12th NCERT Application of Derivatives Exercise 6.5 Questions 1 to 12
Do or do not
There is no try

Question (1)

Find the maximum and minimum value, if any, of the following functions given by
(i) f(x) = (2x - 1)2 + 3
(ii) f(x) = 9x2 + 12x + 2
(iii) f(x) = -(x - 1)2 + 10
(iv) g(x) = x3 + 1

Solution

(i) f(x) = (2x - 1)2 + 3
For all x, (2x-1)2 ≥ 0
minimum value of f(x) = 0 + 3 = 3
(ii) f(x) = 9x2 + 12x + 2
f(x) = (9x2 + 12x + 4)-2
f(x) = (3x + 2)2 - 2
∀ x ∈ R (3x + 2)2 ≥ 0
∴ minimum value of f(x) = 0-2 = -2
(iii) f(x) = -(x-a)2 + 10
∀ x ∈ R (x-a)2 ≥ 0
∴ maximum value of f(x) = 0+10=10
(iv) g(x) = x3 + 1
∀ x ∈ R; x3 ∈ R
No, it will not have minimum or maximum value

Question (2)

Find the maximum and minimum value , if any, of the following function given by
(i) f(x) = |x + 2| -1
(ii) g(x) = -|x+1| + 3
(iii) h(x) = sin(2x) + 5
(iv) f(x) = |sin4x + 3|
(v) h(x) = x + 1, x ∈ (-1, 1)

Solution

(i) f(x) = |x + 2| -1
|x+2| ≥ 0 ⇒ minimum value of |x+2| = 0
∴ minimum of f(x) = 0 - 1 = -1
No maximum value

(ii) g(x) = -|x+1| + 3
|x+1| ≥ 0 ⇒ maximum of - |x+1| = 0
∴ maximum of g(x) = 0+3 =3
No minimum value

(iii) h(x) = sin(2x) + 5
Range of sin is [-1, 1]
sin2x ∈ [ -1, 1]
minimum value of sin2x = -1 ⇒ h(x) = -1+5 = 4
maximum value of sin2x = 1 ⇒ maximum of h(x) = 1+5 = 6

(iv) f(x) = |sin4x + 3|
Range of sin4x is [-1, 1]
-1 ≤ sin4x ≤ 1
-1 + 3 ≤ sin4x+3 ≤ 1 + 3
2 ≤ sin4x+3 ≤ 4
sin4x+3 ∈ [2, 4]
|sin4x+3| ∈ [2, 4]
minimum value = 2
maximum value =4

(v) h(x) = x + 1, x ∈ (-1, 1)
-1 < x < 1
-1+1 < x+1 < 1+1
0 < x+1 < 2
(x+1) ∈ (0, 2)
No maximum and no minimum

Question (3)

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
(ii) g(x) = x3 - 3x
(iii) h(x) = sinx + cos x, 0< x < (π/2)
(iv) f(x) = sin x - cos x, 0< x < 2π
(v) f(x) = x3 - 6x2 + 9x + 15
$(vi)\quad g\left( x \right) = \frac{x}{2} + \frac{2}{x}$ $(vii)\quad g\left( x \right) = \frac{1}{{{x^2} + 2}}$ (viii) f(x) = x√(1-x) , x > 0

Solution

(i) f(x) = x2
f'(x) = 2x
f"(x) = 2
Now f'(x) = 0
⇒ 2x =0 ⇒ x = 0
f"(0) = 2 > 0 ⇒ f is minimum at x = 0
f(0) = 02 = 0
⇒ local minimum value 0 at x = 0

(ii) g(x) = x3 - 3x
g'(x) = 3x2 - 3
g"(x) = 6x
Now g'(x) = 0
3x2 - 3 = 0
3(x2 -1) = 0
(x+1) (x-1) = 0
x = 1 or x = -1

g"(-1) = 6(-1) = -6 < 0
⇒ g is maximum at x = -1
g(-1) = x3 -3x = (-1)3 - 3(-1) = 2
g"(1) = 6(1) = 6 > 0
⇒ g is minimum at x = 1
g(1) = 13 - 3(1) = 1-3=-2
maximum value of g(x) = 2 at x = -1
minimum value of g(x) = -2 at x = 1

