Ncert Solutions Maths class 12th

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New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

**The curves y = 4x² + 2x – 8 and y = x³ – x + 13 touch each other at the point ___.**

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e y = 4 x^{2} + 2 x - 8 \dots (i) \\ a n d y = x^{3} - x + 13 \dots (i i) \\ D i f f e r e n t i a t i n g e q . (i) w . r . t . x, w e h a v e \\ \frac{d y}{d x} = 8 x + 2 \Rightarrow m_{1} = 8 x + 2 [m_{1} i s s l o p e o f c u r v e (i)] \\ D i f f e r e n t i a t i n g e q . (i i) w . r . t . x, w e g e t \\ \frac{d y}{d x} = 3 x^{2} - 1 \Rightarrow m_{2} = 3 x^{2} - 1 [m_{2} i s s l o p e o f c u r v e (i i)] \\ I f t h e t w o c u r v e s t o u c h e a c h o t h e r, t h e n m_{1} = m_{2} \\ ∴ 8 x + 2 = 3 x^{2} - 1 \\ \Rightarrow 3 x^{2} - 8 x - 3 = 0 \Rightarrow 3 x^{2} - 9 x + x - 3 = 0 \\ \Rightarrow 3 x (x - 3) + 1 (x - 3) = 0 \Rightarrow (x - 3) (3 x + 1) = 0 \\ ∴ x = 3, \frac{- 1}{3} \\ P u t t i n g x = 3 i n e q (i), w e g e t \\ y = 4 {(3)}^{2} + 2 (3) - 8 = 36 + 6 - 8 = 34 \\ S o, t h e r e q u i r e d point i s (3, 34) \\ N o w f o r x = \frac{- 1}{3} \\ y = 4 {(\frac{- 1}{3})}^{2} + 2 (\frac{- 1}{3}) - 8 = 4 \times \frac{1}{9} - \frac{2}{3} - 8 \\ = \frac{4}{9} - \frac{2}{3} - 8 = \frac{4 - 6 - 72}{9} = \frac{- 74}{9} \\ ∴ O t h e r r e q u i r e d point i s (\frac{- 1}{3}, \frac{- 74}{9}) . \\ H e n c e, t h e r e \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

**The maximum value of (1/x)^x is:

(A) e

(B) $e^{e}$

(C) $\frac{1}{e^{e}}$

(D)** $e^{\frac{1}{e}}$

22 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e f (x) = {(\frac{1}{x})}^{x} \\ T a k i n g l o g o f b o t h s i d e s, w e h a v e \\ l o g f (x) = x l o g \frac{1}{x} \\ l o g f (x) = x l o g x^{- 1} \Rightarrow l o g [f (x)] = - [x l o g x] \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t \\ \frac{1}{f (x)} f^{'} (x) = - [x . \frac{1}{x} + l o g x . 1] \\ \Rightarrow f^{'} (x) = - f (x) [1 + l o g x] = - {(\frac{1}{x})}^{x} [1 + l o g x] \\ F o r l o c a l maxima a n d l o c a l minima f^{'} (x) = 0 \\ - {(\frac{1}{x})}^{x} [1 + l o g x] = 0 \Rightarrow {(\frac{1}{x})}^{x} [1 + l o g x] = 0 \\ {(\frac{1}{x})}^{x} \neq 0 ∴ 1 + l o g x = 0 \\ \Rightarrow l o g x = - 1 \Rightarrow x = e^{- 1} \Rightarrow x = \frac{1}{e} \\ S o, x = \frac{1}{e} i s t h e s t a t i o n a r y point. \\ N o w, f^{'} (x) = - {(\frac{1}{x})}^{x} [1 + l o g x] \\ f^{''} (x) = - [{(\frac{1}{x})}^{x} (\frac{1}{x}) + (1 + l o g x) . \frac{d}{d x} {(x)}^{x}] \\ f^{''} (x) = - [{(e)}^{1 / e} (e) + (1 + l o g \frac{1}{e}) \frac{d}{d x} {(\frac{1}{e})}^{1 / e}] \\ x = \frac{1}{e} = - e^{\frac{1}{e} + 1} < 0 maxima \\ ∴ Maximum value of the function at x = \frac{1}{e} i s \\ f (\frac{1}{e}) = {(\frac{1}{1 / e})}^{1 / e} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

