Suppose that `f(x)` is differentiable invertible function `f^(prime)(x)!=0a n dh^(prime)(x)=f(x)dot` Given that `f(1)=f^(prime)(1)=1,h(1)=0` and `g(x)` is inverse of `f(x)` . Let `G(x)=x^2g(x)-x h(g(x))AAx in Rdot` Which of the following is/are correct? `G^(prime)(1)=2` b. `G^(prime)(1)=3` c.`G^(1)=2` d. `G^(1)=3`

Options

A. G''(1) = 2
B. G'(1) = 3
C. G''(1) = 2
D. G''(1) = 3

Answer

A, D

Explanation

`h'(x)=f(x)`
`h''(x)=f'(x)`
`h(1)=0,f(1)=f'(1)=h'(1)=h''(1)=1=g(1)`
'g' is inverse of 'f'
`therefore" "f(g(x)) = x`
`rArr" "f'(g(x)).g'(x)=1`
`rArr" "f'(g(1)).g'(1)=1`
`rArr" "f'(1).g'(1)=1`
`rArr" "g'(1)=1`
`" "G(x)=x^(2)g(x)-xh(g(x))`
`G'(x)=2xg(x)+x^(2)g'(x)-g(g(x))-xh'(g(x)).g'(x)`
`=2xg(x)+x^(2)g'(x)-h(g(x))-x^(2)g'(x)`
`(because h'(x)=f(x)" "therefore h'(g(x))=f(g(x))=x)`
`=2xg(x)-h(g(x))`
`G''(x)=2g(x)+2xg'(x)-h'(g(x)).g'(x)`
`=2g(x)+2xg'(x)-f(g(x)).g'(x)`
`2g(x)+xg'(x)`
`G'(1)=2g(1)-h(g(1))=2g(1)-h(1)=2-0=2`
`G''(1)=2g(1)+g'(1)=3`