JEE Mathematics 2024 Question with Solution

Given: The piecewise function is

and it is differentiable on $\mathbb{R}$.

Find: The value of $48(a+b)$.

For differentiability, the function must be both continuous and differentiable at the boundary points $x=2$ and $x=-2$.

Using continuity at $x=2$:

Using continuity at $x=-2$ gives the same equation:

Now differentiate each part:

Using differentiability at $x=2$:

Using differentiability at $x=-2$ also gives the same value:

Substitute $a=-\frac{1}{16}$ into the continuity equation:

Now compute:

Therefore, the value of $48(a+b)$ is $15$.

Given: The definition changes at $|x|=2$, so the critical points are $x=2$ and $x=-2$.

Find: Determine $a$ and $b$, then evaluate $48(a+b)$.

At $x=2$, continuity requires

At $x=-2$, continuity requires

So continuity gives the single equation

For differentiability at $x=2$:

For differentiability at $x=-2$:

This confirms the same value of $a$.

Now solve for $b$:

Finally,

Hence, the required numerical value is $15$.