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

RBSE Solutions For Class 12 Maths Chapter 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
RBSE Solutions For Class 12 Maths Chapter 8.5
At x = 2, maximum value of function
= 2(2)3 – 15(2)2 + 36 × 2 +10
= 16 – 60 + 72 + 10
= 38
RBSE Solutions For Class 12 Maths Chapter 8 Application Of Derivatives
(b) (x – 1) (x – 2) (x – 3)
Let y = (x – 1) (x – 2) (x – 3)
Then, = x3 – 6x2 + 11x – 6
Ex 8.5 Class 12 RBSE Application Of Derivatives
Exercise 8.5 Class 12 RBSE Application Of Derivatives
(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
RBSE Solutions For Class 12 Maths Application Of Derivatives
12 Math RBSE Solutions Application Of Derivatives
(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)
= 5x[x(x – 3) -1 (x – 3)]
= 5x(x – 3)(x – 1)
For maxima or minima
RBSE Solutions For Class 12th Maths Application Of Derivatives

Ex 8.5 Class 12 RBSE 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

RBSE Solutions For Class 12 Maths Chapter 8 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
RBSE Solutions For Class 12 Math Application Of Derivatives
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
12th Maths RBSE Solution Application Of Derivatives
12th RBSE Solution Maths Application Of Derivatives
∴ Hence, maximum value of function is 2π at x = 2π and minimum value of function is 0 at x = 0.
Class 12 Math RBSE Solutions Application Of Derivatives
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

Exercise 8.5 Class 12 RBSE 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)
RBSE Solution Class 12 Maths Application Of Derivatives
Class 12th Maths RBSE Solution Application Of Derivatives
So, at x = \(\frac { 3\pi }{ 2 } \) function will be maxima and maximum value of function
RBSE Solutions Of Class 12 Maths Application Of Derivatives
Www RBSE Solutions Com Class 12 Maths Application Of Derivatives
Math Class 12 RBSE Solutions Application Of Derivatives
RBSE Class 12 Maths Application Of Derivatives
RBSE Solution Class 12 Math Application Of Derivatives
RBSE Solutions For Class 10 Maths Chapter 8 In Hindi Application Of Derivatives
RBSE Solution Class 12th Maths Application Of Derivatives
Hence, on putting x = e in given function we get maximum value of function which is (e)1/e.
12th Math RBSE Solutions Application Of Derivatives

RBSE Solutions For Class 12 Maths Question 5.
Prove that value of \(\frac { x }{ 1+xtanx } \) is maximam at x = cos x.
Solution:
RBSE Solutions 12th Maths Application Of Derivatives
Class 12 Maths RBSE Solutions Application Of Derivatives

12 Math RBSE Solutions 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
RBSE Solutions 12 Maths Application Of Derivatives
12 Maths RBSE Solution Application Of Derivatives
Hence, at cos x = \(\frac { 1 }{ 3 } \) function has maximum value.

RBSE Solutions For Class 12th Maths Question 7.
Prove that ,y = sinP θ cosq θ is maximum at tan θ = \(\sqrt { \frac { p }{ q } } \)
Solution:
RBSE Solutions Class 12 Maths Application Of Derivatives
RBSE Solution 12 Math Application Of Derivatives
+ (p cot θ – q tan θ) × 0
= -y {(q + p) + (p + q)} + 0
= -2y (p + q)
= -2 (sinθ cosθ) (p + q)
= -ve
So, y will be maximum
Hence, y will be of maximum value if
RBSE Solution Class 12th Math Application Of Derivatives

RBSE Solutions for Class 12 Maths