(iii) h(x) = sinx+cosx 0 < x < π/2
h'(x) = cosx + (-sinx) = cos x-sinx
h"(x) = (-sinx) - (cosx) = -sinx - cos x
Now h'(x) = 0
cosx - sinx = 0
sinx = cosx
tanx = 1
∴ x = tan-1(1) = π/4 as x ∈ (0, π/2)
$h"\left( {\frac{\pi }{4}} \right) = - \sin \left( {\frac{\pi }{4}} \right) - \cos \left( {\frac{\pi }{4}} \right)$ $h"\left( {\frac{\pi }{4}} \right) = - \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}$ $h"\left( {\frac{\pi }{4}} \right) = - \frac{2}{{\sqrt 2 }} = - \sqrt 2 < 0$ ⇒ h(x) is maximum at x = π/4
$h\left( {\frac{\pi }{4}} \right) = \sin \frac{\pi }{4} + \cos \frac{\pi }{4}$ $h\left( {\frac{\pi }{4}} \right) = \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} = \frac{2}{{\sqrt 2 }} = \sqrt 2$ local maximum value = √2 at x = π/4

(iv) f(x) = sinx - cosx; 0 < x < 2π
f'(x) = cos x - (-sinx)
f'(x) = cos x + sin x
f"(x) = -sinx + cosx
Now f'(x) = 0
cosx + sinx = 0
sinx = -cosx
tanx = -1
$x = \frac{3}{4}\pi$ $or$ $x = - \frac{\pi }{4}or\frac{{7\pi }}{4}$ $f"\left( {\frac{{3\pi }}{4}} \right) = - \sin \frac{{3\pi }}{4} + \cos \frac{{3\pi }}{4}$ $f"\left( {\frac{{3\pi }}{4}} \right) = - \sin \left( {\pi - \frac{\pi }{4}} \right) + \cos \left( {\pi - \frac{\pi }{4}} \right)$ $f"\left( {\frac{{3\pi }}{4}} \right) = - \sin \left( {\frac{\pi }{4}} \right) - \cos \left( {\frac{\pi }{4}} \right)$ $f"\left( {\frac{{3\pi }}{4}} \right) = - \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} = - \sqrt 2 < 0$ ⇒ f(x) has maximum value at x = 3π/4 $f\left( {\frac{{3\pi }}{4}} \right) = \sin \frac{{3\pi }}{4} - \cos \frac{{3\pi }}{4}$ $f\left( {\frac{{3\pi }}{4}} \right) = \frac{1}{{\sqrt 2 }} - \left( { - \frac{1}{{\sqrt 2 }}} \right) = \sqrt 2$ $f"\left( {\frac{{ - \pi }}{4}} \right) = - \sin \left( {\frac{{ - \pi }}{4}} \right) + \cos \left( {\frac{{ - \pi }}{4}} \right)$ $f"\left( {\frac{{ - \pi }}{4}} \right) = \sin \left( {\frac{\pi }{4}} \right) + \cos \left( {\frac{\pi }{4}} \right)$ $f"\left( {\frac{{ - \pi }}{4}} \right) = \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} > 0$ ⇒ f(x) is minimum at x= -π/4 $f\left( {\frac{{ - \pi }}{4}} \right) = \sin \left( {\frac{{ - \pi }}{4}} \right) - \cos \left( {\frac{{ - \pi }}{4}} \right)$ $f\left( {\frac{{ - \pi }}{4}} \right) = - \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} = - \sqrt 2$ ∴ local minimum value = -√2 at x = -π/4 = 7π/4
local maxima value = √2 at x = 3π/4

(v) f(x) = x3 - 6x2 + 9x + 15
f'(x) = 3x2 -12x + 9
f"(x) = 6x - 12
Now f'(x) = 0
⇒ 3x2 -12x + 9 = 0
3(x2 -4x + 3) = 0
(x - 3) ( x - 1) = 0
x = 1 or x = 3
f"(1) = 6(1) -12 = -6 < 0
⇒ f has maximum value at x = 1
f(1) = 13 - 6(1)2 + 9(1) + 15
f(1) = 1 - 6 + 9 + 15 = 19
f"(3) = 6(3) - 12 = 6 > 0
⇒ f is minimum at x = 3
f(3) = (3)3 - 6(3)2 + 9(3) + 15
f(3) = 27 -54 + 27 +15 = 15
local minimum value = 15 at x = 3
local maximum value =19 at x = 1

