JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Application of Derivatives

The function

f (x) = x^{3} - 6 x^{2} + a x + b

is such that

f (2) = f (4) = 0

.
Consider two statements.
(S1) there exists

x_{1}, x_{2} \in (2, 4), x_{1} < x_{2}

, such that

f^{'} (x_{1}) = - 1

and

f^{'} (x_{2}) = 0

.
(S2) there exists

x_{3}, x_{4} \in (2, 4), x_{3} < x_{4}

, such that

f

is decreasing in

(2, x_{4})

, increasing in

(x_{4}, 4)

and

2 f^{'} (x_{3}) = 3 f (x_{4})

Answer: (A)

Solution

From

f (2) = f (4) = 0

2 a + b = 16

and

4 a + b = 32

, giving

a = 8

b = 0

, so

f (x) = x^{3} - 6 x^{2} + 8 x

and

f^{'} (x) = 3 x^{2} - 12 x + 8

. Solving

f^{'} (x) = 0

gives

x = 2 \pm \frac{2}{3}

, with

x_{2} = 2 + \frac{2}{3} \in (2, 4)

; solving

f^{'} (x) = - 1

gives

x = 3 \in (2, 4)

, and

3 < 2 + \frac{2}{3}

, so Statement I holds. On

(2, 4)

f

decreases on

(2, 2 + \frac{2}{3})

and increases on

(2 + \frac{2}{3}, 4)

, so

x_{4} = 2 + \frac{2}{3}

and

f (x_{4}) = - \frac{16}{3 3}

. The value

\frac{3}{2} f (x_{4}) = - \frac{8}{3}

lies between

f^{'} (2) = - 4

and

f^{'} (x_{4}) = 0

, so some

x_{3} \in (2, x_{4})

satisfies

2 f^{'} (x_{3}) = 3 f (x_{4})

. Both statements are true.

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option