Question

Assertion :Let $f\left(x\right)={\int }_0^{\pi /2}({\sin }^{6}x+{\cos }^{6}x)dx$ then $\frac{\pi }{8}<f\left(x\right)<\frac{\pi }{2}$ Reason: If $m$ and $M$ are the smallest & greatest values, of a function f(x) such that $a<f\left(x\right)<b$

then $m(b-a)<{\int }_a^{b}f\left(x\right)dx<M(b-a)$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Given: $f\left(x\right)={\int }_0^{\frac{\pi }{2}}({\sin }^{6}x+{\cos }^{6}x)dx$

$f\left(x\right)={\int }_0^{\frac{\pi }{2}}({\sin }^{2}x+{\cos }^{2}x)-({\sin }^{4}x+{\cos }^{4}x-{\sin }^{2}x{\cos }^{2}x)dxf\left(x\right)={\int }_0^{\frac{\pi }{2}}{({\sin }^{2}x+{\cos }^{2}x)}^2-2{\sin }^{2}x+{\cos }^{2}x-{\sin }^{2}x{\cos }^{2}x\left)dxf\right(x)={\int }_0^{\frac{\pi }{2}}(1-3{\sin }^{2}x{\cos }^{2}x)dxf(x)={\int }_0^{\frac{\pi }{2}}dx-{\int }_0^{\frac{\pi }{2}}\frac{3}{4}\left({\sin }^{2}2x\right)dxf(x)={\int }_0^{\frac{\pi }{2}}dx-\frac{3}{8}{\int }_0^{\frac{\pi }{2}}(1-\cos 4x)dxf(x)={\int }_0^{\frac{\pi }{2}}(1-\frac{3}{8})dx-\frac{3}{8}{\int }_0^{\frac{\pi }{2}}\left(\cos 4x\right)dxf(x)=\frac{5}{8}{\int }_0^{\frac{\pi }{2}}dx-\frac{3}{8}{\int }_0^{\frac{\pi }{2}}\left(\cos 4x\right)dxf(x)=\frac{5}{8}{\left[x\right]}_0^{\frac{\pi }{2}}-\frac{3}{8}{\left[\frac{\sin 4x}{4}\right]}_0^{\frac{\pi }{2}}f(x)=\frac{5}{8}\times \frac{\pi }{2}=\frac{5\pi }{16}$

$Max({\sin }^{6}x+{\cos }^{6}x)=1=M(b-a)M=\frac{\pi }{2}Min({\sin }^{6}x+{\cos }^{6}x)=\frac{1}{4}m(b-a)m=\frac{\pi }{8}$

Hence both assertion and reason are correct and reason is correct explanation for assertion.