Let `g: R -> R` be a differentiable function with `g(0) = 0,,g'(1)!=0`.Let `f(x)={x/|x|g(x), 0 !=0 and 0,x=0` and `h(x)=e^(|x|)` for all `x in R`. Let `(foh)(x)` denote `f(h(x)) and (hof)(x)` denote `h(f(x))`. Then which of the following is (are) true?

A. f is differentiable at x = 0
B. h is differentiable at x = 0
C. f o h is differentiable at x = 0
D. h o f is differentiable at x = 0

Options

A. f is differentiable at ` x = 0`
B. h is differentiable at ` x = 0 `
C. foh is differentiable at x = 0
D. hofis differentiable at ` x = 0 `

Answer

A, D

Explanation

We have,
`h(x) = e^(|x|) = {{:(e^(-x),,"," x lt 0 ), ( e ^(x),, "," x ge 0 ):}`
` therefore " " ` (LHD of h (x) at x = 0 ) = ` lim_( x to 0 ^(-)) ( h (x) - h (0))/( x - 0 )`
`" " = lim_( x to 0 ^(-)) (e ^(-x) - 1 )/( x )`
` " " = - lim_( x to 0 ^(-)) ( e ^(x) - 1 )/(- x ) = - 1 `
and, (RHD of h (x) at x = 0 ) = `lim_( x to 0 ^(+)) ( h (x) - h (0))/(x - 0 )`
` " " = lim_(x to 0 ^( +)) ( e ^(x) - 1 )/( x ) = 1 `
` therefore `( LHD at f (x) at x = 0 ) `ne ` ( RHD of h (x ) at x = 0 )
So, h (x) is not differentiable at x = 0 and hence option (b) is not true.
We have,
`f(x) = {{:((x)/(|x|)g (x)",", x ne 0 ) , ( 0",", x = 0):} or , f (x) = {{:( g (x),"," x gt 0 ) , ( 0, "," x = 0 ) ,(- g (x),"," x lt 0 ):}`
` therefore ` (LHD of f(x) at x = 0 ) ` = lim_(x to 0 ^(-)) (f (x) - f (0))/(x - 0 )`
` " " = lim_( x to 0 ^(-)) (- g (x))/(x)" " [ because f (0) = 0 ] `
`" " = - lim_( x to 0 ^(-)) (g (x) - g (0)) /(x - 0 ) " "[ because g (0) =0 ]`
` " " = - g ' (0) = 0 `
` therefore ` (LHD of f(x) at x = 0 ) ` = lim_( x to 0 ^(+)) (f (x) - f (0))/( x - 0 )`
` " " = lim_( x to 0 ^(+)) (f (x))/( x )`
` " " = lim_( x to 0 ^(+)) (g(x) - g (0))/(x - 0 ) = g'(0) = 0 `
`therefore ` ( LHD of f(x) at x = 0 ) = ( RHD of f(x) at x = 0 )
So, f (x) is differentiable at x = 0.
So, option (a) is true
We have,
` f(x) = {{:((x)/(|x|) g (x), "," x ne 0 ) , ( 0 ,"," x = 0 ):} `
` rArr f (x) = {{:(- g (x),"," x lt 0 ), ( 0 , "," x = 0 ) ,( g (x), "," x gt 0 ):}`
and , ` h (x) = e ^(|x|) gt 0 ` for all ` x in R`.
` f (h (x)) = g ( e ^( |x|)) ` for all ` x in R`.
`rArr f ( h (x)) = {{:(g (e ^(-x)), "," x lt 0 ) , ( g (1), "," x = 0 ), ( g (e^(x)), ","x gt 0 ):} `
(LHD of foh(x) at x = 0 =`lim_( x to 0 ^(-)) (f ( h(x)) - f ( h (0)))/( x - 0 )`
`" " = lim_( x to 0 ^(-)) (g ( e ^(-x)) - g (1))/( x )`
` " " = lim_( x to 0 ^(-)) ( g ( e ^( - x )) - g (1))/(e ^(- x ) - 1 ) xx ( e^(-x) - 1 )/( x )`
` " " = g ' (1) xx (-1) = - g ' (1)`
(RHD of foh(x) at x = 0 ) `= lim _( x to 0 ^(+)) (f ( h (x)) - f ( h (0)))/(x - 0 )`
`" " = lim_( x to 0 ^(+)) ( g( e ^(x)) - g (1))/( e ^(x) - 1) xx ( e ^(x) - 1 ) /( x )`
` " " = g ' (1) xx 1 = g ' (1)`
` therefore ` (LHD of foh(x) at x =0 ) =`lim_( x to 0 ^(+)) (f ( h (x)) - f ( h (0)))/(x - 0 ) `
` " " = lim_(x to 0 ^(+)) (g ( e ^(x)) - g (1))/( e ^(x) - 1 ) xx ( e ^(x) - 1 ) /( x )`
`therefore ` ( LHD of foh(x) at x = 0 )` ne ` ( RHD of foh (x) at x = 0 )
So, foh is not differentiable at x = 0. Hence, option (c) is not true.
Again,
` h (x) = e ^( |x|) and f(x) = {{:( (x)/(|x|) g (x), ", "x ne 0 ) , ( 0, "," x = 0):} `
` rArr hof (x ) = h ( f (x)) = {{:( e ^(|f(x)|, "," x ne 0 ) , ( e ^(0) = 1, "," x = 0 ):}`
` rArr hof (x) = {{:( e ^( | f (x)|, "," x ne 0 ), (1, "," x = 0 ):}`
` therefore ` (LHD of hof at x = 0 )
`" " = lim_( x to 0 ^(-))( hof (x ) - hof(0))/(x - 0 ) `
`= lim_( x to 0 ^(-)) ( e ^( | g(x)|) - 1 )/( x )`
` = lim_( x to 0 ^(-)) (e ^(|g (x)|) - 1 ) /(| g (x)|) xx (| g (x)|)/( x )`
` lim_( x to 0 ^(-)) ( e ^(| g (x)| - 1 ) /(|g (x)|) xx |(g (x) - g (0))/( x - 0 )| xx( |x -0|)/(x) `
` = 1 xx g ' (0) xx (-1) = 0 `
( RHD of hof at x = 0 )
`= lim_( x to 0 ^(+)) ( hof (x) - hof(0))/( x - 0 ) `
` = lim _( x to 0 ^(+))( h (f(x)) - h ( f (0)))/(x)`
` = lim_( x to 0 ^(+)) = ( e ^(|g(x)|) - 1 ) /( |g (x)|) xx (|g(x)|)/(x)`
` = lim_(x to 0 ^(+ )) ( e ^(|g(x)|) - 1 )/(| g (x)|) xx |(g (x) - g(0))/(x - 0 ) | xx ( |x|)/(x) `
`= 1 xx|g'(0)|xx 1 = 0 `
So, hof is differentiable at x = 0 . Consequently, option (d) is true.