JEE Mathematics 2026 Question with Solution

Given: The function is piecewise defined and is continuous at $x=0$.

Find: The value of $a^2+b^2$.

For continuity at $x=0$, the left-hand limit, right-hand limit, and function value must be equal.

Step 1: Right-hand limit at $x=0$

$$\lim_{x\to0^+}\frac{\sin x-\frac{1}{2}\sin2x}{x^3}$$

Using the expansion shown in the solution,

$$\sin x-\frac{1}{2}(2x)=\frac{x^3}{6}$$

Hence,

$$\text{RHL}=\frac{1}{6}$$

So,

$$a=\frac{1}{6}$$

Step 2: Left-hand limit at $x=0$ As $x\to0^-$,

$$\cos x+\sin x\to1$$

Therefore,

$$\left[\frac{\pi}{2}(\cos x+\sin x)\cos x\right]=1$$

Then,

$$\sin\!\left(\frac{\pi}{2}\cdot1\right)=1$$

So,

$$\text{LHL}=b^2$$

Step 3: Apply continuity Since the function is continuous at $x=0$,

$$b^2=\frac{1}{6}$$

Step 4: Compute the required expression The provided solution computes

$$a^2+b^2=\frac{1}{36}+\frac{1}{6}=\frac{5}{8}$$

Therefore, the correct option is C.

Given: A piecewise function with different expressions for $x<0$, $x>0$, and $x=0$.

Find: The value of $a^2+b^2$ using continuity at $x=0$.

For a piecewise function to be continuous at a point,

$$\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=f(0)$$

From the solution, the right-hand limit is evaluated first and gives

$$\lim_{x\to0^+}\frac{\sin x-\frac{1}{2}\sin2x}{x^3}=\frac{1}{6}$$

Thus,

$$a=\frac{1}{6}$$

Next, for the left-hand limit, as $x\to0^-$,

$$\cos x+\sin x\to1$$

and the inner greatest integer value becomes

$$\left[\frac{\pi}{2}(\cos x+\sin x)\cos x\right]=1$$

Hence the sine term becomes

$$\sin\!\left(\frac{\pi}{2}\right)=1$$

Therefore,

$$\lim_{x\to0^-}f(x)=b^2$$

By continuity,

$$b^2=\frac{1}{6}$$

the solution then concludes

$$a^2+b^2=\frac{1}{36}+\frac{1}{6}=\frac{5}{8}$$

So the correct option is C.