Question
Let the function $$f\left( x \right)$$ be defined as below.
\[f\left( x \right) = \left\{ \begin{array}{l}
{\sin ^{ - 1}}\lambda + {x^2},\,0 < x < 1\\
2x,\,x \ge 1
\end{array} \right.\]
$$f\left( x \right)$$ can have a minimum at $$x=1$$ if the value of $$\lambda $$ is :
A.
1
B.
$$-1$$
C.
0
D.
none of these
Answer :
none of these
Solution :
$$\eqalign{
& \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{2\left( {1 + h} \right) - 2}}{h} = \mathop {\lim }\limits_{h \to 0} 2 = 2 \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 - h} \right) - f\left( 1 \right)}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^{ - 1}}\lambda + {{\left( {1 - h} \right)}^2} - 2}}{{ - h}} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^{ - 1}}\lambda - 1 - 2h + {h^2}}}{{ - h}} = 2{\text{ if }}{\sin ^{ - 1}}\lambda = 1,\,{\text{i}}{\text{.e}}{\text{.,}}\,\lambda = \sin \,1 \cr
& \therefore \,f\left( x \right)\,{\text{is differentiable at }}x = 1\,\,{\text{if }}\lambda = {\text{sin}}\,{\text{1}}.{\text{ Then}} \cr
& f'\left( x \right) = 2x,\,0 < x < 1{\text{ and }}f'\left( x \right) = 2,\,x \geqslant 1 \cr
& \therefore f'\left( {1 - \in } \right) = 2\left( {1 - \in } \right) > 0{\text{ and }}f'\left( {1 + \in } \right) = 2 > 0 \cr} $$
So, $$f\left( x \right)$$ does not have a maximum or a minimum at $$x=1.$$