Ncert Solutions Maths class 12th

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New answer posted•a year ago

Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

For the function $f (x) = x + \frac{1}{x}$ , $x \in [1, 3]$ , the value of $c$ for the Mean Value Theorem is

(A) 1

(B) 3

(C) 2

(D) None of these

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted•a year ago

Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

The value of $c$ in Rolle's theorem for the function $f (x) = x^{3} - 3 x$ in the interval $[0, \sqrt3]$ is

(A) 1

(B) -1

(C) $\frac{3}{2}$

(D) $\frac{1}{3}$

13 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted•a year ago

Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

If $x = t^{2}, y = t^{3}$ , then $\frac{d^{2} y}{d x^{2}}$ is

(A) $\frac{3}{2}$

(B) $\frac{3}{4 t}$

(C) $\frac{3}{2 t}$

(D) $\frac{3}{4}$

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t x = t^{2} a n d y = t^{3} \\ D i f f e r e n t i a t i n g b o t h f u n c t i o n s w . r . t . t \\ \Rightarrow \frac{d x}{d t} = 2 t a n d \frac{d y}{d t} = 3 t^{2} \\ ∴ \frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{3 t^{2}}{2 t} = \frac{3 t}{2} \\ D i f f e r e n t i a t i n g a g a i n w . r . t . x \\ \frac{d}{d x} (\frac{d y}{d x}) = \frac{3}{2} . \frac{d t}{d x} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{3}{2} . \frac{1}{2 t} = \frac{3}{4 t} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

Kindly consider the following

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

If $y =\sqrtsinx+y$ , then $\frac{d y}{d x}$ is equal to

(A) $\frac{x}{2 y - 1}$

(B) $\frac{x}{1 - 2 y}$

(C) $\frac{x}{1 - 2 y}$

(D) $\frac{x}{2 y - 1}$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted•a year ago

Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

If $y = l o g_{} (\frac{1 - x^{2}}{1 + x^{2}})$ , then $\frac{d y}{d x}$ is equal to

(A) $\frac{4 x^{3}}{1 - x^{4}}$

(B) $\frac{- 4 x}{1 - x^{4}}$

(C) $\frac{1}{4 - x^{4}}$

(D) $\frac{- 4 x^{3}}{1 - x^{4}}$

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = l o g (\frac{1 - x^{2}}{1 + x^{2}}) \\ \Rightarrow y = l o g (1 - x^{2}) - l o g (1 + x^{2}) [? l o g \frac{x}{y} = l o g x - l o g y] \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow \frac{d y}{d x} = \frac{1}{1 - x^{2}} . \frac{d}{d x} (1 - x^{2}) - \frac{1}{1 + x^{2}} . \frac{d}{d x} (1 + x^{2}) \\ \Rightarrow = \frac{- 2 x}{1 - x^{2}} - \frac{2 x}{1 + x^{2}} = \frac{- 2 x - 2 x^{3} - 2 x + 2 x^{3}}{(1 - x^{2}) (1 + x^{2})} = \frac{- 4 x}{(1 - x^{4})} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

Let $f (x) = ∣ s i n x ∣$ . Then

(A) $f$ is everywhere differentiable

(B) $f$ is everywhere continuous but not differentiable at $x = n π, n \in$ .

(C) $f$ is everywhere continuous but not differentiable at $x = (2 n + 1) \frac{π}{2}, n \in$ .

(D) None of these

11 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = | s i n x | \\ L e t g (x) = | s i n x | a n d t (x) = | x | \\ ∴ f (x) = t o g (x) = t [g (x)] = t (s i n x) = | s i n x | \\ w h e r e g (x) a n d t (x) b o t h a r e c o n t i n u o u s . \\ ∴ f (x) = g o t (x) i s c o n t i n u o u s b u t t (x) i s n o t d i f f e r e n t i a b l e a t x = 0 . \\ S o, f (x) i s n o t c o n t i n u o u s a t s i n x = 0 \Rightarrow x = n π, n \in Z . \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

If $f (x) = {\begin{matrix} m x + 1, & x \leq \frac{π}{2} \\ s i n x + n, & x > \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2}$ , then

(A) $m = 1, n = 0$

(B) $m = \frac{n π}{2} + 1$

(C) $n = \frac{m π}{2}$

(D) $m = n = \frac{π}{2}$

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = {\begin{matrix} m x + 1, & x \leq \frac{π}{2} \\ s i n x + n, & x > \frac{π}{2} \end{matrix} i s c o n t i n u o u s a t x = \frac{π}{2} \\ R . H . L . = \underset{x \to {\frac{π}{2}}^{+}}{l i m} (s i n x + n) = \underset{h \to 0}{l i m} [s i n (\frac{π}{2} + h) + n] \\ = \underset{h \to 0}{l i m} c o s h + n = 1 + n \\ A l s o, L . H . L . = \underset{x \to {\frac{π}{2}}^{-}}{l i m} (m x + 1) = \underset{h \to 0}{l i m} [m (\frac{π}{2} - h) + 1] \\ = \frac{m π}{2} + 1 \\ W h e n f (x) i s c o n t i n u o u s a t x = \frac{π}{2} . \\ ∴ L . H . L . = R . H . L . \\ \frac{m π}{2} + 1 = 1 + n \Rightarrow n = \frac{m π}{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

If $f (x) = \frac{x^{2} - 1}{x} s i n x$ , where $x \neq 0$ , then the value of the function $f$ at $x = 0$ , so that the function is continuous at $x = 0$ , is

(A) 0

(B) -1

(C) 1

(D) None of these

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = x^{2} s i n \frac{1}{x} w h e r e x \neq 0 . \\ S o, t h e V a l u e o f t h e f u n c t i o n f a t x = 0, s o t h a t f (x) i s c o n t i n u o u s i s 0 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

The function $f (x) = e^{| x |}$ is

(A) Continuous everywhere but not differentiable at $x = 0$

(B) Continuous and differentiable everywhere

(C) Not continuous at $x = 0$

(D) None of these

13 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = e^{| x |} \\ W e k n o w t h a t modulus f u n c t i o n i s c o n t i n u o u s b u t n o t d i f f e r e n t i a b l e i n i t s d o m a i n . \\ L e t g (x) = | x | a n d t (x) = e^{| x |} \\ ∴ f (x) = g o t (x) = g [t (x)] = e^{| x |} \\ Since g (x) a n d t (x) b o t h a r e c o n t i n u o u s a t x = 0 b u t f (x) i s n o t d i f f e r e n t i a b l e a t x = 0 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) \end{array}$

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Ncert Solutions Maths class 12th

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