Definite Integration — JEE Main practice

1. Split the numerator into even ($\pi$) and odd ($4x^{11}$) parts, and pair $x$ with $-x$ over the symmetric interval — since $|x|=|-x|$, the denominator is the same for both. $I=\int_0^{\pi/6}\left[\frac{\pi+4x^{11}}{1-\sin(x+\pi/6)}+\frac{\pi-4x^{11}}{1-\sin(x+\pi/6)}\right]dx$

2. The odd terms cancel, leaving twice the constant-numerator term. $I=2\pi\int_0^{\pi/6}\frac{dx}{1-\sin(x+\pi/6)}$

3. Rationalize by multiplying numerator and denominator by $1+\sin\theta$, using $1-\sin^2\theta=\cos^2\theta$. $\frac{1}{1-\sin\theta}=\frac{1+\sin\theta}{\cos^2\theta}=\sec^2\theta+\tan\theta\sec\theta$

4. Substitute this identity with $\theta=x+\pi/6$. $I=2\pi\int_0^{\pi/6}\left[\sec^2\left(x+\frac{\pi}{6}\right)+\tan\left(x+\frac{\pi}{6}\right)\sec\left(x+\frac{\pi}{6}\right)\right]dx$

5. Integrate directly, since $\frac{d}{dx}(\tan\theta)=\sec^2\theta$ and $\frac{d}{dx}(\sec\theta)=\sec\theta\tan\theta$. $I=2\pi\left[\tan\left(x+\frac{\pi}{6}\right)+\sec\left(x+\frac{\pi}{6}\right)\right]_0^{\pi/6}$

6. Evaluate at the limits (argument becomes $\pi/3$ at $x=\pi/6$, and $\pi/6$ at $x=0$). $\tan\frac{\pi}{3}+\sec\frac{\pi}{3}=\sqrt3+2,\qquad \tan\frac{\pi}{6}+\sec\frac{\pi}{6}=\sqrt3$

7. Subtract and multiply by $2\pi$. $I=2\pi\left[(\sqrt3+2)-\sqrt3\right]=2\pi(2)=4\pi$

Hence, the answer is Option B.

1. Differentiate both sides of the given equation, using the Fundamental Theorem of Calculus on the left and the product rule on the right. $\frac{d}{dx}\left[6\int_1^x f(t)\,dt\right]=6f(x)$ $\frac{d}{dx}\left[3xf(x)+x^3-4\right]=3f(x)+3xf'(x)+3x^2$

2. Equate the two derivatives and simplify. $6f(x)=3f(x)+3xf'(x)+3x^2\implies f(x)-xf'(x)=x^2$

3. Divide through by $x^2$, recognizing the left side as the derivative of $\frac{f(x)}{x}$. $\frac{f(x)-xf'(x)}{x^2}=1\implies \frac{d}{dx}\left(\frac{f(x)}{x}\right)=-1$

4. Integrate both sides. $\frac{f(x)}{x}=-x+k\implies f(x)=-x^2+kx$

5. Use the boundary condition — setting $x=1$ in the original equation gives $f(1)=1$, which fixes $k$. $1=-1+k\implies k=2$

6. So the function is fully determined. $f(x)=-x^2+2x$

7. Compute $f(2)$ and $f(3)$ and subtract. $f(2)=0,\qquad f(3)=-3$ $f(2)-f(3)=0-(-3)=3$

Hence, the answer is Option B.

1. Break the interval $[-\pi/2,\pi/2]$ at the integer points $-1,0,1$, since $[x]$ is constant between consecutive integers.

2. On $[-\pi/2,-1)$, $[x]=-2$, so the integrand is constant. $\frac{1}{[x]+4}=\frac12$

3. On $[-1,0)$, $[x]=-1$. $\frac{1}{[x]+4}=\frac13$

4. On $[0,1)$, $[x]=0$. $\frac{1}{[x]+4}=\frac14$

5. On $[1,\pi/2]$, $[x]=1$. $\frac{1}{[x]+4}=\frac15$

6. Integrate each piece over its own subinterval and add the results. $I=\frac12\left(-1+\frac{\pi}{2}\right)+\frac13(1)+\frac14(1)+\frac15\left(\frac{\pi}{2}-1\right)$

7. Simplify each term. $I=\frac{\pi}{4}-\frac12+\frac13+\frac14+\frac{\pi}{10}-\frac15$

8. Combine over a common denominator of $60$. $I=\frac{21\pi-7}{60}=\frac{7}{60}(3\pi-1)$

Hence, the answer is Option B.