JEE Main 2024 (Online) 30th January Evening ShiftQuestion 1
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 $27\int_{1}^{3}\left(\left(f'(t)\right)^{2}+1\right)f''(t)\,dt=\alpha+\beta\sqrt{3}$ where $\alpha,\beta$ are integers, then the value of $\alpha+\beta$ equals
Definite Integration
Solution
Concept. $(f'^2+1)f''$ is the derivative of $\frac{f'^3}3+f'$, so the integral needs only $f'(1)$ and $f'(3)$. Formulas used. (1) Slope of tangent $=\tan(\text{angle with }x\text{-axis})$(2) $\int_1^3\big(f'^2+1\big)f''\,dt=\Big[\frac{f'^3}3+f'\Big]_1^3$Step 1 - Slopes. By (1): $f'(1)=\tan\frac\pi6=\frac1{\sqrt3}$, $f'(3)=\tan\frac\pi4=1$. Step 2 - Use (2). $$\Big(\frac13+1\Big)-\Big(\frac{1}{9\sqrt3}+\frac1{\sqrt3}\Big)=\frac43-\frac{10}{9\sqrt3}=\frac43-\frac{10\sqrt3}{27}.$$ Step 3 - Multiply by 27. $36-10\sqrt3$, so $\alpha=36$, $\beta=-10$ and $\alpha+\beta=26$. Answer: (A)