If $I=\int_{1}^{2} \frac{dx}{\sqrt{2 x^{3}-9 x^{2}+12 x+4}}$, then:

(B) $\frac{1}{9}<I^{2}<\frac{1}{8}$

(A) $\frac{1}{16}<I^{2}<\frac{1}{9}$

(C) $\frac{1}{8}<I^{2}<\frac{1}{4}$

(D) $\frac{1}{6}<I^{2}<\frac{1}{2}$

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Definite Integrals

I = \int_{1}^{2} \frac{d x}{2 x ^{3} - 9 x ^{2} + 12 x + 4}

, then:

Answer: (B)

Let

f (x) = \frac{1}{2 x ^{3} - 9 x ^{2} + 12 x + 4}

. The cubic under the root,

g (x) = 2 x^{3} - 9 x^{2} + 12 x + 4

, has

g^{'} (x) = 6 (x - 1) (x - 2)

, which is negative on

(1, 2)

, so

g

decreases from

g (1) = 9

g (2) = 8

and

f

increases from

f (1) = \frac{1}{3}

f (2) = \frac{1}{2 2}

. The integral over this unit-length interval therefore lies between those values,

\frac{1}{3} < I < \frac{1}{2 2}

, and squaring gives

\frac{1}{9} < I^{2} < \frac{1}{8}

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Single correct option