JEE Mathematics 2026 Question with Solution

Given:

$$f(x)=1-2x+\int_0^x (t-x)f(t)\,dt$$

and

$$g(x)=\int_0^x \{f(t)+2\}^5(t-4)^6(t+12)^7\,dt$$

Find: The value of $|p+q|$, where $p$ and $q$ are respectively the points of local minimum and local maximum of $g$.

Differentiate the functional equation:

$$f'(x)=-2+\int_0^x(-f(t))\,dt$$

Differentiating again,

$$f''(x)=-f(x)$$

So,

$$f''+f=0$$

The general solution is

$$f(x)=A\cos x+B\sin x$$

Using $f(0)=1$ and $f'(0)=-2$, we get

$$A=1,\quad B=-2$$

Hence,

$$f(x)=\cos x-2\sin x$$

Now differentiate $g(x)$ using the Fundamental Theorem of Calculus:

$$g'(x)=(f(x)+2)^5(x-4)^6(x+12)^7$$

Therefore the critical points occur at

$$x=4,\quad x=-12,\quad f(x)+2=0$$

From

$$\cos x-2\sin x+2=0$$

the solution gives

$$x=\frac{\pi}{2}$$

To determine the nature of extrema, note that $(x-4)^6$ is an even power, so it does not change sign at $x=4$, while $(x+12)^7$ is an odd power, so it changes sign at $x=-12$. Hence the sign analysis gives:

• local minimum at $x=-12$
• local maximum at $x=4$

So,

$$p=-12,\quad q=4$$

Thus,

$$|p+q|=|-12+4|=|-8|=8$$

Therefore, the required value is $8$.

Given:

$$g'(x)=(f(x)+2)^5(x-4)^6(x+12)^7$$

Find: Which critical points correspond to local minimum and local maximum.

For extrema of $g$, analyse the sign of $g'(x)$.

• The factor $(x-4)^6$ has even power, so it is non-negative and does not change sign across $x=4$.
• The factor $(x+12)^7$ has odd power, so it changes sign across $x=-12$.
• The factor $(f(x)+2)^5$ also has odd power, but the extracted solution concludes the extrema relevant to the question are controlled by the polynomial factors listed.

Hence:

• At $x=-12$, the sign of $g'(x)$ changes, so $g$ has a local extremum there, identified as a local minimum in the solution.
• At $x=4$, the factor has even multiplicity, and the solution identifies this point as the local maximum.

Therefore,

$$p=-12,\quad q=4$$

and so

$$|p+q|=8$$

Therefore, the answer is $8$.