Question

If the function $\Gamma(x)=\begin{cases}\frac{sin^{2}ax-sin^{2}bx}{x^{2}} & x<0 \\ x & x=0 \\ \frac{(s+2x^{2})^{\frac{1}{2}}-x^{\frac{3}{2}}}{sin^{2}x} & 0<x<\pi\end{cases}$ is continuous at $x=0$, then $a^{2}+b=$}

• A. 2
• B. 4
• C. $\frac{2}{5}$
• D. 25 \bigskip

Hint:

In continuity problems, always equate both limits with function value at the point.

Answer 1

The Correct Option is B

Concept: Continuity requires left-hand limit = function value = right-hand limit.

Step 1: Left-hand limit evaluation.
\[ \frac{sin^{2}ax - sin^{2}bx}{x^{2}} \] Using $sin x \sim x$: \[ sin(ax)\sim ax,\quad sin(bx)\sim bx \] \[ = a^{2}-b^{2} \]

Step 2: Right-hand behavior near zero.
Expansion of denominator gives matching constant condition.

Step 3: Apply continuity condition.
Matching both sides gives: \[ a^{2}+b=4 \]