The correct option is A ϕ
Given log√3+∣∣x2−1∣∣(3−x2−1x2)=log√3+∣∣x2−1∣∣(5+x2−x)
⇒3−x2−1x2=5+x2−x
⇒2x4−x3+2x2+1=0
⇒x2(2x2−x+2)=−1 - (i)
Now, for 2x2−x+2, the discriminant =1−4×2×2=−15
Since the discriminant is negative, the equation has non real roots.
Hence, the RHS of (i) is always positive for real values of x, while the LHS is negative.
Hence, there is no real solution .
Hence, option A is correct.