = \underset{x \to 0}{l i m} \frac{[2 s i n α x . c o s β x + s i n 2 α x] . x}{2 s i n (α + β) x . s i n (α - β) x} \\ = \underset{x \to 0}{l i m} \frac{[2 s i n α x . c o s β x + 2 s i n α x . c o s α x] . x}{2 s i n (α + β) x . s i n (α - β) x} \\ = \underset{x \to 0}{l i m} \frac{2 s i n α x (c o s β x + c o s α x) . x}{2 s i n (α + β) x . s i n (α - β) x} \\ = \underset{x \to 0}{l i m} \frac{s i n α x [2 c o s (\frac{α + β}{2}) x . c o s (\frac{α - β}{2}) x] . x}{s i n (α + β) x . s i n (α - β) x} \\ = \underset{x \to 0}{l i m} \frac{s i n α x [2 c o s (\frac{α + β}{2}) x . c o s (\frac{α - β}{2}) x] . x}{2 s i n (\frac{α + β}{2}) x . c o s (\frac{α + β}{2}) x . 2 s i n (\frac{α - β}{2}) x . c o s (\frac{α - β}{2}) x} \\ = \underset{x \to 0}{l i m} \frac{s i n α x . x}{2 s i n (\frac{α + β}{2}) x s i n (\frac{α - β}{2}) x} \\ = \underset{x \to 0}{l i m} \frac{1}{2} \frac{\frac{s i n α x}{α x} . (α x) . x}{[\frac{s i n (\frac{α + β}{2}) x}{(\frac{α + β}{2}) x} * (\frac{α + β}{2}) . x] [\frac{s i n (\frac{α - β}{2}) x}{(\frac{α - β}{2}) x} * (\frac{α - β}{2}) . x]} \\ = \frac{1}{2} . \frac{α x^{2}}{(\frac{α + β}{2}) x (\frac{α - β}{2}) x} = \frac{1}{2} . [\frac{α}{(\frac{α + β}{2}) (\frac{α - β}{2})}] \\ = \frac{1}{2} . \frac{4 α}{α^{2} - β^{2}} = \frac{2 α}{α^{2} - β^{2}} \end{array}$

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Evaluate each of the following limits in Exercises 47 to 53.

$l i m_{y \to 0} \frac{(x + y) s e c (x + y) - x s e c x}{y^{}}$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{y \to 0}{l i m} \frac{(x + y) s e c (x + y) - x s e c x}{y} \\ = \underset{y \to 0}{l i m} \frac{x s e c (x + y) + y s e c (x + y) - x s e c x}{y} \\ = \underset{y \to 0}{l i m} \frac{[x s e c (x + y) - x s e c x]}{y} + \underset{y \to 0}{l i m} \frac{y s e c (x + y)}{y} \\ = \underset{y \to 0}{l i m} \frac{x [s e c (x + y) - s e c x]}{y} + \underset{y \to 0}{l i m} s e c (x + y) \\ = \underset{y \to 0}{l i m} \frac{x [\frac{1}{c o s (x + y)} - \frac{1}{c o s x}]}{y} + \underset{y \to 0}{l i m} s e c (x + y) \\ = \underset{y \to 0}{l i m} x [\frac{c o s x - c o s (x + y)}{y . c o s x . c o s (x + y)}] + \underset{y \to 0}{l i m} s e c (x + y) \\ = \underset{y \to 0}{l i m} x [\frac{- 2 s i n (\frac{x + x + y}{2}) . s i n (\frac{x - x - y}{2})}{y . c o s x . c o s (x + y)}] + \underset{y \to 0}{l i m} s e c (x + y) \\ = \underset{y \to 0}{l i m} x [\frac{- 2 s i n (x + \frac{y}{2}) . s i n (\frac{- y}{2})}{y . c o s x . c o s (x + y)}] + \underset{y \to 0}{l i m} s e c (x + y) \\ = \underset{\begin{array}{l} y \to 0 \\ ∴ \frac{y}{2} \to 0 \end{array}}{l i m} x [\frac{2 s i n (x + \frac{y}{2}) . s i n (\frac{y}{2})}{c o s x . c o s (x + y) . (\frac{y}{2}) . 2}] + \underset{y \to 0}{l i m} s e c (x + y) \\ T a k i n g w e h a v e \\ = x [s i n x . \frac{1}{c o s x . c o s x}] + s e c x \\ = x s e c x t a n x + s e c x = s e c x (x t a n x + 1) \end{array}$

