Let y=f(x) be a thrice differentiable function in (-5,5).

Question

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 $$\pi / 6$$ and $$\pi / 4$$, respectively with positive $x$-axis. If $$27 \int_\limits1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3}$$ where $$\alpha, \beta$$ are integers, then the value of $$\alpha+\beta$$ equals

Accepted Answer

$$\begin{aligned}
& y=f(x) \Rightarrow \frac{d y}{d x}=f^{\prime}(x) \
& \left.\frac{d y}{d x}\right)_{(1, f(1))}=f^{\prime}(1)=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} \Rightarrow f^{\prime}(1)=\frac{1}{\sqrt{3}} \
& \left.\frac{d y}{d x}\right)_{(3, f(3))}=f^{\prime}(3)=\tan \frac{\pi}{4}=1 \Rightarrow f^{\prime}(3)=1 \
& 27 \int_\limits1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3} \
& I=\int_\limits1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t \
& f^{\prime}(t)=z \Rightarrow f^{\prime \prime}(t) d t=d z \
& z=f^{\prime}(3)=1 \
& z=f^{\prime}(1)=\frac{1}{\sqrt{3}}
\end{aligned}$$

$$\begin{aligned}
& I=\int_\limits{1 / \sqrt{3}}^1\left(z^2+1\right) d z=\left(\frac{z^3}{3}+z\right)_{1 / \sqrt{3}}^1 \
& =\left(\frac{1}{3}+1\right)-\left(\frac{1}{3} \cdot \frac{1}{3 \sqrt{3}}+\frac{1}{\sqrt{3}}\right) \
& =\frac{4}{3}-\frac{10}{9 \sqrt{3}}=\frac{4}{3}-\frac{10}{27} \sqrt{3} \
& \alpha+\beta \sqrt{3}=27\left(\frac{4}{3}-\frac{10}{27} \sqrt{3}\right)=36-10 \sqrt{3} \
& \alpha=36, \beta=-10 \
& \alpha+\beta=36-10=26
\end{aligned}$$

Solutions · 1