$(vi)\quad g\left( x \right) = \frac{x}{2} + \frac{2}{x}$ $g'\left( x \right) = \frac{1}{2} + 2\left( {\frac{{ - 1}}{{{x^2}}}} \right) = \frac{1}{2} - \frac{2}{{{x^2}}}$ $g"\left( x \right) = - 2\left( {\frac{{ - 2}}{{{x^3}}}} \right) = \frac{4}{{{x^3}}}$ Now g'(x) = 0 $\frac{1}{2} - \frac{2}{{{x^2}}} = 0$ x2 - 4 = 0
(x + 2) (x - 2) = 0
x = -2 OR x = 2
as x > 0
x = -2 not possible
$g"(2) = \frac{4}{{{2^3}}} = \frac{4}{8} > 0$ ⇒ g(x) is minimum at x = 2
$g(2) = \frac{2}{2} + \frac{2}{2} = 2$ local minimum value 2 at x = 2
local maximum value not possible as x > 0

$(vii)\quad g\left( x \right) = \frac{1}{{{x^2} + 2}}$ ${g'}\left( x \right) = \frac{1}{{{{\left( {{x^2} + 2} \right)}^2}}} \times \left( { - 2x} \right)$ $g"\left( x \right) = \frac{{{{\left( {{x^2} + 2} \right)}^2}\frac{d}{{dx}}\left( { - 2x} \right) - \left( { - 2x} \right)\frac{d}{{dx}}{{\left( {{x^2} + 2} \right)}^2}}}{{{{\left( {{x^2} + 2} \right)}^4}}}$ $g"\left( x \right) = \frac{{{{\left( {{x^2} + 2} \right)}^2}\left( { - 2} \right) + 2x\left( 2 \right)\left( {{x^2} + 2} \right)\left( {2x} \right)}}{{{{\left( {{x^2} + 2} \right)}^4}}}$ $g"\left( x \right) = \frac{{\left( {{x^2} + 2} \right)\left[ { - 2\left( {{x^2} + 2} \right) + 8x} \right]}}{{{{\left( {{x^2} + 2} \right)}^4}}}$ $g"\left( x \right) = \frac{{ - 2{x^2} - 4 + 8x}}{{{{\left( {{x^2} + 2} \right)}^3}}}$ Now g'(x) = 0
$\Rightarrow \frac{{ - 2x}}{{{{\left( {{x^2} + 2} \right)}^2}}} = 0$ -2 x = 0 , x = 0.
$g"\left( 0 \right) = \frac{{ - 0 - 4 + 0}}{{{{\left( 2 \right)}^3}}} = \frac{{ - 4}}{8} < 0$ ⇒ g(x) is maximum at x = 0
$g\left( 0 \right) = \frac{1}{{0 + 2}} = \frac{1}{2}$ ∴ local maximum value = ½ at x = 0

