JEE MAIN - Mathematics (2021 - 16th March Morning Shift - No. 11)
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 :
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, $$\infty $$)
($$-$$3, 1)
$$\left[ { - {4 \over 3},2} \right]$$
($$-$$$$\infty $$, $$-$$1]
Explanation
$$f(x) = (4a - 3)(x + \ln 5) + 2(a - 7)\left( {{{\cos {x \over 2}} \over {\sin {x \over 2}}}.{{\sin }^2}{x \over 2}} \right)$$
$$f(x) = (4a - 3)(x + \ln 5) + (a - 7)\sin x$$
$$f'(x) = (4a - 3) + (a - 7)\cos x = 0$$
$$\cos x = {{ - (4a - 3)} \over {a - 7}}$$
$$ - 1 \le - {{4a - 3} \over {a - 7}} \le 1$$
$$ - 1 \le {{4a - 3} \over {a - 7}} \le 1$$
$${{4a - 3} \over {a - 7}} - 1 \le 0$$ and $${{4a - 3} \over {a - 7}} + 1 \ge 0$$
$$ \Rightarrow {{ - 4} \over 3} \le a \le 2$$
$$f(x) = (4a - 3)(x + \ln 5) + (a - 7)\sin x$$
$$f'(x) = (4a - 3) + (a - 7)\cos x = 0$$
$$\cos x = {{ - (4a - 3)} \over {a - 7}}$$
$$ - 1 \le - {{4a - 3} \over {a - 7}} \le 1$$
$$ - 1 \le {{4a - 3} \over {a - 7}} \le 1$$
$${{4a - 3} \over {a - 7}} - 1 \le 0$$ and $${{4a - 3} \over {a - 7}} + 1 \ge 0$$
$$ \Rightarrow {{ - 4} \over 3} \le a \le 2$$
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