QJEEM23Maef2b: JEE Main 2023 Mathematics - Class 12 JEEM23Maef2b Solution: If , then ∑ x ∈ s ( sin ⁡ ( ( x 2 + x + 5 ) π 2 ) − cos ⁡ ( ( x …

Given equation is

$\sin^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right)-\sin^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right)=\frac{\pi}{4}.$

Let's denote:

$A = \sin^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right),$

$B = \sin^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right).$

So, we have the equation $A - B = \frac{\pi}{4}$ .

We can also write this as $A = B + \frac{\pi}{4}$ .

This gives us

$\sin(A) = \sin\left(B + \frac{\pi}{4}\right).$

We can use the identity $\sin(a + b) = \sin a \cos b + \cos a \sin b$ and rewrite this equation as:

$\frac{x+1}{\sqrt{(x+1)^2+1}} = \frac{x}{\sqrt{x^2+1}} \cos\left(\frac{\pi}{4}\right) + \sqrt{1-\left(\frac{x}{\sqrt{x^2+1}}\right)^2} \sin\left(\frac{\pi}{4}\right).$

After simplifying, we get:

$\frac{x+1}{\sqrt{x^2 + 2x + 2}} = \frac{1}{\sqrt{2}}\left(\frac{x}{\sqrt{x^2 + 1}} + \sqrt{1 - \frac{x^2}{x^2 + 1}}\right).$

Let's square both sides to remove the square roots:

On the left side, squaring gives:

$\left(\frac{x+1}{\sqrt{x^2 + 2x + 2}}\right)^2 = \frac{(x+1)^2}{x^2 + 2x + 2}.$

On the right side, squaring gives:

$\left(\frac{1}{\sqrt{2}}\left(\frac{x}{\sqrt{x^2 + 1}} + \sqrt{1 - \frac{x^2}{x^2 + 1}}\right)\right)^2 = \frac{1}{2}\left(\frac{x^2}{x^2+1} + 2\frac{x}{\sqrt{x^2+1}}\sqrt{1-\frac{x^2}{x^2+1}} + 1 - \frac{x^2}{x^2+1}\right).$

$\therefore$ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = $\frac{1}{2}\left(\frac{x^2}{x^2+1} + 2\frac{x}{\sqrt{x^2+1}}\sqrt{1-\frac{x^2}{x^2+1}} + 1 - \frac{x^2}{x^2+1}\right)$

$\Rightarrow$ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = ${1 \over 2}\left( {2 \times {x \over {\sqrt {{x^2} + 1} }}\sqrt {{{{x^2} + 1 - {x^2}} \over {{x^2} + 1}}} + 1} \right)$

$\Rightarrow$ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = ${1 \over 2}\left( {2 \times {x \over {\sqrt {{x^2} + 1} }}{1 \over {\sqrt {{x^2} + 1} }} + 1} \right)$

$\Rightarrow$ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = ${1 \over 2}\left( {2 \times {x \over {{x^2} + 1}} + 1} \right)$

$\Rightarrow$ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = ${1 \over 2}\left( {{{2x + {x^2} + 1} \over {{x^2} + 1}}} \right)$

$\Rightarrow$ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = ${1 \over 2}\left( {{{{{\left( {x + 1} \right)}^2}} \over {{x^2} + 1}}} \right)$

$\begin{aligned}& \Rightarrow \frac{x+1}{\sqrt{x^2+2x+2}}=\frac{x+1}{\sqrt{2} \sqrt{x^2+1}} \\\\& \Rightarrow x=-1 \text { OR } \sqrt{x^2+2x+2}=\sqrt{2} \cdot \sqrt{x^2+1} \\\\& \Rightarrow x=0, x=2 \text { (Rejected) } \\\\& S=\{0,-1\}\end{aligned}$

$\begin{aligned}& \sum_{x \in R}\left(\sin \left(\left(x^2+x+5\right) \frac{\pi}{2}\right)-\cos \left(\left(x^2+x+5\right) \pi\right)\right) \\\\& =\left[\sin \left(\frac{5 \pi}{2}\right)-\cos (5 \pi)\right]+\left[\sin \left(\frac{5 \pi}{2}\right)-\cos (5 \pi)\right] \\\\& = (1 -(-1)) + (1 -(-1))\\\\& = 2 + 2 \\\\& = 4\end{aligned}$