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 :
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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