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$x c o s x$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t y = x c o s x \dots (i) \\ \Rightarrow y + Δ y = (x + Δ x) c o s (x + Δ x) \dots (i i) \\ S u b t r a c t i n g e q n . (i) f r o m e q n . (i i) \\ y + Δ y - y = (x + Δ x) c o s (x + Δ x) - x c o s x \\ \Rightarrow Δ y = x c o s (x + Δ x) + Δ x c o s (x + Δ x) - x c o s x \\ D i v i d i n g b o t h s i d e s b y Δ x a n d t a k e t h e limit w e g e t \\ \underset{Δ x \to 0}{l i m} \frac{Δ y}{Δ x} = \underset{Δ x \to 0}{l i m} \frac{x c o s (x + Δ x) - x c o s x + Δ x c o s (x + Δ x)}{Δ x} \\ \frac{d y}{d x} = \underset{Δ x \to 0}{l i m} \frac{x [c o s (x + Δ x) - c o s x]}{Δ x} + \underset{Δ x \to 0}{l i m} \frac{Δ x c o s (x + Δ x)}{Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{x [- 2 s i n \frac{(x + Δ x + x)}{2} . s i n \frac{(x + Δ x - x)}{2}]}{Δ x} + \underset{Δ x \to 0}{l i m} c o s (x + Δ x) \\ = \underset{\begin{array}{l} Δ x \to 0 \\ ∴ \frac{Δ x}{2} \to 0 \end{array}}{l i m} \frac{x [- 2 s i n (x + \frac{Δ x}{2}) . s i n \frac{Δ x}{2}]}{2 * \frac{Δ x}{2}} + \underset{Δ x \to 0}{l i m} c o s (x + Δ x) \\ ∴ \frac{Δ x}{2} \to 0 T a k i n g limits w e h a v e \\ = x [- s i n x] + c o s x [? \underset{\frac{Δ x}{2} \to 0}{l i m} \frac{s i n \frac{Δ x}{2}}{\frac{Δ x}{2}} = 1] \\ = - x s i n x + c o s x \end{array}$

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$x^{\frac{2}{3}}$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t f (x) = x^{2 / 3} \dots (i) \\ \Rightarrow f (x + Δ x) = {(x + Δ x)}^{2 / 3} \dots (i i) \\ S u b t r a c t i n g e q n . (i) f r o m e q n . (i i) \\ f (x + Δ x) - f (x) = {(x + Δ x)}^{2 / 3} - x^{2 / 3} \\ D i v i d i n g b o t h s i d e s b y Δ x a n d t a k e t h e w e g e t \\ \underset{Δ x \to 0}{l i m} \frac{f (x + Δ x) - f (x)}{Δ x} = \underset{Δ x \to 0}{l i m} \frac{{(x + Δ x)}^{2 / 3} - x^{2 / 3}}{Δ x} \\ f' (x) = \underset{Δ x \to 0}{l i m} \frac{x^{2 / 3} {[1 + \frac{Δ x}{x}]}^{2 / 3} - x^{2 / 3}}{Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{x^{2 / 3} [{(1 + \frac{Δ x}{x})}^{2 / 3} - 1]}{Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{x^{2 / 3} [(1 + \frac{2}{3} . \frac{Δ x}{x} + \dots) - 1]}{Δ x} [\begin{array}{l} E x p a n d i n g b y B i n o m i a l t h e o r e m a n d r e j e c t i n g \\ t h e h i g h e r p o w e r s o f Δ x a s Δ x \to 0 \end{array}] \\ = \underset{Δ x \to 0}{l i m} \frac{x^{2 / 3} . \frac{2}{3} . \frac{Δ x}{x}}{Δ x} = \frac{2}{3} x^{2 / 3 - 1} = \frac{2}{3} x^{- 1 / 3} . \end{array}$

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$\frac{a x + b}{c x + d}$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t f (x) = \frac{a x + b}{c x + d} \dots (i) \\ \Rightarrow f (x + Δ x) = \frac{a (x + Δ x) + b}{c (x + Δ x) + d} \dots (i i) \\ S u b t r a c t i n g e q n . (i) f r o m e q n . (i i) \\ f (x + Δ x) - f (x) = \frac{a (x + Δ x) + b}{c (x + Δ x) + d} - \frac{a x + b}{c x + d} \\ D i v i d i n g b o t h s i d e s b y Δ x a n d t a k e t h e w e g e t \\ \underset{Δ x \to 0}{l i m} \frac{f (x + Δ x) - f (x)}{Δ x} = \underset{Δ x \to 0}{l i m} \frac{\frac{a (x + Δ x) + b}{c (x + Δ x) + d} - \frac{a x + b}{c x + d}}{Δ x} \\ f' (x) = \underset{Δ x \to 0}{l i m} \frac{(a x + a Δ x + b) (c x + d) - (a x + b) (c x + c Δ x + d)}{[c (x + Δ x) + d] (c x + d) . Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{a c x^{2} + a c Δ x . x + b c x + a d x + a d Δ x + b d - a c x^{2} - a c Δ x . x - a d x - b c x - b c . Δ x - b d}{(c x + c Δ x + d) (c x + d) Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{(a d - b c) Δ x}{(c x + c Δ x + d) (c x + d) . Δ x} = \underset{Δ x \to 0}{l i m} \frac{(a d - b c)}{(c x + c . Δ x + d) (c x + d)} \\ T a k i n g limit, w e h a v e \\ = \frac{(a d - b c)}{(c x + d) (c x + d)} = \frac{(a d - b c)}{{(c x + d)}^{2}} \end{array}$

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