Method of DifferentiationeasyFree

Method of Differentiation — JEE Maths practice question

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

Question
If ddx{x2a2−x2+a22sin⁡−1xa}=f(x)\dfrac{d}{dx}\left\{\dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}\right\} = f(x), then y=(f(x))2+x2y = (f(x))^2 + x^2 is:
AConstantcorrect
BLine inclined to the x-axis
CParabola
DCircle
Solution
Step 1: Substitute x=asin⁡θx = a\sin\theta Then the bracket becomes a24sin⁡2θ+a22θ\frac{a^2}{4}\sin 2\theta + \frac{a^2}{2}\theta, and
f(x)=a22(1+cos⁡2θ)dθdx=a22⋅2cos⁡2θ⋅1acos⁡θ=acos⁡θ=a2−x2f(x) = \frac{a^2}{2}(1 + \cos 2\theta)\frac{d\theta}{dx} = \frac{a^2}{2} \cdot 2\cos^2\theta \cdot \frac{1}{a\cos\theta} = a\cos\theta = \sqrt{a^2 - x^2}
 Step 2: Form the required expression
y=(f(x))2+x2=(a2−x2)+x2=a2y = (f(x))^2 + x^2 = (a^2 - x^2) + x^2 = a^2
 Since a2a^2 is independent of xx, yy is a constant function. Answer: (1)
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