(viii) f(x) = x√(1-x) , x > 0 $f'\left( x \right) = x\frac{d}{{dx}}\sqrt {1 - x} + \sqrt {1 - x} \frac{d}{{dx}}x$ $f'\left( x \right) = x\frac{1}{{2\sqrt {1 - x} }} \times \left( { - 1} \right) + \sqrt {1 - x} \left( 1 \right)$ $f'\left( x \right) = \frac{{ - x}}{{2\sqrt {1 - x} }} + \sqrt {1 - x}$ $f'\left( x \right) = \frac{{ - x + 2\left( {1 - x} \right)}}{{2\sqrt {1 - x} }}$ $f'\left( x \right) = \frac{{2 - 3x}}{{2\sqrt {1 - x} }}$ $f"\left( x \right) = \frac{1}{2}\left[ {\frac{{\sqrt {1 - x} \frac{d}{{dx}}\left( {2 - 3x} \right) - \left( {2 - 3x} \right)\frac{d}{{dx}}\sqrt {1 - x} }}{{{{\left( {\sqrt {1 - x} } \right)}^2}}}} \right]$ $f"\left( x \right) = \frac{1}{{2\left( {1 - x} \right)}}\left[ {\left( {\sqrt {1 - x} } \right)\left( { - 3} \right) - \left( {2 - 3x} \right)\frac{1}{{2\sqrt {1 - x} }}\left( { - 1} \right)} \right]$ $f"\left( x \right) = \frac{1}{{2\left( {1 - x} \right)}}\left[ {\frac{{ - 6\left( {1 - x} \right) + 2 - 3x}}{{2\sqrt {1 - x} }}} \right]$ $f"\left( x \right) = \frac{1}{{4{{\left( {1 - x} \right)}^{\frac{3}{2}}}}}\left[ { - 6 + 6x + 2 - 3x} \right]$ $f"\left( x \right) = \frac{{3x - 4}}{{4{{\left( {1 - x} \right)}^{\frac{3}{2}}}}}$ Now f'(x) = 0
$\frac{{2 - 3x}}{{2\sqrt {1 - x} }} = 0$ 2 - 3x = 0
x = 2/3 $f"\left( {\frac{2}{3}} \right) = \frac{{\require{cancel} \cancel{3}^1\left( {\frac{2}{\cancel{3}}} \right) - 4}}{{4{{\left( {1 - \frac{2}{3}} \right)}^{\frac{3}{2}}}}}$ $f"\left( {\frac{2}{3}} \right) = \frac{{2 - 4}}{{4{{\left( {\frac{1}{3}} \right)}^{\frac{3}{2}}}}} < 0$ ⇒ f(x) is maximum at x = 2/3
$f\left( {\frac{2}{3}} \right) = \frac{2}{3}\sqrt {1 - \frac{2}{3}}$ $f\left( {\frac{2}{3}} \right) = \frac{2}{3}\sqrt {\frac{1}{3}} = \frac{2}{{3\sqrt 3 }}$ ∴ local maximum value = 2/(3√3) at x = 2/3

Question (4)

Prove that the following functions do not have maxima or minima:
(i) f(x) = ex
(ii) g(x) = log x
(iii) h(x) = x3 + x2 + x +1

Solution

(i) f(x) = ex
f'(x) = ex and f"(x) = ex
Now f'(x) = 0
ex = 0
it is not possible for any value of x
so it does not have max or minimum value

(ii) g(x) = log x
$g'\left( x \right) = \frac{1}{x}$ $g"\left( x \right) = \frac{{ - 1}}{{{x^2}}}$ Now g'(x) = 0
(1/x) = 0
not possible
so no value of x for which it have minimum or minimum

(iii) h(x) = x3 + x2 + x + 1
h'(x) = 3x2 + 2x +1
h"(x) = 6x + 2
Now h'(x) = 0
3x2 + 2x + 1 = 0
Comparing to standard quadratic equation ax2 bx + c = 0
we get a = 3, b = 2, c = 1
b2 - 4ac = 22 - 4(3)(1) = -8 < 0
no real roots
so no value of x

Question (5)

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]
(ii) f(x) = sin x + cos x, x ∈ [0, π]
$(iii)f\left( x \right) = 4x - \frac{1}{2}{x^2},x \in \left[ { - 2,\frac{9}{2}} \right]$
(iv) f(x) = (x-1)2 + 3, x ∈ [-3, 1]

Solution

(i) f(x) = x3, x ∈ [-2, 2]
f'(x) = 3x2
f"(x) = 6x
Now f'(x) = 0
3x2 = 0
⇒ x = 0
f"(0) = 0 ⇒ no local maximum or minimum
f(-2) = (-2)3 = -8
f(2) = 23 = 8
Absolute max = max{-8, 8} = 8
Absolute minimum = min{-8, 8} = -8

