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Application of Derivatives Class 12th

81. At what points in the interval [0, 2π], does the function sin 2x attain its maximum value?

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Vishal Baghel

Contributor-Level 10

We have, f(x) = sin 2x, x ∈ [0, 2π],

f(x) = 2cos 2x.

At f(x) = 0.

2 cos 2x = 0

cos 2x = 0

$\Rightarrow 2 x = (2 x + 1) \frac{π}{2}, x = 0, 1, 2, 3 .$

$\Rightarrow x = (2 x + 1) \frac{π}{4} .$

$x = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}, \in [0, 2 π]$

$∴ f (\frac{π}{4}) = s i n \frac{2 π}{4} = s i n \frac{π}{2} = 1$ .

$f (\frac{3 π}{4}) = s i n 2 (\frac{3 π}{4}) =$ $s i n \frac{3 π}{2}$ $f (\frac{7 π}{4})$

$= s i n (π + \frac{π}{2}) =$ $= s i n 2 * \frac{7 π}{4}$

$= - s i n \frac{π}{2}$ $= s i n \frac{7 π}{2}$

= 1. $= s i n 3 π + \frac{π}{2}$

$f (\frac{5 π}{4}) = s i n 2 * (\frac{5 π}{4}) = s i n \frac{5 π}{2} = s i n (2 x + \frac{π}{4})$ $= - s i n \frac{π}{2}$

$= s i n \frac{π}{4} = 1$ = 1.

f(0) = sin 2(0) = sin 0 = 0

f(2π) = sin 2(2π) = sin 4π = 0

Hence, the points of maximum $x f x are.$

$(\frac{π}{4}, 2)and (\frac{5 π}{4}, 1) .$

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Application of Derivatives Class 12th

80. Find both the maximum value and the minimum value of 3x⁴ – 8x³ + 12x² – 48x + 25 on the interval [0, 3].

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Vishal Baghel

Contributor-Level 10

We have, f (x) = 3x⁴- 8x³ + 12x²- 48x + 25, x ∈ [0, 3].

f (x) = 12x³- 24x² + 24x - 48.

At f (x) = 0.

12x³- 24x² + 24x - 48 = 0.

x³- 2x² + 2x - 4 = 0

x² (x - 2) + 2 (x - 2) = 0

(x - 2) + (x² + 2) = 0

x = 2 ∈ [0, 3] or x = $\pm\sqrt-2$ which is not possible as

∴f (x) = 3 (2)⁴- 8 (2)³ + 12 (2)²- 4 (2) + 25.

=48 - 64 + 48 - 96 + 25.

= -39.

f (0) =3 (0)⁴- 8 (0)³ + 12 (0)²- 48 (0) + 25.

= 25.

f (3) = 3 (3)⁴- 8 (3)³ + 12 (3)²- 48 (3) + 25.

= 243 - 216 + 108 - 144 + 25

= 16.

Maximum value of f (x) = 25 at x = 0.

and minimum value of f (x) = -39 at x = 2.

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Application of Derivatives Class 12th

79. Find the ma proximumfit that a company can make, if the profit function is given by p(x) = 41 – 72x – 18x²

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Vishal Baghel

Contributor-Level 10

We have, p (x) = 41 -f2x - 18x².

P (x) = - 72 - 36x

P (x) = -36

At extreme point,

- 72 - 36x = 0

$\Rightarrow$ $x = \frac{- 72}{36} = - 2$ .

At x = - 2, p" (x) = - 36 < 0.

∴x = -2 is a point of local maximum and the value of local

Maximum is given by P (2) = 41 - 72 (- 2) - 18 (- 2)²

41 + 144 - 72 = 113 units.

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Application of Derivatives Class 12th

78. Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i) f(x) =x³ , x ∈ [– 2, 2]

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Vishal Baghel

Contributor-Level 10

(i) We have,

f(x) = x³ , x ∈ [– 2, 2].

f(x) = 3x².

At, f(x) = 0

3x² = 0

x = 0 <--[-2, 2].

We shall absolute the value of f at x = 0 and points of interval [ -2, 2]. So,

f(0) = 0

f(- 2) = (- 2)³ = 8

f(2) = 2³ = 8.

∴ Absolute maximum value of f(x) = 8 at x = 2

and absolute minimum value of f(x) = -8 at x = -2.

