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

The function $f (x) = c o t x$ is discontinuous on the set

(A) ${x = n π : n \in}$

(B) ${x = 2 n π : n \in}$

(C) ${x = (2 n + 1) \frac{π}{2} : n \in}$

(D) ${x = \frac{n π}{2} : n \in}$

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:

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

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New answer posted•a year ago

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

The set of points where the function $f (x) =\sqrt2 - s i n x$ is differentiable is

(A) R

(B) R $- {\frac{1}{2}}$

(C) $(0, \infty)$

(D) None of these

9 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:

$\begin{array}{l} G i v e n t h a t f (x) = | 2 x - 1 | s i n x \\ C l e a r l y, f (x) i s n o t d i f f e r e n t i a b l e a t x = \frac{1}{2} \\ R . H . L . = f^{'} (\frac{1}{2}) = \underset{h \to 0}{l i m} \frac{f (\frac{1}{2} + h) - f (\frac{1}{2})}{h} \\ = \underset{h \to 0}{l i m} \frac{| 2 (\frac{1}{2} + h) - 1 | s i n (\frac{1}{2} + h) - 0}{h} \\ = \underset{h \to 0}{l i m} \frac{| 2 h | s i n (\frac{1 + 2 h}{2})}{h} = 2 s i n (\frac{1}{2}) \\ A l s o, L . H . L . = f^{'} (\frac{1}{2}) = \underset{h \to 0}{l i m} \frac{f (\frac{1}{2} - h) - f (\frac{1}{2})}{- h} \\ = \underset{h \to 0}{l i m} \frac{| 2 (\frac{1}{2} - h) - 1 | [- s i n (\frac{1}{2} - h)] - 0}{- h} \\ = \underset{h \to 0}{l i m} \frac{| - 2 h | [- s i n (\frac{1}{2} - h)]}{- h} = - 2 s i n (\frac{1}{2}) \\ ∴ L . H . L . \neq R . H . L . \\ S o, t h e g i v e n f u n c t i o n f (x) i s n o t d i f f e r e n t i a b l e a t x = \frac{1}{2} . \\ ∴ f (x) i s d i f f e r e n t i a b l e i n R - {\frac{1}{2}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) \end{array}$

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New answer posted•a year ago

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

The function $f (x) = \frac{4 - x^{2}}{4 x - x^{3}}$ is

(A) Discontinuous at only one point

(B) Discontinuous at exactly two points

(C) Discontinuous at exactly three points

(D) None of these

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:

$\begin{array}{l} G i v e n t h a t f (x) = \frac{4 - x^{2}}{4 x - x^{3}} \\ F o r d i s c o n t i n u o u s f u n c t i o n \\ \Rightarrow 4 x - x^{3} = 0 \\ \Rightarrow 4 (x - x^{2}) = 0 \\ \Rightarrow x (2 - x) (2 + x) = 0 \\ \Rightarrow x = 0, x = 2, x = - 2 \\ H e n c e, t h e g i v e n f u c t i o n i s d i s c o n t i n u o u s e x a c t l y a t t h r e e points . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) \end{array}$

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New answer posted•a year ago

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

Choose the correct answer from given four options in each of the Exercises from 83 to 96.

If $f (x) = 2 x$ and $g (x) = \frac{x^{2}}{2} + 1$ , then which of the following can be a discontinuous function:

(A) $f (x) + g (x)$

(B) $f (x) - g (x)$

(C) $f (x) \cdot g (x)$

(D) $\frac{g (x)}{f (x)}$

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:

$\begin{array}{l} W e k n o w t h a t t h e algebraic polynomials are continuous functions everywhere. \\ ∴ f (x) + g (x) i s c o n t i n u o u s [\begin{array}{l} ? S u m, d i f f e r e n c e a n d p r o d u c t o f t w o c o n t i n u o u s \\ f u n c t i o n s i s a l s o c o n t i n u o u s \end{array}] \\ f (x) - g (x) i s c o n t i n u o u s \\ f (x) . g (x) i s c o n t i n u o u s \\ \frac{g (x)}{f (x)} i s o n l y c o n t i n u o u s i f g (x) \neq 0 \\ ∴ \frac{f (x)}{g (x)} = \frac{2 x}{\frac{x^{2}}{2} + 1} = \frac{4 x}{x^{2} + 2} \\ H e r e, \frac{g (x)}{f (x)} = \frac{\frac{x^{2}}{2} + 1}{2 x} = \frac{x^{2} + 2}{4 x} w h i c h i s d i s c o n t i n u o u s a t x = 0 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) \end{array}$

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New answer posted•a year ago

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

Using the Mean Value Theorem, prove that there is a point on the curve $y = 2 x^{2} - 5 x + 3$ between the points $A (1, 0)$ and $B (2, 1)$ , where the tangent is parallel to the chord $A B$ . Also, find that point.

