Let f(x)=2cos-1x+4cot-1x-3x2-2x+10,x∈[-1,1]. If [a, b] is the range of the function then 4a- b is equal to:

f′(x)=−21−x2−41+x2−6x−2=−211−x2+21+x2+3x+1f′(x)<0⇒f(x) is a decreasing. function f(1)=π+5f(−1)=5π+9 Range :[a,b]≡[π+5,5π+9] a =π+5,b=5π+9⇒4a−b=11−π

Let f(x)=2cos-1x+4cot-1x-3x2-2x+10,x∈[-1,1]. If [a, b] is the range of the function then 4a- b is equal to:

f′(x)=−21−x2−41+x2−6x−2=−211−x2+21+x2+3x+1f′(x)<0⇒f(x) is a decreasing. function f(1)=π+5f(−1)=5π+9 Range :[a,b]≡[π+5,5π+9] a =π+5,b=5π+9⇒4a−b=11−π

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Let $f (x) = 2 \cos^{- 1} x + 4 \cot^{- 1} x - 3 x^{2} - 2 x + 10, x \in [- 1, 1]$ . If [a, b] is the range of the function then 4a- b is equal to:

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$11 - π$

$11 + π$

$15 - π$

answer is B.

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Detailed Solution

$\begin{array}{l} f^{'} (x) = \frac{- 2}{\sqrt{1 - x^{2}}} - \frac{4}{1 + x^{2}} - 6 x - 2 \\ = - 2 [\frac{1}{\sqrt{1 - x^{2}}} + \frac{2}{1 + x^{2}} + 3 x + 1] \\ f^{'} (x) < 0 \Rightarrow f (x) is a decreasing. function \\ f (1) = π + 5 \\ f (- 1) = 5 π + 9 \\ Range : [a, b] \equiv [π + 5, 5 π + 9] \\ a = π + 5, b = 5 π + 9 \Rightarrow 4 a - b = 11 - π \end{array}$