(ii) f (x) = sin x + cos x , x ∈ [0, π]

A.(ii)

We have, f(x) = sin x + cos x , x ∈ [0, π]

f(x) = cos x - sin x.

atf(x) = 0

$\Rightarrow$ cosx - sin x = 0

$\Rightarrow$ sinx = cos x

$\Rightarrow \frac{s i n^{°} x}{c o r x} = 1 t a n x = 1 \Rightarrow t a n x = t a n \frac{π}{4}$

$\Rightarrow x = \frac{π}{4} \in [0, π]$

(iii) f(x) = 4x $- \frac{1}{2} x^{2}, x \in [- 2, \frac{9}{2}]$

A.(iii)

We have, f(x) = 4x $- \frac{1}{2} x^{2}, x \in [- 2, \frac{9}{2}]$

f(x) = 4 - x

atf(x) = 0

$\Rightarrow$ 4- x = 0

$\Rightarrow$ x = 4 $\in$ $[-2, \frac{9}{2}]$

$∴ f (4) = 4 (4) - \frac{1}{2} {(4)}^{2} = 16 - 8 = 8 .$

$f (- 2) = 4 (- 2) - \frac{1}{2} {(- 2)}^{2} = - 8 - 2 = - 10 .$

$f (\frac{9}{2}) = 4 (\frac{9}{2}) - \frac{1}{2} {(\frac{9}{2})}^{2} = 18 - \frac{81}{8} = \frac{94 - 81}{8} = \frac{63}{8} .$

= 7.87.5

Hence, absolute maximum value of f(x) = 8 at x = 4

and absolute minimum value of f(

...more

(i) We have,

f(x) = x³ , x ∈ [– 2, 2].

f(x) = 3x².

At, f(x) = 0

3x² = 0

x = 0 <--[-2, 2].

We shall absolute the value of f at x = 0 and points of interval [ -2, 2]. So,

f(0) = 0

f(- 2) = (- 2)³ = 8

f(2) = 2³ = 8.

∴ Absolute maximum value of f(x) = 8 at x = 2

and absolute minimum value of f(x) = -8 at x = -2.

(ii) f (x) = sin x + cos x , x ∈ [0, π]

A.(ii)

We have, f(x) = sin x + cos x , x ∈ [0, π]

f(x) = cos x - sin x.

atf(x) = 0

$\Rightarrow$ cosx - sin x = 0

$\Rightarrow$ sinx = cos x

$\Rightarrow \frac{s i n^{°} x}{c o r x} = 1 t a n x = 1 \Rightarrow t a n x = t a n \frac{π}{4}$

$\Rightarrow x = \frac{π}{4} \in [0, π]$

(iii) f(x) = 4x $- \frac{1}{2} x^{2}, x \in [- 2, \frac{9}{2}]$

A.(iii)

We have, f(x) = 4x $- \frac{1}{2} x^{2}, x \in [- 2, \frac{9}{2}]$

f(x) = 4 - x

atf(x) = 0

$\Rightarrow$ 4- x = 0

$\Rightarrow$ x = 4 $\in$ $[-2, \frac{9}{2}]$

$∴ f (4) = 4 (4) - \frac{1}{2} {(4)}^{2} = 16 - 8 = 8 .$

$f (- 2) = 4 (- 2) - \frac{1}{2} {(- 2)}^{2} = - 8 - 2 = - 10 .$

$f (\frac{9}{2}) = 4 (\frac{9}{2}) - \frac{1}{2} {(\frac{9}{2})}^{2} = 18 - \frac{81}{8} = \frac{94 - 81}{8} = \frac{63}{8} .$

= 7.87.5

Hence, absolute maximum value of f(x) = 8 at x = 4

and absolute minimum value of f(x) = -10 at x = -2.

(iv) f(x) = (x - 1)² + 3, x∈[ -3, 1]

A.(iv)

We have, f(x) = (x - 1)² + 3, x∈[ -3, 1]

f(x) = 2(x -1)

atf(x) = 0

2(x - 1) = 0

x = 1∈ [ -3, 1]

∴f(1) = (1 - 1)² + 3 = 3

f(- 3) = (- 3 - 1)² + 3 = 16 + 3 = 1

∴ Absolute maximum value of f(x) = 19 at x = -3.