**The function f(x) = xˣ has a stationary point at:

(A) x = e

(B) x = 1/e

(C) x = 1

(D)** x =√e

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 Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} W e h a v e f (x) = x^{x} \\ T a k i n g l o g o f b o t h s i d e s, w e h a v e \\ l o g f (x) = x l o g x \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t \\ \frac{1}{f (x)} f^{'} (x) = x . \frac{1}{x} + l o g x . 1 \\ \Rightarrow f^{'} (x) = f (x) [1 + l o g x] = x^{x} [1 + l o g x] \\ T o f i n d s t a t i o n a r y point, f^{'} (x) = 0 \\ ∴ x^{x} [1 + l o g x] = 0 \\ x^{x} \neq 0 ∴ 1 + l o g x = 0 \\ \Rightarrow l o g x = - 1 \Rightarrow x = e^{- 1} \Rightarrow x = \frac{1}{e} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

**The maximum slope of the curve y = –x³ + 3x² + 9x – 27 is:

(A) 0

(B) 12

(C) 16

(D)** 32

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = - x^{3} + 3 x^{2} + 9 x - 27 \\ S o, \frac{d y}{d x} = - 3 x^{2} + 6 x + 9 \\ ∴ S l o p e o f t h e g i v e n c u r v e, \\ m = - 3 x^{2} + 6 x + 9 \\ \frac{d m}{d x} = - 6 x + 6 \\ F o r l o c a l ? ? ? maxima a n d l o c a l minima, \frac{d m}{d x} = 0 \\ ∴ - 6 x + 6 = 0 \Rightarrow x = 1 \\ Now \frac{d^{2} m}{d x^{2}} = - 6 < 0 maxima \\ ∴ M aximum v a l u e o f t h e s l o p e at x =1 is \\ m_{x = 1} = - 3 {(1)}^{2} + 6 (1) + 9 = - 3 + 6 + 9 = 12 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

At $x = \frac{5 π}{6},$ **the function f(x) = 2sin³x + 3cos³x is:

(A) Maximum

(B) Minimum

(C) Zero

(D)** Neither maximum nor minimum

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e f (x) = 2 s i n 3 x + 3 c o s 3 x \\ S o, f^{'} (x) = 2 c o s 3 x . 3 - 3 s i n 3 x . 3 = 6 c o s 3 x - 9 s i n 3 x \\ f^{''} (x) = - 6 s i n 3 x . 3 - 9 c o s 3 x . 3 \\ = - 18 s i n 3 x - 27 c o s 3 x \\ Now f^{''} (\frac{5 π}{6}) = - 18 s i n 3 (\frac{5 π}{6}) - 27 c o s 3 (\frac{5 π}{6}) \\ = - 18 s i n (\frac{5 π}{2}) - 27 c o s (\frac{5 π}{2}) \\ = - 18 s i n (2 π + \frac{π}{2}) - 27 c o s (2 π + \frac{π}{2}) \\ = - 18 s i n \frac{π}{2} - 27 c o s \frac{π}{2} = - 18.1 - 27.0 \\ = - 18 < 0 maxima \\ M aximum v a l u e o f f (x) a t x = \frac{5 π}{6} \\ f (\frac{5 π}{6}) = 2 s i n 3 (\frac{5 π}{6}) + 3 c o s 3 (\frac{5 π}{6}) = 2 s i n (\frac{5 π}{2}) + 3 c o s (\frac{5 π}{2}) \\ = 2 s i n (2 π + \frac{π}{2}) + 3 c o s (2 π + \frac{π}{2}) = 2 s i n \frac{π}{2} + 3 c o s \frac{π}{2} = 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

The maximum value of sinx –cosx is:

