If 24 \int\limits_0^\frac\pi4 \bigg[\sin | 4x - \frac\pi12 | + [2 \sin x] \bigg] dx = 2\pi + \alpha , where [\cdot] denotes the greatest integer function, then \alpha is equal to \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2025 · 29 Jan · Shift 2 · Q46 with Solution

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Correct answer: 12

We need to evaluate $24\int_0^{\pi/4}\left[\sin\left|4x-\frac{\pi}{12}\right|+[2\sin x]\right]dx=2\pi+\alpha.$ Let $I=\int_0^{\pi/4}\left[\sin\left|4x-\frac{\pi}{12}\right|+[2\sin x]\right]dx.$ Then we compute the two parts separately.

First part: $I_1=\int_0^{\pi/4}\sin\left|4x-\frac{\pi}{12}\right|dx.$

The expression inside modulus changes sign at $4x-\frac{\pi}{12}=0\quad\Rightarrow\quad x=\frac{\pi}{48}.$ So split the integral: $I_1=\int_0^{\pi/48}\sin\left(\frac{\pi}{12}-4x\right)dx+\int_{\pi/48}^{\pi/4}\sin\left(4x-\frac{\pi}{12}\right)dx.$

Now, $\int \sin(a-4x)dx=\frac{1}{4}\cos(a-4x),$ so

=\frac14\left[\cos\left(\frac{\pi}{12}-4x\right)\right]_0^{\pi/48} =\frac14\left(\cos 0-\cos\frac{\pi}{12}\right).$$ Thus, $$=\frac14\left(1-\cos\frac{\pi}{12}\right).$$ Also, $$\int \sin(4x-a)dx=-\frac14\cos(4x-a),$$ so $$\int_{\pi/48}^{\pi/4}\sin\left(4x-\frac{\pi}{12}\right)dx =-\frac14\left[\cos\left(4x-\frac{\pi}{12}\right)\right]_{\pi/48}^{\pi/4}.$$ At $x=\pi/48$, angle $=0$; at $x=\pi/4$, angle $=\pi-\pi/12=11\pi/12$. Hence $$= -\frac14\left(\cos\frac{11\pi}{12}-\cos 0\right) =\frac14\left(1-\cos\frac{11\pi}{12}\right).$$ Since $$\cos\frac{11\pi}{12}=-\cos\frac{\pi}{12},$$ this becomes $$\frac14(1+\cos\tfrac{\pi}{12}).$$ Therefore, $$I_1=\frac14\left(1-\cos\frac{\pi}{12}\right)+\frac14\left(1+\cos\frac{\pi}{12}\right)=\frac12.$$ --- 3. Second part: $$I_2=\int_0^{\pi/4}[2\sin x]dx.$$ For $x\in[0,\pi/4]$, $$2\sin x\in[0,\sqrt2).$$ So the greatest integer value changes when $2\sin x=1$, i.e. $$\sin x=\frac12 \Rightarrow x=\frac\pi6.$$ Thus, - for $0\le x<\pi/6$, $[2\sin x]=0$, - for $\pi/6\le x\le \pi/4$, $[2\sin x]=1$. Hence $$I_2=\int_0^{\pi/6}0\,dx+\int_{\pi/6}^{\pi/4}1\,dx =\frac\pi4-\frac\pi6=\frac\pi{12}.$$ --- 4. Total integral: $$I=I_1+I_2=\frac12+\frac\pi{12}.$$ Therefore, $$24I=24\left(\frac12+\frac\pi{12}\right)=12+2\pi.$$ Given that $$24I=2\pi+\alpha,$$ we get $$\alpha=12.$$ --- 5. Comparison with stored answer: Stored correct answer = $12$. Our derived answer is also $12$. So the stored answer is correct.

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View written solutionFree

Correct answer: 12

We need to evaluate $24\int_0^{\pi/4}\left[\sin\left|4x-\frac{\pi}{12}\right|+[2\sin x]\right]dx=2\pi+\alpha.$ Let $I=\int_0^{\pi/4}\left[\sin\left|4x-\frac{\pi}{12}\right|+[2\sin x]\right]dx.$ Then we compute the two parts separately.

First part: $I_1=\int_0^{\pi/4}\sin\left|4x-\frac{\pi}{12}\right|dx.$

Now, $\int \sin(a-4x)dx=\frac{1}{4}\cos(a-4x),$ so

=\frac14\left[\cos\left(\frac{\pi}{12}-4x\right)\right]_0^{\pi/48} =\frac14\left(\cos 0-\cos\frac{\pi}{12}\right).$$ Thus, $$=\frac14\left(1-\cos\frac{\pi}{12}\right).$$ Also, $$\int \sin(4x-a)dx=-\frac14\cos(4x-a),$$ so $$\int_{\pi/48}^{\pi/4}\sin\left(4x-\frac{\pi}{12}\right)dx =-\frac14\left[\cos\left(4x-\frac{\pi}{12}\right)\right]_{\pi/48}^{\pi/4}.$$ At $x=\pi/48$, angle $=0$; at $x=\pi/4$, angle $=\pi-\pi/12=11\pi/12$. Hence $$= -\frac14\left(\cos\frac{11\pi}{12}-\cos 0\right) =\frac14\left(1-\cos\frac{11\pi}{12}\right).$$ Since $$\cos\frac{11\pi}{12}=-\cos\frac{\pi}{12},$$ this becomes $$\frac14(1+\cos\tfrac{\pi}{12}).$$ Therefore, $$I_1=\frac14\left(1-\cos\frac{\pi}{12}\right)+\frac14\left(1+\cos\frac{\pi}{12}\right)=\frac12.$$ --- 3. Second part: $$I_2=\int_0^{\pi/4}[2\sin x]dx.$$ For $x\in[0,\pi/4]$, $$2\sin x\in[0,\sqrt2).$$ So the greatest integer value changes when $2\sin x=1$, i.e. $$\sin x=\frac12 \Rightarrow x=\frac\pi6.$$ Thus, - for $0\le x<\pi/6$, $[2\sin x]=0$, - for $\pi/6\le x\le \pi/4$, $[2\sin x]=1$. Hence $$I_2=\int_0^{\pi/6}0\,dx+\int_{\pi/6}^{\pi/4}1\,dx =\frac\pi4-\frac\pi6=\frac\pi{12}.$$ --- 4. Total integral: $$I=I_1+I_2=\frac12+\frac\pi{12}.$$ Therefore, $$24I=24\left(\frac12+\frac\pi{12}\right)=12+2\pi.$$ Given that $$24I=2\pi+\alpha,$$ we get $$\alpha=12.$$ --- 5. Comparison with stored answer: Stored correct answer = $12$. Our derived answer is also $12$. So the stored answer is correct.

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