And absolute minimum value of f(x) = 3 at x = 1.

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Application of Derivatives Class 12th

77. Prove that the following functions do not have maxima or minima:

(i) f(x) = e^x

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Vishal Baghel

Contributor-Level 10

(i) We have, f (x) = e^x

$\Rightarrow$ f (x) = e^x.

$\Rightarrow f^{'} (x) = e^{x .}$

At, extreme points,

f (x) = 0

$\Rightarrow$ e^x = 0 which has no real 'a' value

∴f (x) has with maximum or minima

(ii) g (x) = log x

A (ii)

We have, g (x) = log x,

$\Rightarrow$ g (x) = $\frac{1}{x}$

At extreme points,

g (x) = 0

$\to \frac{1}{x} = 0 .$

$\Rightarrow$ 1 = 0 which is not true.

∴g (x) was value minima or maxima

(iii) h (x) = x³ + x² + x + 1.

A (iii)

We have, h (x) = x³ + x² + x + 1.

h (x) = 3x² + 2x + 1

At extreme points,

h (x) = 0

$\Rightarrow$ 3x² + 2x + 1 = 0

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Application of Derivatives Class 12th

76. Find the local maxima and local minima, if any, of the following functions.

Find also the local maximum and the local minimum values, as the case may be:

(i) f(x) = x².

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Vishal Baghel

Contributor-Level 10

(i) we have, f(x) = x².

f(x) = 2x.

andf(x) = 2.

AR extreme point, f(x) = 0

2x = 0

x = 0.

When x = 0, f(0) = 2 > 0.

∴x = 0 is a point of local minima and value of local minimum is given by f(0) = 0² = 0.

(ii) g(x) = x³ 3x

A(ii)

we have, g(x) = x³- 3x

$\Rightarrow$ g'(x) = 3x²- 3

$\Rightarrow$ g''(x) = 6x.

At extreme point,

g'(x) = 0

3x²- 3 = 0.

3(x²- 1) = 0 ⇒ 3(x - 1)(x + 1) = 0.

x = 1 or x = -1.

At x = 1, g"(1) = 6.1 = 6 > 0.

So, x = 1 is a point of local minima and value of local minimum is given by g(1) = 1³- 3.1 = 1 - 3 = - 2.

And at x = -1, g"( -1) = 6 ( -1) = 6 < 0.

So, x = -1 is a point of local minima and value of local minimum is given by

g(- 1) = (- 1)³- 3(- 1) = 1 + 3 = 2.

...more

(i) we have, f(x) = x².

f(x) = 2x.

andf(x) = 2.

AR extreme point, f(x) = 0

2x = 0

x = 0.

When x = 0, f(0) = 2 > 0.

∴x = 0 is a point of local minima and value of local minimum is given by f(0) = 0² = 0.

(ii) g(x) = x³ 3x

A(ii)

we have, g(x) = x³- 3x

$\Rightarrow$ g'(x) = 3x²- 3

$\Rightarrow$ g''(x) = 6x.

At extreme point,

g'(x) = 0

3x²- 3 = 0.

3(x²- 1) = 0 ⇒ 3(x - 1)(x + 1) = 0.

x = 1 or x = -1.

At x = 1, g"(1) = 6.1 = 6 > 0.

So, x = 1 is a point of local minima and value of local minimum is given by g(1) = 1³- 3.1 = 1 - 3 = - 2.

And at x = -1, g"( -1) = 6 ( -1) = 6 < 0.

So, x = -1 is a point of local minima and value of local minimum is given by

g(- 1) = (- 1)³- 3(- 1) = 1 + 3 = 2.

(iii) $h (x) = s i n x + c o s x . 0 < x < \frac{π}{2}$

A(iii)

we have, h(x) = sin x + cos x.

$\Rightarrow$ h'(x) = cos x - sin x

$\Rightarrow$ h'(x) = -sin x - cos x.

At extreme points,

h'(x) = 0

cosx - sin x = 0

cosx = sin x

$\Rightarrow \frac{c o s x}{c o s x} = \frac{s i n x}{c o s x}$

$\Rightarrow$ 1 = tan x

$\Rightarrow t a n x = t a n \frac{π}{4}$

$\Rightarrow x = \frac{π}{4} .$

(iv) $f (x) = s i n x - c o s x, 0 < x < 2 π$

A(iv)

we have, f(x) = sin x cos x.

