JEE Mathematics 2025 Question with Solution

Given:

$$f(x)=\begin{cases} -3ax^2-2, & x<1 \\ a^2+bx, & x\geq 1 \end{cases}$$

The function is differentiable for all $x \in \mathbb{R}$ with $a>1$.

Find: The value of $\alpha+\beta$ if the enclosed area with the line $y=-20$ is $\alpha+\beta\sqrt{3}$.

For differentiability at $x=1$, the function must be continuous and must have equal derivatives from both sides.

Continuity at $x=1$ gives

$$-3a(1)^2-2=a^2+b$$

so

$$a^2+b=-3a-2$$

For $x<1$,

$$f'(x)=-6ax$$

and at $x=1$ the left derivative is

$$-6a$$

For $x\geq 1$,

$$f'(x)=b$$

Hence differentiability gives

$$b=-6a$$

Substitute this into the continuity equation:

$$a^2-6a=-3a-2$$ $$a^2-3a+2=0$$ $$(a-1)(a-2)=0$$

Since $a>1$, we get

$$a=2$$

and therefore

$$b=-12$$

So the function becomes

$$f(x)=\begin{cases} -6x^2-2, & x<1 \\ 4-12x, & x\geq 1 \end{cases}$$

Now find the intersections with the line $y=-20$.

For $x<1$,

$$-6x^2-2=-20$$ $$x^2=3$$ $$x=\pm\sqrt{3}$$

Only $x=-\sqrt{3}$ satisfies $x<1$.

For $x\geq 1$,

$$4-12x=-20$$ $$x=2$$

Therefore the enclosed region lies from $x=-\sqrt{3}$ to $x=2$, split at $x=1$.

Area

$$=\int_{-\sqrt{3}}^{1}(f(x)+20)\,dx+\int_{1}^{2}(f(x)+20)\,dx$$

For $x\in[-\sqrt{3},1]$,

$$f(x)+20=-6x^2-2+20=-6x^2+18$$

So

$$\int_{-\sqrt{3}}^{1}(-6x^2+18)\,dx$$ $$=\left[-2x^3+18x\right]_{-\sqrt{3}}^{1}$$ $$=(-2+18)-\left(-2(-3\sqrt{3})+18(-\sqrt{3})\right)$$ $$=16-(6\sqrt{3}-18\sqrt{3})$$ $$=16+12\sqrt{3}$$

For $x\in[1,2]$,

$$f(x)+20=4-12x+20=24-12x$$

So

$$\int_{1}^{2}(24-12x)\,dx$$ $$=\left[24x-6x^2\right]_{1}^{2}$$ $$=(48-24)-(24-6)$$ $$=24-18=6$$

Hence total area is

$$(16+12\sqrt{3})+6=22+12\sqrt{3}$$

So

$$\alpha=22, \qquad \beta=12$$

and

$$\alpha+\beta=34$$

Therefore, the required value is $34$.

Given: A piecewise function and the condition that it is differentiable everywhere.

Find: The enclosed area expression and then $\alpha+\beta$.

The first step is to determine the constants by matching both function value and derivative at the joining point $x=1$.

From continuity:

$$-3a-2=a^2+b$$

From differentiability:

$$b=-6a$$

Substituting,

$$a^2-6a=-3a-2$$ $$a^2-3a+2=0$$ $$(a-1)(a-2)=0$$

Since $a>1$, take

$$a=2$$

and hence

$$b=-12$$

Thus,

$$f(x)=\begin{cases} -6x^2-2, & x<1 \\ 4-12x, & x\geq 1 \end{cases}$$

Now the phrase "area of the region enclosed" means we must locate the finite bounded region between the curve and the line $y=-20$.

For the left branch:

$$-6x^2-2=-20$$ $$x^2=3$$

This gives $x=\pm\sqrt{3}$, but only $x=-\sqrt{3}$ belongs to the branch $x<1$ in the bounded region connected to the join.

For the right branch:

$$4-12x=-20$$ $$x=2$$

Hence the bounded region runs from $x=-\sqrt{3}$ to $x=2$.

The area above the line $y=-20$ is obtained by integrating $f(x)-(-20)=f(x)+20$.

Left part:

$$A_1=\int_{-\sqrt{3}}^{1}(-6x^2+18)\,dx$$ $$A_1=\left[-2x^3+18x\right]_{-\sqrt{3}}^{1}$$ $$A_1=16+12\sqrt{3}$$

Right part:

$$A_2=\int_{1}^{2}(24-12x)\,dx$$ $$A_2=\left[24x-6x^2\right]_{1}^{2}=6$$

Total area:

$$A=A_1+A_2=22+12\sqrt{3}$$

Therefore,

$$\alpha=22, \quad \beta=12$$

so

$$\alpha+\beta=34$$

The correct numerical answer is $34$.