JEE Main 2021 — Mathematics Algebra

For domain ${\mathrm{log}}_{5}({\mathrm{log}}_{3}(18x-{x}^{2}-77))>0$log5​(log3​(18x−x2−77))>0

$${\mathrm{log}}_{3}(18x-{x}^{2}-77)>1$$log3​(18x−x2−77)>1

$$\Rightarrow 18x-{x}^{2}-77>3$$⇒18x−x2−77>3

$$\Rightarrow {x}^{2}-18x+80<0$$⇒x2−18x+80<0

$$\Rightarrow x\in (8,10)$$⇒x∈(8,10)

$\Rightarrow a=8$⇒a=8 and $b=10$b=10

$$I={\int }_{a}^{b}\frac{{\mathrm{sin}}^{3}x}{{\mathrm{sin}}^{3}x+{\mathrm{sin}}^{3}(a+b-x)}dx$$I=∫ab​sin3x+sin3(a+b−x)sin3x​dx

Apply $\displaystyle {\int }_{a}^{b}f(x)dx={\int }_{a}^{b}f(a+b-x)dx$∫ab​f(x)dx=∫ab​f(a+b−x)dx

$$I={\int }_{a}^{b}\frac{{\mathrm{sin}}^{3}(a+b-x)}{{\mathrm{sin}}^{3}x+{\mathrm{sin}}^{3}(a+b-x)}dx$$I=∫ab​sin3x+sin3(a+b−x)sin3(a+b−x)​dx

\Rightarrow 2I=(b-a)\Rightarrow I=\frac{b-a}{2}(\because a=8&b=10)

$$\Rightarrow I=\frac{10-8}{2}=1$$⇒I=210−8​=1