$\Rightarrow$ f(x) = cos x + sin x

$\Rightarrow$ f(x) = sin x + cos x.

At extreme points,

f(x) = 0

$\Rightarrow$ cosx + sin x = 0.

$\Rightarrow \frac{s i n x}{c o s x} = \frac{- c o s x}{c o s x}$

$\Rightarrow t a n x = - 1 = - t a n \frac{π}{4} = t a n (π - \frac{π}{4}) = t a n (2 π - \frac{π}{4})$

$∴ x = (π - \frac{π}{4})$ IInd quadrate or $(2 π - \frac{π}{4})$ IIIth quadrate

$\Rightarrow x = \frac{4 π - π}{4} \frac{8 x - π}{4}$

$\Rightarrow x = \frac{3 π}{4} o r x = \frac{f π}{4} .$

$A t x = \frac{3 π}{4}, f^{''} (\frac{3 π}{4}) = - s i n \frac{3 π}{4} + c o s \frac{3 π}{4}$

$= - s i n (π - \frac{π}{4}) + c o s (π - \frac{π}{4})$

$= - s i n \frac{π}{4} - c o t \frac{π}{4}$

(v) f(x) = x³- 6x² + 9x + 15.

A(v)

we have, f(x) = x³- 6x² + 9x + 15.

$\Rightarrow$ f(x) = 3x²- 12x + 9.

$\Rightarrow$ f'(x) = 6x - 12.

At extreme point, f'(x) = 0.

$\Rightarrow$ 3x²- 12x + 9 = 0.

$\Rightarrow$ x²- 4x + 3 = 0

$\Rightarrow$ x²-x - 3x + 3 = 0

$\Rightarrow$ x(x - 1) -3(x - 1) = 0

$\Rightarrow$ (x - 1)(x - 3) = 0

$\Rightarrow$ x = 1 or x = 3.

At, x = 1,f"(1) = 6 * 1 - 12 = 6 - 12 = - 6 < 0

∴x = 1 is a point of local maxima and the value of local

Maximum is given by f(1) = 1³- 6(1)² + 9(1) + 15

= 1 - 6 + 9 + 15

= 19.

And at x = 3,f"(3) = 6 * 3 - 12 = 18 - 12 = 6 > 0

∴x = 3 is a point of local minima and the value of

local minimum is given by f (3) = 3³- 6(3)² + 9(3) + 15.

= 27 - 54 + 27 + 15.

= 15.

(vi) $g (x) = \frac{x}{2} + \frac{2}{x}, x > 0$

A(vi)

We have, $g (x) = \frac{x}{2} + \frac{2}{x}, x > 0$

$\Rightarrow g^{'} (x) = \frac{1}{2} - \frac{2}{x^{2}}$

$\Rightarrow g g^{' z} (x) = \frac{x^{2} - 4}{2 x^{2}} = \frac{(x - 2) (x + 2)}{2 x^{2}}$

$g^{''} (x) = \frac{4}{x^{3}} .$

At, extreme point, g'(x) = 0

$\Rightarrow \frac{(x + 2) (x - 2)}{x^{2}} = 0$

$\Rightarrow$ x = -2 or x = 2

Given, that x > 0, hence we have x = 2.

At, x = 2, ${\overset{?}{g}}^{'} (x) = \frac{4}{2^{3}} = \frac{4}{8} = \frac{1}{2} > 0$

∴x = 2 is a point of local minima and value of local

Minimum is given by g(2) = $\frac{2}{2} + \frac{2}{2} = 1 + 1 = 2 .$

(vii) $f (x) = \frac{1}{x^{2} + 2}$

A(vii)

Given, $f (x) = \frac{1}{x^{2} + 2}$

$\Rightarrow g^{'} (x) = - \frac{1}{{(x^{2} + 2)}^{2}} \frac{d}{d x} (x^{2} + 2) = \frac{- 2 x}{{(x^{2} + 2)}^{2}} .$

$\Rightarrow g^{''} (x) = - [\frac{{(x^{2} + 2)}^{2} \frac{d}{d x} 2 x - 2 x \frac{d}{d x} {(x^{2} + 2)}^{2}}{{(x^{2} + 2)}^{4}}]$

