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Maths NCERT Exemplar Solutions Class 12th Chapter Six

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Maths NCERT Exemplar Solutions Class 12th Chapter Six 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

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alok kumar singh

Contributor-Level 10

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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Maths NCERT Exemplar Solutions Class 12th Chapter Six Application of Derivatives

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

(A) 0

(B) 12

(C) 16

(D) 32

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alok kumar singh

Contributor-Level 10

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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Maths NCERT Exemplar Solutions Class 12th Chapter Six 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

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alok kumar singh

Contributor-Level 10

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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Maths NCERT Exemplar Solutions Class 12th Chapter Six Application of Derivatives

The maximum value of sinx –cosx is:

(A) 1/4
(B) 1/2
(C) √2
(D) 2√2

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted

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Maths NCERT Exemplar Solutions Class 12th Chapter Six 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

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alok kumar singh

Contributor-Level 10

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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Maths NCERT Exemplar Solutions Class 12th Chapter Six Application of Derivatives

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

(A) 126

(B) 0

(C) 135

(D) 160

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alok kumar singh

Contributor-Level 10

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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Maths NCERT Exemplar Solutions Class 12th Chapter Six Application of Derivatives

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

(A) –1

(B) 0

(C) 1

(D) 2

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alok kumar singh

Contributor-Level 10

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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Maths NCERT Exemplar Solutions Class 12th Chapter Six Application of Derivatives

The function f(x) = tanx – x:

(A) Always increases

(B) Always decreases

(C) Never increases

(D) Sometimes increases and sometimes decreases

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alok kumar singh

Contributor-Level 10

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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Maths NCERT Exemplar Solutions Class 12th Chapter Six Application of Derivatives

Which of the following functions is decreasing in (0,) π/2 )?

(A) sin2x

(B) tanx

(C) cosx

(D) cos3x

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t f (x) = c o s x \\ S o, f^{'} (x) = - s i n x \Rightarrow f^{'} (x) < 0 i n (0, \frac{π}{2}) \\ S o, f (x) = c o s x i s decreasing i n (0, \frac{π}{2}) \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Six Application of Derivatives

The function f(x) = 4 sin³x – 6 sin²x + 12 sinx + 100 is strictly decreasing in:

(A) Increasing in (π, 3π/2)

(B) Decreasing in ( π/2,π)

(C) Decreasing in (-π/2,π/2 )

(D) Decreasing in (0, π/2)

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n f u n c t i o n i s f (x) = 4 s i n^{3} x - 6 s i n^{2} x + 12 s i n x + 100 \\ f^{'} (x) = 12 s i n^{2} x . c o s x - 12 s i n x c o s x + 12 c o s x \\ = 12 c o s x [s i n^{2} x - s i n x + 1] \\ = 12 c o s x [s i n^{2} x + (1 - s i n x)] \\ ? 1 - s i n x \geq 0 a n d s i n^{2} x \geq 0 \\ ∴ s i n^{2} x + 1 - s i n x \geq 0 (w h e n c o s x > 0) \\ H e n c e, f^{'} (x) > 0 w h e n c o s x > 0 i . e ., x \in (- \frac{π}{2}, \frac{π}{2}) \\ S o, f (x) i s increasing w h e r e x \in (- \frac{π}{2}, \frac{π}{2}) f^{'} (x) < 0 \\ w h e n c o s x < 0 i . e ., x \in (\frac{π}{2}, \frac{3 π}{2}) \\ H e n c e, f (x) i s decreasing w h e n x \in (\frac{π}{2}, \frac{3 π}{2}) \\ A s (\frac{π}{2}, π) \in (\frac{π}{2}, \frac{3 π}{2}) \\ S o, f (x) i s decreasing i n (\frac{π}{2}, π) \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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