Let y=f(x) be a thrice differentiable function in (-5,5). Let the tangents to the curve y=f(x) at (1,f(1)) and (3,f(3)) make angles π 6 and π 4,…

Given:

$$y=f(x)$$y=f(x)

$$\Rightarrow \frac{dy}{dx}={f}^{'}(x)$$⇒dxdy​=f′(x)

$${(\frac{dy}{dx})}_{(1,f(1))}={f}^{'}(1)=\mathrm{tan}\frac{\pi }{6}=\frac{1}{\sqrt{3}}$$(dxdy​)(1,f(1))​=f′(1)=tan6π​=3​1​

$$\Rightarrow {f}^{'}(1)=\frac{1}{\sqrt{3}}...(i)$$⇒f′(1)=3​1​...(i)

$${(\frac{\mathrm{dy}}{\mathrm{dx}})}_{(3,f(3))}={f}^{'}(3)=\mathrm{tan}\frac{\pi }{4}=1$$(dxdy​)(3,f(3))​=f′(3)=tan4π​=1

$$\Rightarrow {f}^{'}(3)=1...(ii)$$⇒f′(3)=1...(ii)

It is given that, $\displaystyle 27{\int }_{1}^{3}({({f}^{'}(t))}^{2}+1){f}^{"}(t)dt=\alpha +\beta \sqrt{3}$27∫13​((f′(t))2+1)f"(t)dt=α+β3​

Let, $\displaystyle I={\int }_{1}^{3}({({f}^{'}(t))}^{2}+1){f}^{"}(t)dt$I=∫13​((f′(t))2+1)f"(t)dt

Putting, ${f}^{'}(t)=z\Rightarrow {f}^{"}(t)dt=dz$f′(t)=z⇒f"(t)dt=dz

$$z={f}^{'}(3)=1$$z=f′(3)=1

$$z={f}^{'}(1)=\frac{1}{\sqrt{3}}$$z=f′(1)=3​1​

$$\Rightarrow I={\int }_{\frac{1}{\sqrt{3}}}^{1}({z}^{2}+1)dz$$⇒I=∫3​1​1​(z2+1)dz

$$\Rightarrow I={(\frac{{z}^{3}}{3}+z)}_{\frac{1}{\sqrt{3}}}^{1}$$⇒I=(3z3​+z)3​1​1​

$$\Rightarrow I=(\frac{1}{3}+1)-(\frac{1}{3}\cdot \frac{1}{3\sqrt{3}}+\frac{1}{\sqrt{3}})$$⇒I=(31​+1)−(31​⋅33​1​+3​1​)

$$\Rightarrow I=\frac{4}{3}-\frac{10}{9\sqrt{3}}$$⇒I=34​−93​10​

$$\Rightarrow I=\frac{4}{3}-\frac{10}{27}\sqrt{3}$$⇒I=34​−2710​3​

$$\Rightarrow \alpha +\beta \sqrt{3}=27(\frac{4}{3}-\frac{10}{27}\sqrt{3})$$⇒α+β3​=27(34​−2710​3​)

$$\Rightarrow \alpha +\beta \sqrt{3}=36-10\sqrt{3}$$⇒α+β3​=36−103​

$$\Rightarrow \alpha =36,\beta =-10$$⇒α=36,β=−10

$$\Rightarrow \alpha +\beta =36-10$$⇒α+β=36−10

$$\Rightarrow \alpha +\beta =26$$⇒α+β=26