The range of $a \in R$ for which the function $f(x)=(4 a-3)\left(x+\log _{e} 5\right)+2(a-7) \cot \left(\frac{x}{2}\right) \sin ^{2}\left(\frac{x}{2}\right), x \neq 2 n \pi, n \in N$ has critical points is:

(D) $\left[-\frac{4}{3}, 2\right]$

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Application of Derivatives

The range of

a \in R

for which the function

f (x) = (4 a - 3) (x + lo g_{e} 5) + 2 (a - 7) cot (\frac{x}{2}) sin^{2} (\frac{x}{2}), x \neq = 2 nπ, n \in N

has critical points is:

Answer: (D)

Since

cot \frac{x}{2} sin^{2} \frac{x}{2} = sin \frac{x}{2} cos \frac{x}{2} = \frac{1}{2} sin x

, the function is

f (x) = (4 a - 3) (x + lo g_{e} 5) + (a - 7) sin x

f^{'} (x) = (4 a - 3) + (a - 7) cos x

. Critical points exist exactly when

f^{'} (x) = 0

is solvable, that is when

cos x = \frac{3 - 4 a}{a - 7} \in [- 1, 1]

From

\frac{3 - 4 a}{a - 7} \leq 1

\frac{5 ( 2 - a )}{a - 7} \leq 0

, which holds for

a \leq 2

a > 7

; from

\frac{3 - 4 a}{a - 7} \geq - 1

\frac{3 a + 4}{a - 7} \leq 0

, which holds for

- \frac{4}{3} \leq a < 7

. Both together give

- \frac{4}{3} \leq a \leq 2

So the function has critical points for every

a

[- \frac{4}{3}, 2]

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option