JEE Main 2021 — Mathematics Calculus

We have,

$$f(x)={\mathrm{log}}_{e}(x+\sqrt{{x}^{2}+1}),x\in R$$f(x)=loge​(x+x2+1​),x∈R

$$\Rightarrow f(x)={\mathrm{log}}_{e}(\frac{(x+\sqrt{{x}^{2}+1})(x-\sqrt{{x}^{2}+1})}{x-\sqrt{{x}^{2}+1}})$$⇒f(x)=loge​(x−x2+1​(x+x2+1​)(x−x2+1​)​)

$$\Rightarrow f(x)={\mathrm{log}}_{e}(\frac{-1}{x-\sqrt{{x}^{2}+1}})$$⇒f(x)=loge​(x−x2+1​−1​)

$$\Rightarrow f(x)={\mathrm{log}}_{e}(\frac{1}{\sqrt{{x}^{2}+1}-x})$$⇒f(x)=loge​(x2+1​−x1​)

$$\Rightarrow f(-x)={\mathrm{log}}_{e}(\frac{1}{\sqrt{{x}^{2}+1}+x})$$⇒f(−x)=loge​(x2+1​+x1​)

$$\Rightarrow f(-x)={\mathrm{log}}_{e}{(\sqrt{{x}^{2}+1}+x)}^{-1}$$⇒f(−x)=loge​(x2+1​+x)−1

$$\Rightarrow f(-x)=-{\mathrm{log}}_{e}(\sqrt{{x}^{2}+1}+x)$$⇒f(−x)=−loge​(x2+1​+x)

$$\Rightarrow f(-x)=-f(x)$$⇒f(−x)=−f(x)

Hence, $f(x)$f(x) is an odd function.

Now,

$$g(t)={\int }_{-\pi /2}^{\pi /2}(\mathrm{cos}\frac{\pi }{4}t+f(x))dx$$g(t)=∫−π/2π/2​(cos4π​t+f(x))dx

$$\Rightarrow g(t)=\mathrm{cos}\frac{\pi }{4}t{\int }_{-\pi /2}^{\pi /2}1dx+{\int }_{-\pi /2}^{\pi /2}f(x)dx$$⇒g(t)=cos4π​t∫−π/2π/2​1dx+∫−π/2π/2​f(x)dx

$$\Rightarrow g(t)=\pi \mathrm{cos}\frac{\pi }{4}t+{\int }_{-\pi /2}^{\pi /2}f(x)dx$$⇒g(t)=πcos4π​t+∫−π/2π/2​f(x)dx

$$\Rightarrow g(t)=\pi \mathrm{cos}\frac{\pi }{4}t+0$$⇒g(t)=πcos4π​t+0

$\because f(x)$∵f(x) is an odd function.

$$\therefore g(1)=\frac{\pi }{\sqrt{2}}\Rightarrow \sqrt{2}g(1)=\pi$$∴g(1)=2​π​⇒2​g(1)=π

$$g(0)=\pi$$g(0)=π