JEE 2025 — Mathematics PYQ

$$f(x)=\left[\begin{array}{ccc}-3a{x}^2-2& amp;x& lt;1\\ a^2+bx& amp;x\ge 1& \end{array}\right]$$

$\because f(x)$ is differentiable.

$$f^{\prime }(x)=\left[\begin{array}{ccc}-6ax& amp;x& lt;1\\ b& amp;x& gt;1\end{array}\right]$$

$f(1^{+})=f(1^{-})=f(1)$

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

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

$f^{\prime }(1^{+})=f^{\prime }(1^{-})$

$-6a=b$

Solving (i) and (ii)

$a^2-3a+2=0$

$(a-1)\left(a-2\right)=0\Rightarrow a=1,2$

$\Rightarrow a=2$

$b=-12$

Now, $y=f(x)=\{\begin{array}{llc}-6x^2-2& amp;x& lt;1\\ -12x+4& amp;x\ge 1& \end{array}$

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

$-6x^2=-20$

$6x^2=18$

$x^2=3$

$x=2$

Since, we are in the domain x<1, we have $x=-\sqrt{3}$.

$\therefore$ Required area $=$

$20(\sqrt{3}+1)-{\int }_{-\sqrt{3}}^1-6x^2-2dx+\frac{1}{2}\times 1\times 12$

$=20(\sqrt{3}+1)-{\int }_{-\sqrt{3}}^1-6x^2-2+6$

$$=20\sqrt{3}+26-{\lfloor 2x^3+2x\rfloor }^{1-\sqrt{6}}$$
$$=20\sqrt{3}+26-\lfloor 4+8\sqrt{3}\rfloor$$
$$=20\sqrt{3}+26-8\sqrt{3}-4$$
$$=22+12\sqrt{3}$$
$$\alpha =22$$
$$\beta =12$$
$$\alpha +\beta =34$$