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Q. Let $f(x)=x^3-6 \sin \theta \cdot x^2+\left(2 \sin ^2 \theta+q\right) x-1, q, \theta \in R$ be a cubic polynomial in $x$. If exaclty one tangent can be drawn to the graph of $y=f(x)$ which is parallel to the line $y=x$, then
if q is smallest, then the value of $\left.\frac{ d }{ dx }\left( f ^{-1}( x )\right)\right|_{ x =9}$ is

Application of Derivatives

Solution:

Smallest value of $q =1$ when $\sin \theta=0$
Now,
$f ( x )= x ^3+ x -1 $
$f (2)=9 \Rightarrow f ^{-1}(9)=2 $
$\left.\frac{ d }{ dx }\left( f ^{-1}( x )\right)\right|_{ x =9}=\frac{1}{ f ^{\prime}(2)}=\frac{1}{13}\left\{\Theta f ^{\prime}( x )=3 x ^2+1\right\}$