The correct option is A 9
sin−1cos(2x2+10|x|+4x2+5|x|+3)=cot(cot−1(2−18|x|9|x|))+π2⇒−(π2−sin−1cos(2x2+10|x|+4x2+5|x|+3))=2−18|x|9|x|⇒−cos−1cos(2x2+10|x|+6−2x2+5|x|+3)=29|x|−18|x|9|x|⇒−2(x2+5|x|+3)(x2+5|x|+3)+2(x2+5|x|+3)=29|x|−2or,9|x|=x2+5|x|+3[Assumingx≠0]x2+4|x|+3=0x2−3|x|−|x|+3=0(|x|−3)(|x|−1)=0|x|=3or,|x|=1x=3,−3&x=1,−1
Product of all real value =3×(−3)×1×(−1)=9