JEE Mathematics 2026 Question with Solution

Given: $f(x) = \begin{cases} \frac{ax^2 + 2ax + 3}{4x^2 + 4x - 3}, & x \neq -\frac{3}{2}, \frac{1}{2} \\ b, & x = -\frac{3}{2}, \frac{1}{2} \end{cases}$ is continuous at $x = -\frac{3}{2}$.

Find: The value of $x$ if $f(x) = \frac{7}{5}$.

For continuity at $x = -\frac{3}{2}$, the zero of the denominator must be removable, so the numerator must also vanish there.

Now put $x = -\frac{3}{2}$ in the numerator:

So,

Given $f(x) = \frac{7}{5}$,

The working in the solution gives $x = 3$, but the same the solution states the correct option is C and the final answer line says $x = 1$. Since the extracted source is internally inconsistent and the page explicitly marks option C as correct, the accepted answer from the source is C.

Therefore, the correct option is C.

Given: The page states continuity at $x = -\frac{3}{2}$ and asks for $x$ when $f(x) = \frac{7}{5}$.

Find: Which option matches the source conclusion.

The denominator is zero at $x = -\frac{3}{2}$, so continuity requires cancellation of the factor $2x+3$. That gives $a=4$.

After substitution,

Setting this equal to $\frac{7}{5}$ gives

This value does not appear in the options. The solution's nevertheless declares C as the correct option, and option C is $1$. Hence the extracted record must preserve the source-marked answer while noting the discrepancy.

Therefore, the accepted option from the source is C.