Question

Assertion :The area bounded by the curve $y=\frac{1}{1+{\left(\tan x\right)}^{\sqrt{2}}}$ & x-axis between the ordinates $x=\frac{\pi }{6}$ & $x=\frac{\pi }{3}$ is $\frac{\pi }{12}$ square units. Reason: ${\int }_a^b\frac{f\left(x\right)}{f(a+b-x)+f\left(x\right)}dx=\frac{b-a}{2}$

• 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 curve $y=\frac{1}{1+{\left(\tan x\right)}^{\sqrt{2}}}=\frac{{\left(\cos x\right)}^{\sqrt{2}}}{{\left(\sin x\right)}^{\sqrt{2}}+{\left(\cos x\right)}^{\sqrt{2}}}$
$\therefore$ Area bounded between the integral $x=\frac{\pi }{6}$, $x=\frac{\pi }{3}$ is given by
$\Delta ={\int }_{\pi /6}^{\pi /3}\frac{{\left(\cos x\right)}^{\sqrt{2}}}{{\left(\sin x\right)}^{\sqrt{2}}+{\left(\cos x\right)}^{\sqrt{2}}}dx$ (i)
$\Delta ={\int }_{\pi /6}^{\pi /3}\frac{{\left(\sin x\right)}^{\sqrt{2}}}{{\left(\cos x\right)}^{\sqrt{2}}+{\left(\sin x\right)}^{\sqrt{2}}}dx$
(ii) (using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$)
On adding (i) & (ii)
$\therefore$
$2\Delta ={\int }_{\pi /6}^{\pi /3}dx$ $\Rightarrow$
$\Delta =\frac{1}{2}(\frac{\pi }{3}-\frac{\pi }{6})=\frac{\pi }{12}$