$= \frac{- [2 {(x^{2} + 2)}^{2} - 2 x \cdot 2 (x^{2} + 2) \cdot 2 x]}{{(x^{2} + 2)}^{4}} .$

$= \frac{- [2 {(x^{2} + 2)}^{2} [x^{2} + 2 - 4 x^{2}]]}{{(x^{2} + 2)}^{4}}$

$= \frac{- 2 (x^{2} + 2) (2 - 3 x^{2})}{{(x^{2} + 2)}^{4}} = - \frac{2 (2 - 3 x^{2})}{{(x^{2} + 2)}^{3}} .$

At extreme points, g'(x) = 0.

$\Rightarrow \frac{- 2 x}{{(x^{2} + 2)}^{2}} = 0 .$

$\Rightarrow$ x = 0.

At, x = 0, $g^{''} (x) = \frac{- 2 (2 - 2 * 0^{2})}{{(0^{2} + 2)}^{3}} = \frac{- 2 (2 - 0)}{2^{3}} = - \frac{1}{2} < 0$

∴x = 0 is a point of local maxima and value of local maximum is given by $g (0) = \frac{1}{0^{2} + 2} = \frac{1}{2} .$

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Application of Derivatives Class 12th

75. Find the maximum and minimum values, if any, of the following functions given by

(i) f(x) = |x + 2| - 1

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Vishal Baghel

Contributor-Level 10

(i) we have, f(x) = |x + 2| - 1

We know that, for all $x \in ?, | x + 2 | \geq 0$

$\Rightarrow | x + 2 | - 1 \geq - 1 .$

$\Rightarrow$ f(x) $\geq$ - 1.

∴ Minimum value of f(x) = -1 when x + 2 = 0 $\Rightarrow$ x = - 2.

And maximum value of f(x) does not exist.

(ii) $g (x) = - | x + 1 | + 3$

A(ii)

We have, $g (x) = - | x + 1 | + 3$

For all $x \in, | x + 1 | \geq 0 .$

$\Rightarrow - | x + 1 | \leq 0$

$\Rightarrow - | x + 1 | + 3 \leq 0 + 3$

$\Rightarrow$ g(x) $\leq$ 3.

∴ Maximum value of g(x) = 3 when $| x + 1 | = 0 \Rightarrow x = - 1 .$

And minimum value does not exist.

(iii) h(x) = sin (2x) + 5.

A(iii)

we have, h(x) = sin (2x) + 5.

For all $x \in ?, - 1 \leq s i n 2 x \leq 1 .$ {range of sine function is [-1, 1]}

$\Rightarrow$ -1 + 5 $\leq$ sin 2x + 5 $\leq$ 1 + 5.

$\Rightarrow$ 4 $\leq$ h(x) $\leq$ 6.

∴ Maximum value of h(x) = 6.

Minimum value of h(x) = 4.

(iv) $f (x) = | s i n 4 x + 3 | .$

A(iv)

we have, $f (x) = | s i n 4 x + 3 | .$

As for all $x \in ?, - 1 \leq s i n 4 x \leq 1$

$\Rightarrow$ -1 + 3 $\leq$ sin 4x + 3 $\leq$ 1 + 3

$\Rightarrow | 2 | \leq | s i n 4 x + 3 | \leq | 4 | .$

$\Rightarrow$ 2 $\leq$ f(x) $\leq$ 4.

∴ Maximum value of f(x) = 4.

Minimum value of f(x) = 2.

(

...more

(i) we have, f(x) = |x + 2| - 1

We know that, for all $x \in ?, | x + 2 | \geq 0$

$\Rightarrow | x + 2 | - 1 \geq - 1 .$

$\Rightarrow$ f(x) $\geq$ - 1.

∴ Minimum value of f(x) = -1 when x + 2 = 0 $\Rightarrow$ x = - 2.

And maximum value of f(x) does not exist.

(ii) $g (x) = - | x + 1 | + 3$

A(ii)

We have, $g (x) = - | x + 1 | + 3$

For all $x \in, | x + 1 | \geq 0 .$

$\Rightarrow - | x + 1 | \leq 0$

$\Rightarrow - | x + 1 | + 3 \leq 0 + 3$

$\Rightarrow$ g(x) $\leq$ 3.

