Solution
Step 1: Find the interval (a,b)
Using the identity sin−1(y)+cos−1(y)=2π, we can rewrite the inequality:
\sin^{-1}(\sin \theta) - \left( \frac{\pi}{2} - \sin^{-1}(\sin \theta) \right) > 0
2\sin^{-1}(\sin \theta) > \frac{\pi}{2}
\sin^{-1}(\sin \theta) > \frac{\pi}{4}
In the interval (0,2π), \sin^{-1}(\sin \theta) > \frac{\pi}{4} holds when \frac{\pi}{4} < \theta < \frac{3\pi}{4}.
Therefore, a=4π and b=43π.
The difference is: b−a=43π−4π=2π.
Step 2: Analyze the second equation
The equation is given as:
αx2+βx+[sin−1(x2−6x+10)+cos−1(x2−6x+10)]=0
For the inverse functions to be defined, the argument must satisfy −1≤x2−6x+10≤1.
Completing the square: x2−6x+10=(x−3)2+1.
Since (x−3)2≥0, the minimum value of (x−3)2+1 is 1 (occurring at x=3).
Thus, the only possible value for x is 3.
At x=3, the term in the brackets becomes sin−1(1)+cos−1(1)=2π.
The equation becomes:
9α+3β=−2π⟹3α+β=−6π— (Equation 1)
Step 3: Solve for α
We are given α−β=b−a=2π— (Equation 2).
Adding Equation 1 and Equation 2:
Correct Option: (3) 12π
