If $\int_0^{2 \pi}\left(\sin ^4 x+\cos ^4 x\right) d x=K \int_0^\pi \sin ^2 x d x+L \int_0^{\frac{\pi}{2}}…
If $\int_0^{2 \pi}\left(\sin ^4 x+\cos ^4 x\right) d x=K \int_0^\pi \sin ^2 x d x+L \int_0^{\frac{\pi}{2}} \cos ^2 x d x$ and $K, L \in N$, then the number of possible ordered pairs $(\mathrm{K}, \mathrm{L})$ is
1
2
3
4
Solution
L.H.S. $=\int_0^{2 \pi}\left(\sin ^4 x+\cos ^4 x\right) d x$
$\begin{aligned}
& =\int_0^{2 \pi} \sin ^4 x d x+\int_0^{2 \pi} \cos ^4 x d x=2 \int_0^\pi \sin ^4 x d x+2 \int_0^\pi \cos ^4 x d x \\
& =2 \int_0^{\pi / 2} \sin ^4 x d x+4 \int_0^{\pi / 2} \cos ^4 x d x \\
& =4\left(\frac{3 \times 1}{4 \times 2}\right) \frac{\pi}{2}+4\left(\frac{3 \times 1}{4 \times 2}\right) \frac{\pi}{2}=\frac{3 \pi}{2} \\
& \text { R.H.S. }=K \int_0^\pi \sin ^2 x d x+L \int_0^{\pi / 2} \cos ^2 x d x \\
& =2 K \int_0^{\pi / 2} \sin ^2 x d x+L \int_0^{\pi / 2} \cos ^2 x d x \\
& =2 K\left(\frac{1}{2} \times \frac{\pi}{2}\right)+L\left(\frac{1}{2} \times \frac{\pi}{2}\right)=(K) \frac{\pi}{2}+\left(\frac{L}{2}\right) \frac{\pi}{2} \\
& \therefore \text { For R.H.S. }=\text { L.H.S. } \\
& \frac{3 \pi}{2}=(K) \frac{\pi}{2}+\left(\frac{L}{2}\right) \frac{\pi}{2} \Rightarrow 3=K+\frac{L}{2}
\end{aligned}$ Since, $K, L \in N$ possible values are
$\therefore(K, L)=(2,2),(1,4)$ So, two ordered pairs of $(K, L)$ are possible.