(ii) f(x) = sin x + cos x, x ∈ [0, π]
f'(x) = cosx - sinx
f"(x) = -sinx - cosx
Now f'(x) = 0
cos x - sin x = 0 ⇒ sin x = cos x
⇒ tan x = 1
⇒ x = tan-1(1) = π/4
$f"\left( {\frac{\pi }{4}} \right) = - \sin \frac{\pi }{4} - \cos \frac{\pi }{4}$ $f"\left( {\frac{\pi }{4}} \right) = - \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} < 0$ ⇒ f is maximum at x = π/4 $f\left( {\frac{\pi }{4}} \right) = \sin \frac{\pi }{4} + \cos \frac{\pi }{4}$ $f\left( {\frac{\pi }{4}} \right) = \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} = \sqrt 2$ f(0) = sin0+cos0 = 0+1 = 1
f(π) = sinπ + cosπ = 0 +(-1) = -1
Absolute min = min{-1, 1, √2} = -1 at x=π
Absolute max = max{-1, 1, √2} = √2 at x=π/4

$(iii)f\left( x \right) = 4x - \frac{1}{2}{x^2},x \in \left[ { - 2,\frac{9}{2}} \right]$ $f'\left( x \right) = 4 - \frac{1}{\cancel{2}} \cdot \cancel{2}x$ f'(x) = 4 -x
f"(x) = -1
f'(x) = 0
4-x = 0
x = 4
f is maximum at x=4
f(4) = 4(4) - ½(4)2 = 8
local maximum value = 8 at x = 4
f(-2) = 4(-2) - ½ (-2)2
f(-2) = -8-2 = -10
$f\left( {\frac{9}{2}} \right) = 4\left( {\frac{9}{2}} \right) - \frac{1}{2}{\left( {\frac{9}{2}} \right)^2}$ $f\left( {\frac{9}{2}} \right) = 18 - \frac{{81}}{8}$ $f\left( {\frac{9}{2}} \right) = 18 - 10.125 = 7.875$ Absolute max. = max{-10, 8, 7.875} = 8
Absolute min. = min { -10, 8. 7.875} = -10

(iv) f(x) = (x-1)2 + 3, x ∈ [-3, 1]
f'(x) = 2(x-1)
f"(x) = 2
Now f'(x) = 0
2(x - 1) = 0
⇒ x = 1
f"(1) = 2 > 0
∴ f is minimum at x = 1
f(1) = (1 - 1)2 + 2 = 0 + 3 = 3
local minimum = 3 at x = 1
f(-3) = (-3-1)2 + 3 = 19
Absolute max. = max{19, 3} = 19
Absolute min. = min { 19, 3} = 3

Question (6)

Find the maximum profit that a company can make, if the profit function is given by
p(x) = 41 - 24x - 18x2

Solution

$p\left( x \right) = 41 - 24x - 18{x^2}$ $p'\left( x \right) = - 24 - 36x$ $p"\left( x \right) = - 36$ To calculate maximum profit
$p'\left( x \right) = 0$ $- 24 - 36x = 0$ $x = \frac{{ - 24}}{{36}} = \frac{{ - 2}}{3}$ $p'\left( {\frac{{ - 2}}{3}} \right) = - 36 < 0$ Profit is maximum at $x = \frac{{ - 2}}{3}$
$p\left( {\frac{{ - 2}}{3}} \right) = 41 - \cancel{24}^8\left( {\frac{{ - 2}}{\cancel{3}}} \right) - 18{\left( {\frac{2}{3}} \right)^2}$ $p\left( {\frac{{ - 2}}{3}} \right) = 41 + 16 - \cancel{18}^2 \times \frac{4}{\cancel{9}}$ $p\left( {\frac{{ - 2}}{3}} \right) = 57 - 8 = 49$ ∴ maximum profit is 49

Question (7)

