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New answer posted

10 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{θ \to 0} \frac{1 - c o s 4 θ}{1 - c o s 6 θ}$ is

(a) $\frac{4}{9}$
(b) $\frac{1}{2}$
(c) $- \frac{1}{2}$
(d) -1

0 Follower 7 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{θ \to 0}{l i m} \frac{1 - c o s 4 θ}{1 - c o s 6 θ} \\ = \underset{θ \to 0}{l i m} \frac{2 s i n^{2} 2 θ}{2 s i n^{2} 3 θ} [? 1 - c o s θ = 2 s i n^{2} \frac{θ}{2}] \\ = \underset{θ \to 0}{l i m} \frac{s i n^{2} 2 θ}{s i n^{2} 3 θ} = \underset{θ \to 0}{l i m} {[\frac{s i n 2 θ}{s i n 3 θ}]}^{2} = \underset{\begin{array}{l} θ \to 0 \\ 2 θ \to 0 \\ 3 θ \to 0 \end{array}}{l i m} {[\frac{\frac{s i n 2 θ}{2 θ} * 2 θ}{\frac{s i n 3 θ}{3 θ} * 3 θ}]}^{2} \\ = {[\frac{2 θ}{3 θ}]}^{2} = {(\frac{2}{3})}^{2} = \frac{4}{9} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

10 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$l i m_{x \to 1} \frac{x^{m} - 1}{x^{n} - 1}$ is

(a) 1
(b) $\frac{m}{n}$
(c) $\frac{m}{n} - 1$
(d) $\frac{m^{2}}{n^{2}}$

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 1}{l i m} \frac{x^{m} - 1}{x^{n} - 1} \\ = \underset{x \to 1}{l i m} \frac{\frac{x^{m} - {(1)}^{m}}{x - 1}}{\frac{x^{n} - {(1)}^{n}}{x - 1}} = \frac{m {(1)}^{m - 1}}{n {(1)}^{n - 1}} = \frac{m}{n} [? \underset{x \to a}{l i m} \frac{x^{n} - a^{n}}{x - a} = n . a^{n - 1}] \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

10 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

$l i m_{x \to 0} \frac{x^{2} c o s x}{1 - c o s x}$ is

(a) 2

(b) $\frac{3}{2}$
(c) $- \frac{3}{2}$
(d) 1

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

10 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Choose the correct answer from the given four options in each of the Exercises 54 to 76:

$l i m_{x \to π} \frac{s i n x}{x - π}$ is

(a) 1

(b) 2

(c) -1

(d) 2

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to π}{l i m} \frac{s i n x}{x - π} \\ = \underset{x \to π}{l i m} \frac{s i n (π - x)}{- (π - x)} = - 1 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1 a n d π - x \to 0 \Rightarrow x \to π] \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

10 months ago

Maths 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

0 Follower 4 Views

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

10 months ago

Maths 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 $k$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

10 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

0 Follower 4 Views

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

10 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

0 Follower 6 Views

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

10 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

10 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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