Inverse Trigonometric FunctionsmediumFree

Inverse Trigonometric Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
If the domain of the function f(x)=3cos⁡−1(4x)−πf(x)=\sqrt{3\cos^{-1}(4x)-\pi} is [a,b][a,b], then find the value of (4a+64b)(4a+64b).
Solution
Answer: 7
Step 1: For f(x)f(x) to be defined:
3cos⁡−1(4x)−π≥0  ⇒  cos⁡−1(4x)≥π33\cos^{-1}(4x)-\pi\ge 0\;\Rightarrow\;\cos^{-1}(4x)\ge\dfrac{\pi}{3}
 Also cos⁡−1(4x)\cos^{-1}(4x) requires 4x∈[−1,1]4x\in[-1,1], i.e., cos⁡−1(4x)∈[0,π]\cos^{-1}(4x)\in[0,\pi]. Step 2: Combining:
π3≤cos⁡−1(4x)≤π\dfrac{\pi}{3}\le\cos^{-1}(4x)\le\pi
 Step 3: Apply cos⁡\cos (which is decreasing on [0,π][0,\pi]):
cos⁡π≤4x≤cos⁡π3  ⇒  −1≤4x≤12\cos\pi\le 4x\le\cos\dfrac{\pi}{3}\;\Rightarrow\;-1\le 4x\le\dfrac{1}{2}
−14≤x≤18-\dfrac{1}{4}\le x\le\dfrac{1}{8}
 Step 4: So the domain is [−14,18]=[a,b]\left[-\dfrac{1}{4},\dfrac{1}{8}\right]=[a,b]. Step 5: Compute:
4a+64b=4(−14)+64(18)=−1+8=74a+64b=4\left(-\dfrac{1}{4}\right)+64\left(\dfrac{1}{8}\right)=-1+8=7
 Answer: 77
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