## Rajasthan Board RBSE Class 12 Maths Chapter 8 Application of Derivatives Ex 8.5

Question 1.

Find maximum and minimum values of following function :

(a) 2x^{3} – 15x^{2} + 36x + 10

(b) (x – 1)(x – 2)(x – 3)

(c) sin x + cos 2x

(d) x^{4} – 5x^{4} + 5x^{3} – 1

Solution:

(a) Let y = 2x^{3} – 15x^{2} + 36x + 10

⇒ \(\frac { dy }{ dx } \) = 6x^{2} – 30x + 36

For maxima or minima,

\(\frac { dy }{ dx } \) = 0

⇒ 6x^{2} – 30x + 36 = 0

⇒ 6(x^{2} – 5x + 6) = 0

⇒ (x – 3) (x – 2) = 0

so, x = 2,3

At x = 2, maximum value of function

= 2(2)^{3} – 15(2)^{2} + 36 × 2 +10

= 16 – 60 + 72 + 10

= 38

(b) (x – 1) (x – 2) (x – 3)

Let y = (x – 1) (x – 2) (x – 3)

Then, = x^{3} – 6x^{2} + 11x – 6

(c) Let y = sin x + cos 2x ……(i)

Diff. w.r.t. x,

\(\frac { dy }{ dx } \) = cos x – 2 sin 2x …..(ii)

For maximum or minimum value \(\frac { dy }{ dx } \) = 0

or cos x – 2 sin 2x = 0

⇒ cos x – 4 sin x cos x = 0

⇒ cosx (1 -4 sinx) = 0

(d) Let y = x^{5} – 5x^{4} + 5x^{3} – 1

\(\frac { dy }{ dx } \) = 5x^{4} – 20x^{3} + 15x^{2}

⇒ \(\frac { dy }{ dx } \) 5x^{2} (x^{2} – 4x + 3)

5x^{2}(x^{2} – 3x – x + 3)

= 5x^{2 }[x(x – 3) -1 (x – 3)]

= 5x^{2 }(x – 3)(x – 1)

For maxima or minima

Question 2.

Find maximum and minimum values of following function, if exists :

(a) – | x + 1 | + 3

(b) | x + 2 | – 1

(c) | sin 4x + 3 |

(d) sin 2x + 5

Solution:

(a) Let g(x) = – | x + 1 | + 3

– |x + 1 | < 0, So function has no minimum value

Again minimum value of – | x + 1 | is 0

∴ -| x + 1 | = 0 ⇒ x = -1 then maximum value of g(x) is 3

[∵ g(-1) = – | – 1 + 1 | + 3 = 3]

(b) Let f(x) = | x + 2 | – 1

|x + 2 | > 0, So, f(x) does not have any maximum value.

Again, minimum value of | x + 2 | = 0

∴ | x + 2 | = 0

⇒ x = -2, then minimum value of f(x) is – 1

(∴ f (- 2) = | -2 + 2 | – 1 = 0 – 1 = – 1)

(c) Let f(x) = | sin 4x + 3 |

sin 4x has maximum value as 1.

∴ Maximum value of f(x) is | -1 + 3 | = 4

Again, minimum value of sin 4x

Minimum value of f(x) | -1 + 3 | = | 2 | = 2

(d) Let h(x) = sin 2x + 5

sin (2x) has maximum value as 1

So, h(x) has maximum value as 1 + 5 = 6

Again, sin (2x) has minimum value as – 1

So, minimum value of h (x) will be -1 + 5 = 4

Question 3.

Find maximum and minimum values of following functions in given interval:

(a) 2x^{3} – 24x + 107, x ∈ [1,3] .

(b) 3x^{4} – 2x^{3} – 6x^{2} + 6x + 1, x ∈ [0, 2]

(c) x + sin 2x, x ∈ [0, 2π]

(d) x^{3} – 18x^{2} + 96x, x ∈ [0, 9]

Solution:

(a) Let y = 2x^{3} – 24x + 107, x ∈ [1,3]

\(\frac { dy }{ dx } \) = 6x^{2} – 24

For maxima or minima

\(\frac { dy }{ dx } \) = 0 ⇒ 6x^{2} – 24 = 0

⇒ x = ± 2.

