JEE MAIN - Mathematics (2022 - 26th June Morning Shift - No. 12)
Let $$f(x) = 2{\cos ^{ - 1}}x + 4{\cot ^{ - 1}}x - 3{x^2} - 2x + 10$$, $$x \in [ - 1,1]$$. If [a, b] is the range of the function f, then 4a $$-$$ b is equal to :
11
11 $$-$$ $$\pi$$
11 + $$\pi$$
15 $$-$$ $$\pi$$
Explanation
$$f(x) = 2{\cos ^{ - 1}}x + 4{\cot ^{ - 1}}x - 3{x^2} - 2x + 10\,\forall x \in [ - 1,1]$$
$$ \Rightarrow f'(x) = - {2 \over {\sqrt {1 - {x^2}} }} - {4 \over {1 + {x^2}}} - 6x - 2 < 0\,\forall x \in [ - 1,1]$$
So f(x) is decreasing function and range of f(x) is [f(1), f($$-$$1)], which is [$$\pi$$ + 5, 5$$\pi$$ + 9]
Now $$4a - b = 4(\pi + 5) - (5\pi + 9)$$
$$ = 11 - \pi $$
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