Question

If S={ x ∈ R : sin⁻¹\(( \frac{x + 1}{ √ x^ 2 + 2 x + 2} ) \)−-sin⁻¹\((\frac{x}{√x ^2+1 }) = \frac{π}{ 4} \)}
then \(∑_{x∈R }(sin( (x^2+x+5)\frac{π}{2})−cos( ( x^2+x+5)π ) ) \)is equal to ____.

Answer 1

Correct Answer: 4

Solution:

We are given that

\( S = \left\{ x \in \mathbb{R}: \sin^{-1}\left( \frac{x+1}{\sqrt{x^2 + 2x + 2}} \right) - \sin^{-1}\left( \frac{x}{\sqrt{x^2 + 1}} \right) = -\frac{\pi}{4} \right\} \).

Correct Approach:

Recognize Tangent Form:

Notice that \(\frac{x}{\sqrt{x^2 + 1}}\) and \(\frac{x+1}{\sqrt{(x+1)^2 + 1}}\) resemble the form \(\frac{u}{\sqrt{u^2 + 1}}\). Let \(u = \tan \theta\). Then \(\frac{u}{\sqrt{u^2 + 1}} = \sin \theta\). Therefore, \(\sin^{-1}\left( \frac{u}{\sqrt{u^2 + 1}} \right) = \theta = \tan^{-1} u\).

Apply Tangent Inverse:

Let \(\tan^{-1} x = A\) and \(\tan^{-1} (x+1) = B\). Then, \(\sin^{-1}\left( \frac{x}{\sqrt{x^2 + 1}} \right) = A\) and \(\sin^{-1}\left( \frac{x+1}{\sqrt{x^2 + 2x + 2}} \right) = B\). The given equation becomes: \(B - A = -\frac{\pi}{4}\).

Use Tangent Difference Formula:

We have \(B - A = \tan^{-1}(x+1) - \tan^{-1}(x) = -\frac{\pi}{4}\). Apply the tangent difference formula: \(\tan(B - A) = \frac{\tan B - \tan A}{1 + \tan A \tan B}\). Since \(\tan B = x+1\) and \(\tan A = x\), we get: \(\tan\left( -\frac{\pi}{4} \right) = \frac{(x+1) - x}{1 + x(x+1)}\). \(-1 = \frac{1}{1 + x^2 + x}\).

Solve for x:

\( -1 - x^2 - x = 1 \). \( x^2 + x + 2 = 0 \). The discriminant of this quadratic equation is \( D = b^2 - 4ac = 1^2 - 4(1)(2) = -7 \). Since \( D < 0 \), there are no real solutions for \( x \).

Conclusion:

Therefore, the set \(S\) is empty.

Final Answer:

The set \( S \) is an empty set. \( S = \emptyset \).