∵ x ∈ [1, 3]

∴ x = 2

Now, y_{1} = 2(1)^{3} – 24(1) + 107

= 2 – 24 + 107 = 85

y_{2} = 2(2)^{3} – 24(2) +107

= 16 – 48 + 107 = 75

y_{3} = 2(3)^{3} – 24(3) + 107

= 54 – 72 + 107 = 89

Hence, maximum value is 89 at x = 3 where as minimum value is 75 at x = 2.

(b) Let y = 3x^{4} – 2x^{3} – 6x^{2} + 6x + 1,

x ∈ (0, 2)

\(\frac { dy }{ dx } \) = 12x^{3} – 6x^{2} – 12x + 6

= 6(2x^{3} – x^{2} – 2x + 1)

For maxima or minima

Now y_{1} = 3(1)^{4} – 2(1)^{3} – 6(1)^{2} + 6(1) + 1

⇒ y_{1} = 3 – 2 – 6 + 6 + 1 = 2

⇒ y_{-1} = 3 (- 1)^{4} – 2 (-1)^{3} – 6(- 1)^{2} + 6(- 1) + 1

⇒ y_{-1} = 3 + 2 – 6 – 6 + 1 = -6

⇒ y_{0} = 3(0)^{4} – 2(0)^{3} – 6(0)^{2} + 6(0) + 1

⇒ y_{0} = 1

⇒ y_{2} = 3(2)^{4} – 2(2)^{3} – 6(2)^{2} + 6(2) + 1

= 48 – 16 – 24 + 12 + 1 = 21

∴ Hence, maximum value of function is 2π at x = 2π and minimum value of function is 0 at x = 0.

y_{4} = (0)^{3} – 18(0)^{2} + 96(0)

= 64 – 288 + 384 = 160

y_{8} = (8)^{3} – 18(8)^{2} + 96 × 8

= 512 – 1152 + 768

= 128

y_{9} = (9)^{3} – 18(9)^{2} + 96(9)

= 729 – 1458 + 864= 135

So, at x = 0 minimum value = 0

and at x = 4 maximum value = 160

Question 4.

Find extreme value of following functions :

(a) sin x cos 2x

(b) a sec x + b cosec x, 0 < a < b

(c) (x)^{1/x}, x > 0

(d) \(\frac { 1 }{ x } \) log x, x ∈ (0,α)

Solution:

(a) Let

y = sin x.cos 2x …..(i)

\(\frac { dy }{ dx } \) = sin x.(- sin 2x) + cos 2x.cos x

= – 4 sin^{2} x.cos x + (1 – 2 sin^{2} x).cos x

= – 4 sin^{2} x cos x + cos x – 2 sin^{2} x cos x

= cos x (1 – 6 sin^{2} x) …..(ii)

So, at x = \(\frac { 3\pi }{ 2 } \) function will be maxima and maximum value of function

Hence, on putting x = e in given function we get maximum value of function which is (e)^{1/e}.

Question 5.

Prove that value of \(\frac { x }{ 1+xtanx } \) is maximam at x = cos x.

Solution:

Question 6.

Prove that value of sin^{2} x (1 + cosx) is maximum x = \(\frac { 1 }{ 2 } \)

Solution:

Let y = sin^{2 }x.(1 + cosx) …..(1)

Diff.wrt.x,

⇒ \(\frac { dy }{ dx } \) = sin^{2} x(- sin x) + (1 + cos x)

⇒ \(\frac { dy }{ dx } \) = sinx (-sin^{2} x + 2 cos x + 2 cos^{2} x) …..(2)

For maxima of minima \(\frac { dy }{ dx } \) = 0

Hence, at cos x = \(\frac { 1 }{ 3 } \) function has maximum value.

Question 7.

Prove that ,y = sinP θ cosq θ is maximum at tan θ = \(\sqrt { \frac { p }{ q } } \)

Solution:

+ (p cot θ – q tan θ) × 0

= -y {(q + p) + (p + q)} + 0

= -2y (p + q)

= -2 (sin^{p }θ cos^{q }θ) (p + q)

= -ve

So, y will be maximum

Hence, y will be of maximum value if

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