∴ Maximum value of g(x) = 3 when $| x + 1 | = 0 \Rightarrow x = - 1 .$

And minimum value does not exist.

(iii) h(x) = sin (2x) + 5.

A(iii)

we have, h(x) = sin (2x) + 5.

For all $x \in ?, - 1 \leq s i n 2 x \leq 1 .$ {range of sine function is [-1, 1]}

$\Rightarrow$ -1 + 5 $\leq$ sin 2x + 5 $\leq$ 1 + 5.

$\Rightarrow$ 4 $\leq$ h(x) $\leq$ 6.

∴ Maximum value of h(x) = 6.

Minimum value of h(x) = 4.

(iv) $f (x) = | s i n 4 x + 3 | .$

A(iv)

we have, $f (x) = | s i n 4 x + 3 | .$

As for all $x \in ?, - 1 \leq s i n 4 x \leq 1$

$\Rightarrow$ -1 + 3 $\leq$ sin 4x + 3 $\leq$ 1 + 3

$\Rightarrow | 2 | \leq | s i n 4 x + 3 | \leq | 4 | .$

$\Rightarrow$ 2 $\leq$ f(x) $\leq$ 4.

∴ Maximum value of f(x) = 4.

Minimum value of f(x) = 2.

(v) h(x) = x + 1.,x∈ ( -1, 1).

A(v)

we have, h(x) = x + 1.,x∈ ( -1, 1).

Given, -1

- 1 + 1 <x + 1 < 1 + 1

0 <h (x) < 2.

∴ Maximum value and minimum value of h(x) does not exist.

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Application of Derivatives Class 12th

74.Find the maximum and minimum values, if any, of the following functions

given by.

(i) $f (x) = {(2 x - 1)}^{2} + 3$

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Vishal Baghel

Contributor-Level 10

(i) We have, f(x) = (2x - 1)² + 3.

For all $x \in ?, (2 x - 1)^{2} \geq 0$

(2x - 1)² + 3 ≥ 3.

f(x) ≥ 3.

∴The minimum value of f(x) = 3. When 2x - 1 = 0--> x = $\frac{1}{2}$

Again as $x \to \infty, f (x) \to \infty$ as there is vouppa bound to 'x' value hence, f(x) has no maximum values.

(ii) $f (x) = 9 x^{2} + 12 x + 2$

A(ii)

We have, f(x) = 9² + 12x + 2.

$f (x) = 9 [x^{2} + \frac{12 x}{9} + \frac{2}{9}]$ (Taking 9 common from each team).

$f (x) = 9 [x^{2} + \frac{4 x}{3} + \frac{2}{9}]$

$f (x) = 9 [x^{2} + 2 * \frac{2 x}{3} + {(\frac{2}{3})}^{2} - {(\frac{2}{3})}^{2} + \frac{2}{9}]$

$f (x) = 9 [{(x + \frac{2}{3})}^{2} - \frac{4}{9} + \frac{2}{9}]$

$f (x) = 9 [\cdot {(x + \frac{2}{3})}^{2} - \frac{2}{9}] = 9 {(x + \frac{2}{3})}^{2} - 2$

For all $x \in ?, {(x + \frac{2}{3})}^{2} \geq 0$

$\Rightarrow {(x + \frac{2}{3})}^{2} - 2 \geq \cdot 0 - 2$

$\Rightarrow$ f(x)≥ - 2.

∴The minimum value of f(x) = -2 when $x + \frac{2}{3} = 0$

And as $x \to \infty, f (x) \to \infty$ so f(x) has $\Rightarrow x = - \frac{2}{3} .$

no maximum values.

(iii) f(x) = (x - 1)² + 10

A(iii)

we have, f(x) = - (x - 1)² + 10

For all $x \in ?, (x - 1)^{2} \geq 0 .$

$\Rightarrow$ (x - 1)² ≤ 0

$\Rightarrow$ -(x- 1)² + 10 ≤ 10.

$\Rightarrow$ f(x) ≤ 10.

∴maximum value of f(x) = 10 when x - 1 = 0 $\Rightarrow$ x = 1.

And minimum value of f(x) does n

...more

(i) We have, f(x) = (2x - 1)² + 3.

