Given f(x)=2cos−1x+4cot−1x−3x2−2x+10
f′(x)=1−x2−2−1+x24−6x−2
=−2[1−x21+1+x22+3x+1]
f′(x)<0⇒f(x) is a decreasing function
Now f(1)=π+5 and f(−1)=5π+5
So, range: [a,b]≡[π+5,5π+5]
a=π+5,b=5π+5⇒4a−b=11−π
JEE Main 2022 — Mathematics Calculus
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
Held on 26 Jun 2022 · Verified 6 Jul 2026.
11
11−π
11+π
15−π
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