JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Application of Derivatives

Let

S = (- 1, \infty)

and

f : S \to R

be defined as

f (x) = \int_{- 1}^{x} (e^{t} - 1)^{11} (2 t - 1)^{5} (t - 2)^{7} (t - 3)^{12} (2 t - 10)^{61} d t

Let

p =

Sum of square of the values of

x

, where

f (x)

attains local maxima on

S

, and

q =

Sum of the values of

x

, where

f (x)

attains local minima on

S

.

Then, the value of

p^{2} + 2 q

Answer: 27

Solution

f^{'} (x) = (e^{x} - 1)^{11} (2 x - 1)^{5} (x - 2)^{7} (x - 3)^{12} (2 x - 10)^{61}

Critical points, with multiplicities:

x = 0

(odd, 11),

x = 1/2

(odd, 5),

x = 2

(odd, 7),

x = 3

(even, 12 — no sign change, not an extremum),

x = 5

(odd, 61).

Checking the sign of

f^{'}

across each interval (each factor's sign, multiplied together) shows:

f^{'} > 0

(- 1, 0)

f^{'} < 0

(0, 1/2)

— local maximum at

x = 0

;

f^{'} < 0

(0, 1/2)

f^{'} > 0

(1/2, 2)

— local minimum at

x = 1/2

;

f^{'} > 0

(1/2, 2)

f^{'} < 0

(2, 3)

— local maximum at

x = 2

; sign unchanged across

x = 3

;

f^{'} < 0

(3, 5)

f^{'} > 0

(5, \infty)

— local minimum at

x = 5

.

So maxima at

x = 0, 2

p = 0^{2} + 2^{2} = 4

. Minima at

x = 1/2, 5

q = \frac{1}{2} + 5 = \frac{11}{2}

p^{2} + 2 q = 16 + 11 = 27

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Integer answer