Medium MCQ +4 / -1 PYQ · JEE Mains 2021

The range of a$\in$R for which the

function f(x) = (4a $-$ 3)(x + loge 5) + 2(a $-$ 7) cot$\left( {{x \over 2}} \right)$ sin2$\left( {{x \over 2}} \right)$, x $\ne$ 2n$\pi$, n$\in$N has critical points, is :

  1. A [1, $\infty$)
  2. B ($-$3, 1)
  3. C $\left[ { - {4 \over 3},2} \right]$ Correct answer
  4. D ($-$$\infty$, $-$1]

Solution

$$f(x) = (4a - 3)(x + \ln 5) + 2(a - 7)\left( {{{\cos {x \over 2}} \over {\sin {x \over 2}}}.{{\sin }^2}{x \over 2}} \right)$$<br><br>$f(x) = (4a - 3)(x + \ln 5) + (a - 7)\sin x$<br><br>$f'(x) = (4a - 3) + (a - 7)\cos x = 0$<br><br>$\cos x = {{ - (4a - 3)} \over {a - 7}}$<br><br>$- 1 \le - {{4a - 3} \over {a - 7}} \le 1$<br><br>$- 1 \le {{4a - 3} \over {a - 7}} \le 1$<br><br>${{4a - 3} \over {a - 7}} - 1 \le 0$ and ${{4a - 3} \over {a - 7}} + 1 \ge 0$<br><br>$\Rightarrow {{ - 4} \over 3} \le a \le 2$

About this question

Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations

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