Question
The value of $$\int_1^2 {{{\left[ {f\left\{ {g\left( x \right)} \right\}} \right]}^{ - 1}}.f'\left\{ {g\left( x \right)} \right\}.g'\left( x \right)dx,} $$ where $$g\left( 1 \right) = g\left( 2 \right),$$ is equal to :
A.
1
B.
2
C.
0
D.
none of these
Answer :
0
Solution :
$$\eqalign{
& {\text{Let }}g\left( x \right) = z \cr
& {\text{Then }}I = \int_{g\left( 1 \right)}^{g\left( 2 \right)} {\frac{1}{{f\left( z \right)}}} .f'\left( z \right)dz \cr
& = \left[ {\log \,f\left( z \right)} \right]_{g\left( 1 \right)}^{g\left( 2 \right)} \cr
& = \log \,f\left\{ {g\left( 2 \right)} \right\} - \log \,f\left\{ {g\left( 1 \right)} \right\} \cr
& = 0\,\,\,\,\,\,\,\,\left[ {\because \,g\left( 1 \right) = g\left( 2 \right)} \right] \cr} $$