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

Solution

f'(x) = 12x3 -24x2 +24x - 48
f"(x) = 36x2 - 48x + 24
Now f'(x) = 0
⇒ 12x3 -24x2 +24x - 48 = 0
12[x3 -2x2 + 2x - 4] = 0
x2 ( x - 2) + 2 (x-2) = 0
(x2+2) ( x-2) = 0
x2 +2 = 0 OR x - 2 = 0
x2 = -2 is not possible
∴ x = 2
Now f"(2) = 36(4) - 48(2) +24
f"(2) = 144 - 96 +24
f"(2) = 72 > 0
⇒ f(x) is minimum at x=2
f(2) = 3(2)4 -8(2)3 +12 (2)2 - 48(2) +25
f(2) = 48 - 64 +48 -96 +25
f(2) = -39
Local minimum value = -39 at x = 2
f(0)= 3(0)4 - 8(0)3 +12(0)2 - 48(0) +25
f(0) = 25
f(3) = 3(3)4 - 8(3)3 +12(3)2 - 48(3) +25
f(3) = 243 - 216 +108 -144 +25
f(3) = 376 - 360 = 16
Absolute max = max{-39, 25, 16} = 25 at x = 0
Absolute min = min {-39, 16, 25} = -39 at x = 2

Question (8)

At what points in the interval [0, 2π], does the function sin2x attain its maximum value?

Solution

f'(x) = cos2x×2
f" (x) = 2[-sin2x](2)
f"(x) = -4sin2x
Now f'(x) = 0
2cos2x = 0
cos2x = 0
$\Rightarrow 2x = \frac{\pi }{2},\frac{{3\pi }}{2},\frac{{5\pi }}{2},\frac{{7\pi }}{2}$ $\Rightarrow x = \frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4},\frac{{7\pi }}{4}$ $f"\left( {\frac{\pi }{4}} \right) = - 4\sin \left( {2 \cdot \frac{\pi }{4}} \right) = - 4\sin \left( {\frac{\pi }{2}} \right) = - 4 <0$ ⇒ f is maximum at x = π/4 $f"\left( {\frac{{3\pi }}{4}} \right) = - 4\sin \left( {2 \cdot \frac{{3\pi }}{4}} \right) = - 4\sin \left( {\frac{{3\pi }}{2}} \right) = - 4\left( { - 1} \right) = 4 > 0$ $f"\left( {\frac{{5\pi }}{4}} \right) = - 4\sin \left( {2 \cdot \frac{{5\pi }}{4}} \right) = - 4\sin \left( {\frac{{5\pi }}{2}} \right) = - 4\left( 1 \right) = - 4 < 0$ $f"\left( {\frac{{7\pi }}{4}} \right) = - 4\sin \left( {2 \cdot \frac{{7\pi }}{4}} \right) = - 4\sin \left( {\frac{{7\pi }}{2}} \right) = - 4\left( { - 1} \right) = 4 > 0$ ⇒ f is maximum at x = π/4 and 5π/4

Question (9)

What is the maximum value of the function sinx + cos x?

Solution

f(x) = sinx + cos x
f'(x) = cosx - sinx
f"(x) = -sinx - cosx
Now f'(x) = 0
⇒ cos x - sin x = 0
tan x = 1
x = tan-1 1
x = π/4, 5π/4
$f"\left( {\frac{\pi }{4}} \right) = - \sin \frac{\pi }{4} - \cos \frac{\pi }{4}$ $f\left( {\frac{\pi }{4}} \right) = - \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} = - \sqrt 2 < 0$ ⇒ f is max. at x = π/4
$f\left( {\frac{\pi }{4}} \right) = \sin \frac{\pi }{4} + \cos \frac{\pi }{4}$ $f\left( {\frac{\pi }{4}} \right) = \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} = \sqrt 2$ $f"\left( {\frac{{5\pi }}{4}} \right) = - \cos \left( {\frac{{5\pi }}{4}} \right) - \sin \left( {\frac{{5\pi }}{4}} \right)$ $f"\left( {\frac{{5\pi }}{4}} \right) = - \cos \left( {\pi + \frac{\pi }{4}} \right) - \sin \left( {\pi + \frac{\pi }{4}} \right)$ $f"\left( {\frac{{5\pi }}{4}} \right) = \cos \left( {\frac{\pi }{4}} \right) + \sin \left( {\frac{\pi }{4}} \right)$ $f\left( {\frac{{5\pi }}{4}} \right) = \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} = \sqrt 2 > 0$ ⇒ f is minimum at x = 5π/4
So maximum value of f is √2 at x = π/4

Question (10)

Find the maximum value of 2x3 - 24x + 107 in the interval [1, 3].
Find the maximum value of the same function in [-3, -1]

