(34)

Given, f(x) is continuous and differentiable at x = 1.

$f\left(x\right)=\{\begin{array}{cc}–3a{x}^2–2,& x<1\\ a^2+bx,& x\ge 1\end{array}$

$\Rightarrow f\text{'}\left(x\right)=\{\begin{array}{cc}–6ax,& x<1\\ b,& x\ge 1\end{array}$

L.H.L. at $x=1 \Rightarrow –3a–2$

R.H.L. at $x=1 \Rightarrow a^2+b$

$\therefore$   L.H.L. = R.H.L.          {$\because$ f(x) is continuous}

$\Rightarrow –3a–2=a^2+b$          ... (i)

L.H.D. at $x=1 \Rightarrow –6a$

R.H.D. at $x=1 \Rightarrow b$

$\therefore$   L.H.D. = R.H.D.          [$\because$ f(x) is differentiable]

$\Rightarrow –6a=b$          ... (ii)

From (i) and (ii), we get

$–3a–2=a^2-6a$

$\Rightarrow a^2–3a+2=0 \Rightarrow (a–1)(a–2)=0$

$\Rightarrow a=1 or a=2 \Rightarrow a=2$           ($\because$ a > 1)

From (ii), b = –12

Now, $f\left(x\right)=\{\begin{array}{cc}–6x^2–2,& x<1\\ 4–12x,& x\ge 1\end{array}$

$\therefore$  Area of region $={\int }_{–\sqrt{3}}^1(–6x^2–2+20)dx+{\int }_1^2(4–12x+20)dx$

$={(\frac{–6x^3}{3}–2x+20x)}_{–\sqrt{3}}^1+{(4x–\frac{12x^2}{2}+20x)}_1^2$

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

$\Rightarrow 22+12\sqrt{3}=\alpha +\beta \sqrt{3}$

So, $\alpha =22 \mathrm{and} \beta =12$

$\therefore \alpha +\beta =22+12=34$

(77)

Given, $x^2+y^2=25$

$\Rightarrow x^2+{(x–1)}^2=25$          [$\because$ y = |x –1|]

$\Rightarrow x=4,–3$

$\therefore$   Required area $=25\pi –\left({\int }_{–3}^4\right(\sqrt{25–x^2}–|x–1|\left)dx\right)$

$=25\pi –{[\frac{1}{2}x\sqrt{25–x^2}+\frac{25}{2}{\sin }^{–1}\frac{x}{5}]}_{–3}^4+{\int }_{–3}^1(1–x)dx+{\int }_1^4(x–1)dx$

$=25\pi +\frac{25}{2}–(2\times 3+\frac{25}{2}{\sin }^{–1}(\frac{4}{5})+\frac{3}{2}\times 4+\frac{25}{2}{\sin }^{–1}(\frac{3}{5}\left)\right)$

$=25\pi +\frac{25}{2}–12–\frac{25\pi }{4}$

$=\frac{75\pi }{4}+\frac{1}{2}=\frac{1}{4}(75\pi +2)$

$\therefore \frac{1}{4}(b\pi +c)=\frac{1}{4}(75\pi +2)$

$\Rightarrow b=75, c=2$

$\Rightarrow b+c=77$.