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Limits: Exist Non Zero 2bf

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

Question
If f(0), f′(0), f′′(0)f(0),\,f'(0),\,f''(0) exist and are non-zero, and
lim⁡x→0af(2x)+2bf(x2)−3f(x)sin⁡−1(x2)=3, then\lim_{x\to 0}\dfrac{af(2x)+2bf\left(\dfrac{x}{2}\right)-3f(x)}{\sin^{-1}(x^{2})}=3, \text{ then}
Aa=ba=bcorrect
Bf′′(0)=4f''(0)=4correct
Ca2+b2=2a^2+b^2=2correct
Da2+b2=8a^2+b^2=8
Solution
Step 1: Note that sin⁡−1(x2)∼x2\sin^{-1}(x^{2})\sim x^{2} as x→0x\to 0. So the limit simplifies to
lim⁡x→0af(2x)+2bf(x2)−3f(x)x2=3\lim_{x\to 0}\dfrac{af(2x)+2bf\left(\dfrac{x}{2}\right)-3f(x)}{x^{2}}=3
 Step 2: For the limit to be finite (denominator →0\to 0), the numerator at x=0x=0 must be 00:
af(0)+2bf(0)−3f(0)=0  ⇒  (a+2b−3)f(0)=0af(0)+2bf(0)-3f(0)=0\;\Rightarrow\;(a+2b-3)f(0)=0
 Since f(0)≠0f(0)\neq 0: a+2b=3…(i)a+2b=3\quad\ldots(i) Step 3: Apply L'Hôpital's rule. Differentiate numerator and denominator:
lim⁡x→02af′(2x)+bf′(x2)−3f′(x)2x\lim_{x\to 0}\dfrac{2af'(2x)+bf'\left(\dfrac{x}{2}\right)-3f'(x)}{2x}
 Step 4: For finiteness, numerator at x=0x=0 must be 00:
2af′(0)+bf′(0)−3f′(0)=0  ⇒  (2a+b−3)f′(0)=02af'(0)+bf'(0)-3f'(0)=0\;\Rightarrow\;(2a+b-3)f'(0)=0
 Since f′(0)≠0f'(0)\neq 0: 2a+b=3…(ii)2a+b=3\quad\ldots(ii) Step 5: Solve (i)(i) and (ii)(ii): From (i)(i): a=3−2ba=3-2b. Substituting in (ii)(ii): 2(3−2b)+b=3  ⇒  6−3b=3  ⇒  b=12(3-2b)+b=3\;\Rightarrow\;6-3b=3\;\Rightarrow\;b=1. Then a=3−2(1)=1a=3-2(1)=1. So a=b=1a=b=1. Step 6: Apply L'Hôpital again to find f′′(0)f''(0):
lim⁡x→04af′′(2x)+b2f′′(x2)−3f′′(x)2=3\lim_{x\to 0}\dfrac{4af''(2x)+\dfrac{b}{2}f''\left(\dfrac{x}{2}\right)-3f''(x)}{2}=3
 At x=0x=0:
4af′′(0)+b2f′′(0)−3f′′(0)2=3\dfrac{4af''(0)+\dfrac{b}{2}f''(0)-3f''(0)}{2}=3
 With a=b=1a=b=1:
(4+12−3)f′′(0)2=3  ⇒  32f′′(0)2=3  ⇒  f′′(0)=4\dfrac{(4+\dfrac{1}{2}-3)f''(0)}{2}=3\;\Rightarrow\;\dfrac{\dfrac{3}{2}f''(0)}{2}=3\;\Rightarrow\;f''(0)=4
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