JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Application of Derivatives

m

is the minimum value of

k

for which the function

f (x) = x k x - x^{2}

is increasing in the interval

[0, 3]

and

M

is the maximum value of

f

[0, 3]

when

k = m

, then the ordered pair

(m, M)

is equal to:

Answer: (D)

Solution

For

f (x) = x k x - x^{2}

the derivative is

f^{'} (x) = \frac{3 k x - 4 x ^{2}}{2 k x - x ^{2}}

. For

f

to be increasing throughout

[0, 3]

the numerator

x (3 k - 4 x)

must stay non-negative there, which needs

x \leq \frac{3 k}{4}

to hold even at

x = 3

: that is

3 \leq \frac{3 k}{4}

, so

k \geq 4

and the least such value is

m = 4

. With

k = 4

f (x) = x 4 x - x^{2}

and its largest value on

[0, 3]

occurs at the endpoint

x = 3

, where

f (3) = 3 12 - 9 = 33

. The ordered pair is

(4, 33)

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option