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Functions — JEE Maths practice question

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

Question
The range of the function f(x)=4cos⁡3x−8cos⁡2x+1f(x)=4\cos^{3}x-8\cos^{2}x+1 is:
A[−11,1][-11,1]correct
BR\mathbb{R}
C[−1,1][-1,1]
D[−11,−3][-11,-3]
Solution
Step 1: Since f(x)=4cos⁡3x−8cos⁡2x+1f(x)=4\cos^{3}x-8\cos^{2}x+1, let cos⁡x=t\cos x=t where t∈[−1,1]t\in[-1,1]. So consider g(t)=4t3−8t2+1g(t)=4t^{3}-8t^{2}+1 on [−1,1][-1,1]. Step 2: Find the derivative:
g′(t)=12t2−16t=4t(3t−4)g'(t)=12t^{2}-16t=4t(3t-4)
 g′(t)=0  ⇒  t=0g'(t)=0\;\Rightarrow\;t=0 or t=43t=\dfrac{4}{3}. Since 43∉[−1,1]\dfrac{4}{3}\notin[-1,1], only critical point inside the interval is t=0t=0. Step 3: Evaluate at endpoints and critical point: g(−1)=4(−1)−8(1)+1=−4−8+1=−11g(-1)=4(-1)-8(1)+1=-4-8+1=-11 g(0)=0−0+1=1g(0)=0-0+1=1 g(1)=4−8+1=−3g(1)=4-8+1=-3 Step 4: So on [−1,1][-1,1]: gmin⁡=−11g_{\min}=-11 (at t=−1t=-1) and gmax⁡=1g_{\max}=1 (at t=0t=0). The range is [−11,1][-11,1]. Answer: (1) [−11,1][-11,1]
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