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Official PYQ · 2017Set 2 · Q10Calculus: Limits, Continuity, DifferentiabilityMCQ1 mark

GATE CS 2017 Set 2 Q10 — Calculus: Limits, Continuity, Differentiability

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If f(x)=R sin⁡(πx2)+S,f’(12)=2f(x) = R \: \sin ( \frac{\pi x}{2}) + S, f’\left(\frac{1}{2}\right) = \sqrt{2} and ∫01f(x)dx=2Rπ\int_0^1 f(x) dx = \frac{2R}{\pi}, then the constants RR and SS are

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Answer: C

We are given the function:

f(x)=Rsin⁡(πx2)+Sf(x) = R \sin\left(\frac{\pi x}{2}\right) + S

Step 1: Find RR using the derivative condition

Differentiating f(x)f(x) with respect to xx:

f′(x)=R⋅π2cos⁡(πx2)f'(x) = R \cdot \frac{\pi}{2} \cos\left(\frac{\pi x}{2}\right)

We are given that f′(12)=2f'\left(\frac{1}{2}\right) = \sqrt{2}:

f′(12)=R⋅π2cos⁡(π4)=2f'\left(\frac{1}{2}\right) = R \cdot \frac{\pi}{2} \cos\left(\frac{\pi}{4}\right) = \sqrt{2}

Since cos⁡(π4)=12\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}:

R⋅π2⋅12=2R \cdot \frac{\pi}{2} \cdot \frac{1}{\sqrt{2}} = \sqrt{2} R⋅π2=2  ⟹  R=4πR \cdot \frac{\pi}{2} = 2 \implies R = \frac{4}{\pi}

Step 2: Find SS using the integration condition

Now, we evaluate the definite integral of f(x)f(x) from 00 to 11:

∫01f(x) dx=∫01[Rsin⁡(πx2)+S]dx\int_0^1 f(x) \, dx = \int_0^1 \left[ R \sin\left(\frac{\pi x}{2}\right) + S \right] dx =[−R⋅2πcos⁡(πx2)+Sx]01= \left[ -R \cdot \frac{2}{\pi} \cos\left(\frac{\pi x}{2}\right) + Sx \right]_0^1 =(−R⋅2πcos⁡(π2)+S)−(−R⋅2πcos⁡(0)+0)= \left( -R \cdot \frac{2}{\pi} \cos\left(\frac{\pi}{2}\right) + S \right) - \left( -R \cdot \frac{2}{\pi} \cos(0) + 0 \right)

Since cos⁡(π2)=0\cos\left(\frac{\pi}{2}\right) = 0 and cos⁡(0)=1\cos(0) = 1:

=(0+S)−(−2Rπ)=S+2Rπ= (0 + S) - \left( -\frac{2R}{\pi} \right) = S + \frac{2R}{\pi}

We are given that this integral is equal to 2Rπ\frac{2R}{\pi}:

S+2Rπ=2Rπ  ⟹  S=0S + \frac{2R}{\pi} = \frac{2R}{\pi} \implies S = 0

Conclusion

Thus, the constants are R=4πR = \frac{4}{\pi} and S=0S = 0, which corresponds to Option C.

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