Solution

f(x) = 2x3 - 24x + 107
f'(x) = 6x2 - 24
f"(x) = 12x
Now f'(x) = 0
6x2 - 24 = 0
6(x2 - 4) = 0
(x+2)(x-2) = 0
x = -2 or x = +2
but x ∈ [1, 3] so x = -2 not possible
∴ x = 2
f"(2) = 24 > 0
so f is minimum at x = 2
f(2) = 2(2)3 -24(2) + 107
f(2) = 16 -48 +107 = 75
f(1) = 2 - 24 +107 = 85
f(3) = 54 - 72 + 107 = 89
max value = max {75, 89, 79} = 89 at x = 3
For [-3, -1]
since x ∈ [-3, -1], x = 2 is not possible
then x = -2
f"(-2) = -24 >0
f is maximum at x = -2
f(-2) = -16 + 48 +107
f(-2) = 139
f(-3) = 2(-3)3 - 24(-3) +107 =125
f(-1) = 2(-1)3 - 24(-) +107 =129
max. value = max {125, 129, 139} = 139 at x = -2

Question (11)

It is given that at x=1, the function x4 -62x2 +ax + 9 attend maximum value, on the interval [0, 1]. Find the value of a

Solution

f(x) = x4 -62x2 +ax + 9 f'(x) = 4x3 - 124x + a
f"(x) = 12x2 - 124
Now f'(1) = 0
4(1)3 -124(1) + a = 0
4 - 124 + a = 0
a = 120

Question (12)

Find the maximum and minimum value of x + sin 2x on [0, 2π]

Solution

f(x) = x + sin 2x
f'(x) = 1 + cos2x(2)
f'(x) = 1 + 2cos2x
f"(x) = -2sin2x (2) = -4sin2x
Now f'(x) = 0
1 +2cos2x = 0
cos2x = -1/2
$2x = 2k\pi \pm {\cos ^{ - 1}}\left( {\frac{{ - 1}}{2}} \right)$ $2x = 2k\pi \pm \left[ {\pi - {{\cos }^{ - 1}}\left( {\frac{{ 1}}{2}} \right)} \right]$ $2x = 2k\pi \pm \left[ {\pi - \frac{\pi }{3}} \right]$ $2x = 2k\pi \pm \frac{{2\pi }}{3}$ $x = k\pi \pm \frac{\pi }{3}$ If x = π/3
$f"\left( {\frac{\pi }{3}} \right) = - 4\sin \left( {\frac{{2\pi }}{3}} \right) = - 4 \times \frac{{\sqrt 3 }}{2} = - 2\sqrt 3 < 0$ ⇒ f is maximum at x = π/3 $f\left( {\frac{\pi }{3}} \right) = \frac{\pi }{3} + \sin \left( {\frac{{2\pi }}{3}} \right)$ $f\left( {\frac{\pi }{3}} \right) = \frac{\pi }{3} + \frac{{\sqrt 3 }}{2}$ $f"\left( {\frac{{2\pi }}{3}} \right) = - 4\sin \left( {\frac{{4\pi }}{3}} \right) = - 4 \times \frac{{ - \sqrt 3 }}{2} - 2\sqrt 3 > 0$ ⇒ f is minimum at x = 2π/3 $f\left( {\frac{{2\pi }}{3}} \right) = \frac{{2\pi }}{3} + \sin \left( {\frac{{4\pi }}{3}} \right)$ $f\left( {\frac{{2\pi }}{3}} \right) = \frac{{2\pi }}{3} - \frac{{\sqrt 3 }}{2}$ f(0) = 0+sin2(0) = 0+0=0
f(2π) = 2π+sin(4π) = 2π + 0 = 2π
$\text{max value = max}\left\{ {0,2\pi ,\frac{\pi }{3} + \frac{{\sqrt 3 }}{2},\frac{{2\pi }}{3} - \frac{{\sqrt 3 }}{2}} \right\} = 2\pi$ $\text{min value = min}\left\{ {0,2\pi ,\frac{\pi }{3} + \frac{{\sqrt 3 }}{2},\frac{{2\pi }}{3} - \frac{{\sqrt 3 }}{2}} \right\} = 0$