For all $x \in ?, (2 x - 1)^{2} \geq 0$

(2x - 1)² + 3 ≥ 3.

f(x) ≥ 3.

∴The minimum value of f(x) = 3. When 2x - 1 = 0--> x = $\frac{1}{2}$

Again as $x \to \infty, f (x) \to \infty$ as there is vouppa bound to 'x' value hence, f(x) has no maximum values.

(ii) $f (x) = 9 x^{2} + 12 x + 2$

A(ii)

We have, f(x) = 9² + 12x + 2.

$f (x) = 9 [x^{2} + \frac{12 x}{9} + \frac{2}{9}]$ (Taking 9 common from each team).

$f (x) = 9 [x^{2} + \frac{4 x}{3} + \frac{2}{9}]$

$f (x) = 9 [x^{2} + 2 * \frac{2 x}{3} + {(\frac{2}{3})}^{2} - {(\frac{2}{3})}^{2} + \frac{2}{9}]$

$f (x) = 9 [{(x + \frac{2}{3})}^{2} - \frac{4}{9} + \frac{2}{9}]$

$f (x) = 9 [\cdot {(x + \frac{2}{3})}^{2} - \frac{2}{9}] = 9 {(x + \frac{2}{3})}^{2} - 2$

For all $x \in ?, {(x + \frac{2}{3})}^{2} \geq 0$

$\Rightarrow {(x + \frac{2}{3})}^{2} - 2 \geq \cdot 0 - 2$

$\Rightarrow$ f(x)≥ - 2.

∴The minimum value of f(x) = -2 when $x + \frac{2}{3} = 0$

And as $x \to \infty, f (x) \to \infty$ so f(x) has $\Rightarrow x = - \frac{2}{3} .$

no maximum values.

(iii) f(x) = (x - 1)² + 10

A(iii)

we have, f(x) = - (x - 1)² + 10

For all $x \in ?, (x - 1)^{2} \geq 0 .$

$\Rightarrow$ (x - 1)² ≤ 0

$\Rightarrow$ -(x- 1)² + 10 ≤ 10.

$\Rightarrow$ f(x) ≤ 10.

∴maximum value of f(x) = 10 when x - 1 = 0 $\Rightarrow$ x = 1.

And minimum value of f(x) does not exist.

(iv) g(x) = x³ + 1

A(iv)

we have, g(x) = x³ + 1

For the given fxⁿ, $x \in ?,$

$g (x) \to \infty a s x \to \infty$

and $g (x) - \infty a s x \to - \infty$

Q Maximum and minimum value does not exist.

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4 months ago

Application of Derivatives Class 12th

73. The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is (A) 0.06 x³ m³ (B) 0.6 x³m³ (C) 0.09 x³ m³ (D) 0.9 x³ m³

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Vishal Baghel

Contributor-Level 10

The volume v of a cube with side 'x' metre is v = x³

So, $d v = (\frac{d v}{d x}) Δ x = 3 x^{2} Δ x .$

∴increase in side, Δx = 3% of = $\frac{3 x}{100} .$

∴dv = 3x²π $\frac{3 x}{100} - 0.09 x^{3} m^{3} .$

Hence, option (C) is correct.

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New answer posted

4 months ago

Application of Derivatives Class 12th

72. If f(x) = 3x² + 15x + 5, then the approximate value of f (3.02) is (A) 47.66 (B) 57.66 (C) 67.66 (D) 77.66

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Vishal Baghel

Contributor-Level 10

We have, y = f (x) = 3x² + 15x + 5.

$\frac{d y}{d x} = 6 x + 15$

$\Rightarrow$ dy = (6x + 15) dx

$\Rightarrow$ Δy = (6x + 15) Δx.

Let, x = 3 and Δx = 0.02 then,

Δy = f (x + Δx) - f (x)

$\Rightarrow$ f (x + Δx) = f (x) + Δy = f (x) + (6x + 5) Δx.

$\Rightarrow$ f (3 + 0.02) = 3 (3)² + 15 (3) + 5 + (6 * 3 + 15) (0.02).

$\Rightarrow$ f (3.02) = 27 + 45 + 5 + (18 + 15) (0.02).

= 77 + 0.66

= 77.66

∴ Option (D) is correct.

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