Question
If $$f\left( x \right) = \left( {ab - {b^2} - 2} \right)x + \int_0^x {\left( {{{\cos }^4}\,\theta + {{\sin }^4}\theta \,} \right)} d\theta $$ is a decreasing function of $$x$$ for all $$x\, \in \,R$$ and $$b\, \in \,R,\,b$$ being independent of $$x,$$ then :
A.
$$a\, \in \,\left( {0,\,\sqrt 6 } \right)$$
B.
$$a\, \in \,\left( { - \sqrt 6 ,\,\sqrt 6 } \right)$$
C.
$$a\, \in \,\left( { - \sqrt 6 ,\,0} \right)$$
D.
none of these
Answer :
$$a\, \in \,\left( { - \sqrt 6 ,\,\sqrt 6 } \right)$$
Solution :
$$\eqalign{
& f'\left( x \right) = ab - {b^2} - 2 + {\cos ^4}x + {\sin ^4}x < 0 \cr
& \Rightarrow ab - {b^2} - 2 + {\left( {{{\cos }^2}x + {{\sin }^2}x} \right)^2} - 2{\sin ^2}x.\,{\cos ^2}x < 0 \cr
& {\text{or, }}ab - {b^2} - 1 < \frac{1}{2}{\sin ^2}2x < \frac{1}{2} \cr
& {\text{or, }}2ab - 2{b^2} - 2 < 1 \cr
& {\text{or, }}2{b^2} - 2ab + 3 > 0 \cr} $$
This is true for any $$b\, \in \,R$$ if $$D < 0,$$ that is, $$4{a^2} - 4.2.3 < 0$$
$${\text{or, }}{a^2} < 6\,\,\,\,\,{\text{or, }} - \sqrt 6 < a < \sqrt 6 $$