RBSE Solutions for Class 12 Maths Chapter 8 Application of Derivatives Ex 8.5

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) 2x³ – 15x² + 36x + 10
(b) (x – 1)(x – 2)(x – 3)
(c) sin x + cos 2x
(d) x⁴ – 5x⁴ + 5x³ – 1
Solution:
(a) Let y = 2x³ – 15x² + 36x + 10
⇒ \(\frac { dy }{ dx } \) = 6x² – 30x + 36
For maxima or minima,
\(\frac { dy }{ dx } \) = 0
⇒ 6x² – 30x + 36 = 0
⇒ 6(x² – 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, = x³ – 6x² + 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 = x⁵ – 5x⁴ + 5x³ – 1
\(\frac { dy }{ dx } \) = 5x⁴ – 20x³ + 15x²
⇒ \(\frac { dy }{ dx } \) 5x² (x² – 4x + 3)
5x²(x² – 3x – x + 3)
= 5x² [x(x – 3) -1 (x – 3)]
= 5x² (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) 2x³ – 24x + 107, x ∈ [1,3] .
(b) 3x⁴ – 2x³ – 6x² + 6x + 1, x ∈ [0, 2]
(c) x + sin 2x, x ∈ [0, 2π]
(d) x³ – 18x² + 96x, x ∈ [0, 9]
Solution:
(a) Let y = 2x³ – 24x + 107, x ∈ [1,3]
\(\frac { dy }{ dx } \) = 6x² – 24
For maxima or minima
\(\frac { dy }{ dx } \) = 0 ⇒ 6x² – 24 = 0
⇒ x = ± 2.
∵ x ∈ [1, 3]
∴ x = 2
Now, y₁ = 2(1)³ – 24(1) + 107
= 2 – 24 + 107 = 85
y₂ = 2(2)³ – 24(2) +107
= 16 – 48 + 107 = 75
y₃ = 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 = 3x⁴ – 2x³ – 6x² + 6x + 1,
x ∈ (0, 2)
\(\frac { dy }{ dx } \) = 12x³ – 6x² – 12x + 6
= 6(2x³ – x² – 2x + 1)
For maxima or minima

Now y₁ = 3(1)⁴ – 2(1)³ – 6(1)² + 6(1) + 1
⇒ y₁ = 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
⇒ y₀ = 3(0)⁴ – 2(0)³ – 6(0)² + 6(0) + 1
⇒ y₀ = 1
⇒ y₂ = 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.

y₄ = (0)³ – 18(0)² + 96(0)
= 64 – 288 + 384 = 160
y₈ = (8)³ – 18(8)² + 96 × 8
= 512 – 1152 + 768
= 128
y₉ = (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 sin² x.cos x + (1 – 2 sin² x).cos x
= – 4 sin² x cos x + cos x – 2 sin² x cos x
= cos x (1 – 6 sin² 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 sin² x (1 + cosx) is maximum x = \(\frac { 1 }{ 2 } \)
Solution:
Let y = sin²  x.(1 + cosx) …..(1)
Diff.wrt.x,
⇒ \(\frac { dy }{ dx } \) = sin² x(- sin x) + (1 + cos x)
⇒ \(\frac { dy }{ dx } \) = sinx (-sin² x + 2 cos x + 2 cos² 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

RBSE Solutions for Class 12 Maths