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Functions Part 1

Section: Mathematics

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Question : 45 of 100

Marks: +1, -0

The range of

a \in R

for which the function

f(x)=(4 a-3)(x+\log_{e}5)+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

[16 Mar 2021 Shift 1]

$(-3,1)$
$\left[-\frac{4}{3}, 2\right]$
$[1, \infty)$
$(-\infty,-1]$

Solution:

Given,

f(x)=(4 a-3)(x+\log_{e}5)+2(a-7) \cot \frac{x}{2} \cdot \sin^{2} \frac{x}{2}

\Rightarrow \;\; f(x)=(4 a-3)(x+\log_{e}5)+2(a-7) \cos \frac{x}{2} \sin \frac{x}{2}

\Rightarrow \;\; f(x)=(4 a-3)(x+\log_{e}5)+(a-7) \sin x

\Rightarrow \;\; f'(x)=(4 a-3)(1+0)+(a-7) \cos x

\Rightarrow \;\; f'(x)=(4 a-3)+(a-7) \cos x

When

f'(x)=0,\ (4 a-3)+(a-7) \cos x=0

\Rightarrow \;\; \cos x=\frac{4 a-3}{7-a}

As,

-1 \leq \cos x \leq 1

So,

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

\Rightarrow \;\; \frac{4 a-3}{7-a}+1 \geq 0 \Rightarrow \frac{4 a-3+7-a}{7-a} \geq 0

\Rightarrow \;\; \frac{3 a+4}{7-a} \geq 0

a \in \left[-\frac{4}{3},7\right)

...(i)

Now,

\frac{4 a-3}{7-a}-1 \leq 0

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

\Rightarrow \;\; \frac{5 a-10}{7-a} \leq 0

a \in (-\infty, 2] \cup (7, \infty)

...(ii)

From Eqs. (i) and (ii),

a \in \left[-\frac{4}{3}, 2\right]

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