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∫ e^{5x⁶}(x−α) dx = 0 implies range of α | JEE

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

Question
If ∫01e5x6(x−α) dx=0\displaystyle\int_{0}^{1}e^{5x^{6}}(x-\alpha)\,dx=0, then
A1<α<21<\alpha<2
Bα<0\alpha<0
C0<α<10<\alpha<1correct
Dα=0\alpha=0
Solution
Step 1: Interpret the condition geometrically. The integral being zero means the signed area enclosed by h(x)=e5x6(x−α)h(x)=e^{5x^{6}}(x-\alpha) and the xx-axis over [0,1][0,1] cancels out. For this cancellation, hh must change sign somewhere inside (0,1)(0,1). Step 2: Examine the sign of hh. The factor e5x6>0e^{5x^{6}}>0 for all xx, so the sign of h(x)h(x) is the sign of (x−α)(x-\alpha). Step 3: For (x−α)(x-\alpha) to change sign within (0,1)(0,1), the root x=αx=\alpha must lie strictly inside the interval:
0<α<1.0<\alpha<1.
 Step 4: If α≤0\alpha\le 0 then x−α>0x-\alpha>0 throughout (0,1)(0,1) and the integral is positive; if α≥1\alpha\ge 1 then x−α<0x-\alpha<0 throughout and it is negative. Only 0<α<10<\alpha<1 allows a zero value. Correct answer: (3)
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