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\limits_1^3((f^\prime(t))^2+1) f^\prime \prime(t) dt=\alpha+\beta \sqrt3 where \alpha, \beta are integers, then the value of \alpha+\beta equals

Definite Integration PYQ — JEE Main 2024 · 30 Jan · Shift 2 · Q33 with Solution

View written solutionFree

Correct answer: A

Use the given tangent angles to find derivatives at the endpoints

If the tangent makes angle $\theta$ with the positive $x$ -axis, then its slope is $f'(x)=\tan\theta.$ So, $f'(1)=\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}},$ $f'(3)=\tan\frac{\pi}{4}=1.$

Evaluate the integral

We need to compute $27\int_1^3 \left((f'(t))^2+1\right)f''(t)\,dt.$

Notice that if we set $u=f'(t),$ then $du=f''(t)\,dt.$ Hence,

=\int (u^2+1)\,du =\frac{u^3}{3}+u + C.$$ Therefore, $$\int_1^3 \left((f'(t))^2+1\right)f''(t)\,dt =\left[\frac{(f'(t))^3}{3}+f'(t)\right]_1^3.$$ 3. **Substitute the endpoint values** At $t=3$: $$\frac{(f'(3))^3}{3}+f'(3)=\frac{1^3}{3}+1=\frac{4}{3}.$$ At $t=1$: $$\frac{(f'(1))^3}{3}+f'(1) =\frac{\left(\frac{1}{\sqrt{3}}\right)^3}{3}+\frac{1}{\sqrt{3}}.$$ Now, $$\left(\frac{1}{\sqrt{3}}\right)^3=\frac{1}{3\sqrt{3}},$$ so $$\frac{\left(\frac{1}{\sqrt{3}}\right)^3}{3}= rac{1}{9\sqrt{3}}.$$ Thus, $$\frac{1}{9\sqrt{3}}+\frac{1}{\sqrt{3}}= rac{10}{9\sqrt{3}}.$$ Hence, $$\int_1^3 \left((f'(t))^2+1\right)f''(t)\,dt =\frac{4}{3}-\frac{10}{9\sqrt{3}}.$$ 4. **Multiply by 27** $$27\int_1^3 \left((f'(t))^2+1\right)f''(t)\,dt =27\left(\frac{4}{3}-\frac{10}{9\sqrt{3}}\right).$$ Compute each term: $$27\cdot \frac{4}{3}=36,$$ $$27\cdot \frac{10}{9\sqrt{3}}=\frac{30}{\sqrt{3}}=10\sqrt{3}.$$ So, $$27\int_1^3 \left((f'(t))^2+1\right)f''(t)\,dt=36-10\sqrt{3}.$$ Thus, $$\alpha=36,\quad \beta=-10.$$ Therefore, $$\alpha+\beta=36-10=26.$$ 5. **Check the options** The correct option is: - **A: 26** 6. **Compare with stored correct answer** Stored correct answer is **A**, which matches our result.

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View written solutionFree

Correct answer: A

Use the given tangent angles to find derivatives at the endpoints

If the tangent makes angle $\theta$ with the positive $x$ -axis, then its slope is $f'(x)=\tan\theta.$ So, $f'(1)=\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}},$ $f'(3)=\tan\frac{\pi}{4}=1.$

Evaluate the integral

We need to compute $27\int_1^3 \left((f'(t))^2+1\right)f''(t)\,dt.$

Notice that if we set $u=f'(t),$ then $du=f''(t)\,dt.$ Hence,

=\int (u^2+1)\,du =\frac{u^3}{3}+u + C.$$ Therefore, $$\int_1^3 \left((f'(t))^2+1\right)f''(t)\,dt =\left[\frac{(f'(t))^3}{3}+f'(t)\right]_1^3.$$ 3. **Substitute the endpoint values** At $t=3$: $$\frac{(f'(3))^3}{3}+f'(3)=\frac{1^3}{3}+1=\frac{4}{3}.$$ At $t=1$: $$\frac{(f'(1))^3}{3}+f'(1) =\frac{\left(\frac{1}{\sqrt{3}}\right)^3}{3}+\frac{1}{\sqrt{3}}.$$ Now, $$\left(\frac{1}{\sqrt{3}}\right)^3=\frac{1}{3\sqrt{3}},$$ so $$\frac{\left(\frac{1}{\sqrt{3}}\right)^3}{3}= rac{1}{9\sqrt{3}}.$$ Thus, $$\frac{1}{9\sqrt{3}}+\frac{1}{\sqrt{3}}= rac{10}{9\sqrt{3}}.$$ Hence, $$\int_1^3 \left((f'(t))^2+1\right)f''(t)\,dt =\frac{4}{3}-\frac{10}{9\sqrt{3}}.$$ 4. **Multiply by 27** $$27\int_1^3 \left((f'(t))^2+1\right)f''(t)\,dt =27\left(\frac{4}{3}-\frac{10}{9\sqrt{3}}\right).$$ Compute each term: $$27\cdot \frac{4}{3}=36,$$ $$27\cdot \frac{10}{9\sqrt{3}}=\frac{30}{\sqrt{3}}=10\sqrt{3}.$$ So, $$27\int_1^3 \left((f'(t))^2+1\right)f''(t)\,dt=36-10\sqrt{3}.$$ Thus, $$\alpha=36,\quad \beta=-10.$$ Therefore, $$\alpha+\beta=36-10=26.$$ 5. **Check the options** The correct option is: - **A: 26** 6. **Compare with stored correct answer** Stored correct answer is **A**, which matches our result.

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