5 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:

$\begin{array}{l} G i v e n t h a t y = 2 x^{2} - 5 x + 3 \\ L e t f (x) = 2 x^{2} - 5 x + 3 \\ (i) B e i n g a n p o l y n o m i a l, f (x) i s c o n t i n u o u s i n [1, 2] . \\ (i i) f^{'} (x) = 4 x - 5 w h i c h e x i s t s i n (1, 2) . \\ H e n c e, b y m e a n v a l u e t h e o r e m, t h e r e m u s t e x i s t a \in (1, 2) \\ A (1, 0) a n d B (2, 1) \\ ∴ f^{'} (c) = \frac{f (b) - f (a)}{b - a} \\ \Rightarrow 4 c - 5 = \frac{(8 - 10 + 3) - (2 - 5 + 3)}{2 - 1} \\ \Rightarrow 4 c - 5 = \frac{1 - 0}{1} = 1 \Rightarrow 4 c = 5 + 1 = 6 \\ ∴ c = \frac{6}{4} = \frac{3}{2} \\ ∴ y = 2 {(\frac{3}{2})}^{2} - 5 (\frac{3}{2}) + 3 \\ \Rightarrow = 2 \times \frac{9}{4} - \frac{15}{2} + 3 = \frac{9}{2} - \frac{15}{2} + 3 \\ \Rightarrow = \frac{9 - 15 + 6}{2} = 0 \\ H e n c e, (\frac{3}{2}, 0) i s t h e \\ A (1, 0) a n d B (2, 1) . \end{array}$

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New answer posted•a year ago

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

Find a point on the curve $y = {(x - 3)}^{2}$ , where the tangent is parallel to the chord joining the points $(3, 0)$ and $(4, 1)$ .

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:

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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

5 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:

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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

$f (x) = s i n x - s i n 2 x$ in $[0, π]$ .

5 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:

Upvote

New answer posted•a year ago

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

Kindly consider the following

$f (x) = x^{3} - 2 x^{2} - x + 3$ in $[0, 1]$ .

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = x^{3} - 2 x^{2} - x + 3 i n [0, 1] \\ (i) f (x) i s a n f u n c t i o n, s o i t i s c o n t i n u o u s i n [0, 1] . \\ (i i) f^{'} (x) = 3 x^{2} - 4 x - 1 w h i c h e x i s t s i n (0, 1) \\ S o, f (x) i s d i f f e r e n t i a b l e . \\ A s t h e a b o v e c o n d i t i o n s a r e s a t i s f i e d, t h e n t h e r e m u s t e x i s t a t l e a s t o n e \in (0, 1) s u c h t h a t \\ f^{'} (c) = \frac{f (b) - f (a)}{b - a} \\ \Rightarrow 3 c^{2} - 4 c - 1 = \frac{[{(1)}^{3} - 2 {(1)}^{2} - (1) + 3] - [0 - 0 - 0 - 3]}{1 - 0} \\ \Rightarrow 3 c^{2} - 4 c - 1 = \frac{(1 - 2 - 1 + 3) - (3)}{1} \\ \Rightarrow 3 c^{2} - 4 c - 1 = 1 - 3 \Rightarrow 3 c^{2} - 4 c - 1 = - 2 \\ \Rightarrow 3 c^{2} - 4 c + 1 = 0 \Rightarrow 3 c^{2} - 3 c - c + 1 = 0 \\ \Rightarrow 3 c (c - 1) - 1 (c - 1) = 0 \Rightarrow (c - 1) (3 c - 1) = 0 \\ \Rightarrow c - 1 = 0 ∴ c = 1 \\ 3 c - 1 = 0 ∴ c = \frac{1}{3} \in (0, 1) \\ H e n c e, M e a n V a l u e T h e o r e m i s v e r i f i e d . \end{array}$