The value of 'a' for which
ax2+sin−1x2−2x+2+cos−1x2−2x+2=0 has a real
solution, is
π2
−π2
2π
−2π
We have,
x2−2x+2=(x−1)2+1≥1.
So, the given equation is meaningful for x = 1.
Putting x = 1, we have
a+sin−1(1)+cos−1(1)=0⇒a=−π2