(A) 1/4
(B) 1/2
(C) √2
(D) 2√2

11 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

**The function f(x) = 2x³ – 3x² – 12x + 4 has:

(A) Two points of local maximum

(B) Two points of local minimum

(C) One maximum and one minimum

(D)** No maximum or minimum

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4 \\ S o, f^{'} (x) = 6 x^{2} - 6 x - 12 \\ F o r l o c a l ? ? ? maxima a n d l o c a l minima, f^{'} (x) = 0 \\ ∴ 6 x^{2} - 6 x - 12 = 0 \\ \Rightarrow x^{2} - x - 2 = 0 \\ \Rightarrow x^{2} - 2 x + x - 2 = 0 \Rightarrow x (x - 2) + 1 (x - 2) = 0 \\ \Rightarrow (x + 1) (x - 2) = 0 \\ ∴ x = - 1, 2 \\ S o, x = - 1, 2 a r e t h e points o f l o c a l ? ? ? maxima a n d l o c a l minima. \\ Now f^{''} (x) = 12 x - 6 \\ f^{''} {(x)}_{x = - 1} = 12 (- 1) - 6 = - 12 - 6 = - 18 < 0, maxima \\ f^{''} {(x)}_{x = 2} = 12 (2) - 6 = 24 - 6 = 18 > 0, minima \\ S o t h e function is maximum at x = - 1 and minimum a t x = 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

**The smallest value of the polynomial x³ – 18x² + 96x in [0, 9] is:

(A) 126

(B) 0

(C) 135

(D)** 160

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t f (x) = x^{3} - 18 x^{2} + 96 x \\ S o, f^{'} (x) = 3 x^{2} - 36 x + 96 \\ F o r l o c a l ? ? ? maxima a n d l o c a l minima, f^{'} (x) = 0 \\ ∴ 3 x^{2} - 36 x + 96 = 0 \\ \Rightarrow x^{2} - 12 x + 32 = 0 \\ \Rightarrow x^{2} - 8 x - 4 x + 32 = 0 \Rightarrow x (x - 8) - 4 (x - 8) = 0 \\ \Rightarrow (x - 8) (x - 4) = 0 \\ ∴ x = 8, 4 \in [0, 9] \\ S o, x = 4, 8 a r e t h e points o f l o c a l ? ? ? maxima a n d l o c a l minima. \\ Now we will calculate the absolute maxima o r absolute minima at x=0,4,8,9 \\ ∴ f (x) = x^{3} - 18 x^{2} + 96 x \\ f {(x)}_{x = 0} = 0 - 0 + 0 = 0 \\ f {(x)}_{x = 4} = {(4)}^{3} - 18 {(4)}^{2} + 96 (4) \\ = 64 - 288 + 384 = 448 - 288 = 160 \\ f {(x)}_{x = 8} = {(8)}^{3} - 18 {(8)}^{2} + 96 (8) \\ = 512 - 1152 + 768 = 1280 - 1152 = 128 \\ f {(x)}_{x = 9} = {(9)}^{3} - 18 {(9)}^{2} + 96 (9) \\ = 729 - 1458 + 864 = 1593 - 1458 = 135 \\ S o t h e absolute minimum v a l u e o f &th \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

**If x is real, the minimum value of x² – 8x + 17 is:

(A) –1

(B) 0

(C) 1

(D)** 2

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t f (x) = x^{2} - 8 x + 17 \\ S o, f^{'} (x) = 2 x - 8 \\ F o r l o c a l ? ? ? maxima a n d l o c a l minima, f^{'} (x) = 0 \\ ∴ 2 x - 8 = 0 \Rightarrow x = 4 \\ S o, x = 4 i s t h e point o f l o c a l ? ? ? maxima a n d l o c a l minima. \\ f^{''} (x) = 2 > 0 minima a t x = 4 \\ ∴ f {(x)}_{x = 4} = {(4)}^{2} - 8 (4) + 17 \\ = 16 - 32 + 17 = 33 - 32 = 1 \\ S o t h e minimum v a l u e o f t h e f u n c t i o n i s 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

**The function f(x) = tanx – x:

(A) Always increases

(B) Always decreases

(C) Never increases

(D)** Sometimes increases and sometimes decreases

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t f (x) = t a n x - x \\ S o, f^{'} (x) = s e c^{2} x - 1 \Rightarrow f^{'} (x) > 0 \forall \in R \\ S o, f (x) i s a l w a y s increasing \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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