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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

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

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

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

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$x c o s x$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

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

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

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

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Differentiate each of the functions with respect to ‘x’ in Exercises 43 to 46 using first principle.

$c o s (x^{2} + 1)$

24 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$\frac{1}{a x^{2} + b x + c}$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} L e t y = \frac{1}{a x^{2} + b x + c} \\ \frac{d y}{d x} = \frac{d}{d x} (\frac{1}{a x^{2} + b x + c}) \\ = \frac{(a x^{2} + b x + c) \frac{d}{d x} (1) - 1 . \frac{d}{d x} (a x^{2} + b x + c)}{{(a x^{2} + b x + c)}^{2}} [Using q u o t i e n t r u l e] \\ = \frac{(a x^{2} + b x + c) \times 0 - (2 a x + b)}{{(a x^{2} + b x + c)}^{2}} = \frac{- (2 a x + b)}{{(a x^{2} + b x + c)}^{2}} \end{array}$

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