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Limits — JEE Maths practice question

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

Question

f(0),\,f'(0),\,f''(0)

exist and are non-zero, and

\lim_{x\to 0}\dfrac{af(2x)+2bf\left(\dfrac{x}{2}\right)-3f(x)}{\sin^{-1}(x^{2})}=3, \text{ then}

a=b

correct

f''(0)=4

correct

a^2+b^2=2

correct

a^2+b^2=8

Solution

Step 1: Note that

\sin^{-1}(x^{2})\sim x^{2}

x\to 0

. So the limit simplifies to

\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

\to 0

), the numerator at

x=0

must be

0

af(0)+2bf(0)-3f(0)=0\;\Rightarrow\;(a+2b-3)f(0)=0

Since

f(0)\neq 0

a+2b=3\quad\ldots(i)

Step 3: Apply L'Hôpital's rule. Differentiate numerator and denominator:

\lim_{x\to 0}\dfrac{2af'(2x)+bf'\left(\dfrac{x}{2}\right)-3f'(x)}{2x}

Step 4: For finiteness, numerator at

x=0

must be

0

2af'(0)+bf'(0)-3f'(0)=0\;\Rightarrow\;(2a+b-3)f'(0)=0

Since

f'(0)\neq 0

2a+b=3\quad\ldots(ii)

Step 5: Solve

(i)

and

(ii)

: From

(i)

a=3-2b

. Substituting in

(ii)

2(3-2b)+b=3\;\Rightarrow\;6-3b=3\;\Rightarrow\;b=1

. Then

a=3-2(1)=1

. So

a=b=1

. Step 6: Apply L'Hôpital again to find

f''(0)

\lim_{x\to 0}\dfrac{4af''(2x)+\dfrac{b}{2}f''\left(\dfrac{x}{2}\right)-3f''(x)}{2}=3

x=0

\dfrac{4af''(0)+\dfrac{b}{2}f''(0)-3f''(0)}{2}=3

With

a=b=1

\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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