Let g:R→R be a differentiable function with g0=0,g10=0 and .g11≠0 Let fx=xxgx,x≠00,x=0 And hx=ex for all x∈R . Let fohx denote fhx and hofx denote hfx. Then which of the following is (are) true?

1) f1(x)=limx→0f(x)-f(0)x =limx→0x|x|g(x)-0x =limx→0g(x)|x|Now f1(0+)=limx→0g(x)|x|=g1(0)1=0f1(0-)=limx→0g(x)-x=g1(0)-1=0∴f is differentiable at x=0 2) hx=ex=ex,x≥0e−x,x 0−e−x,x<0⇒h1(0+)=1,h1(0-)=-1∴h is not differentiable at x=0 3) fohx=fexNow limx→0+fohx−foh0x=limx→0+fex−f1x=limx→0+gex−g1x =limx→0+gex−g1ex−1ex−1x=g11 →1Also limx→0−fohx−foh0x=limx→0−fe−x−f1x=limx→0−ge−x−g1x =limx→0−ge−x−g1e−x−1e−x−1x=−g11 →2By 1 and 2, fohx is not differentiable at x=04) hof x=h−gx, x 0 1, x=0⇒hof x=egx,x 01,x=0∴( hof )1(x)=egxg1(x) =0, at x=0∵g1(0)=0∴hofx is differentiable at x=0

Let g:R→R be a differentiable function with g0=0,g10=0 and .g11≠0 Let fx=xxgx,x≠00,x=0 And hx=ex for all x∈R . Let fohx denote fhx and hofx denote hfx. Then which of the following is (are) true?

1) f1(x)=limx→0f(x)-f(0)x =limx→0x|x|g(x)-0x =limx→0g(x)|x|Now f1(0+)=limx→0g(x)|x|=g1(0)1=0f1(0-)=limx→0g(x)-x=g1(0)-1=0∴f is differentiable at x=0 2) hx=ex=ex,x≥0e−x,x 0−e−x,x<0⇒h1(0+)=1,h1(0-)=-1∴h is not differentiable at x=0 3) fohx=fexNow limx→0+fohx−foh0x=limx→0+fex−f1x=limx→0+gex−g1x =limx→0+gex−g1ex−1ex−1x=g11 →1Also limx→0−fohx−foh0x=limx→0−fe−x−f1x=limx→0−ge−x−g1x =limx→0−ge−x−g1e−x−1e−x−1x=−g11 →2By 1 and 2, fohx is not differentiable at x=04) hof x=h−gx, x 0 1, x=0⇒hof x=egx,x 01,x=0∴( hof )1(x)=egxg1(x) =0, at x=0∵g1(0)=0∴hofx is differentiable at x=0

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Let $g : R \to R$ be a differentiable function with $g (0) = 0, g^{1} (0) = 0$ and . $g^{1} (1) \neq 0$ Let $f (x) = \{\begin{matrix} \frac{x}{|x|} g (x), & x \neq 0 \\ 0, & x = 0 \end{matrix}$
And $h (x) = e^{|x|}$ for all $x \in R$ . Let $(f o h) (x)$ denote $f (h (x))$ and $(h o f) (x)$ denote $h (f (x))$ . Then which of the following is (are) true?

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$f ο h$ is differentiable at $x = 0$

) f is differentiable at $x = 0$

h is differentiable at $x = 0$

$h ο f$ is differentiable at $x = 0$

answer is A, D.

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Detailed Solution

$\begin{array}{l} 1) f^{1} (x) = \lim_{x \to 0} \frac{f (x) - f (0)}{x} \\ = \lim_{x \to 0} \frac{\frac{x}{| x |} g (x) - 0}{x} \\ = \lim_{x \to 0} \frac{g (x)}{| x |} \\ N o w f^{1} (0 +) = \lim_{x \to 0} \frac{g (x)}{| x |} = \frac{g^{1} (0)}{1} = 0 \\ f^{1} (0 -) = \lim_{x \to 0} \frac{g (x)}{- x} = \frac{g^{1} (0)}{- 1} = 0 \\ ∴ f is differentiable at x = 0 \end{array}$

$\begin{array}{l} 2) h (x) = e^{|x|} = \{\begin{cases} e^{x}, x \geq 0 \\ e^{- x}, x < 0 \end{cases} \\ \Rightarrow h^{1} (x) = \{\begin{cases} e^{x}, x > 0 \\ - e^{- x}, x < 0 \end{cases} \\ \Rightarrow h^{1} (0 +) = 1, h^{1} (0 -) = - 1 \\ ∴ h is not differentiable at x = 0 \end{array}$

$\begin{array}{l} 3) f o h (x) = f (e^{|x|}) \\ N o w \lim_{x \to 0^{+}} \frac{f o h (x) - f o h (0)}{x} = \lim_{x \to 0^{+}} \frac{f (e^{x}) - f (1)}{x} = \lim_{x \to 0^{+}} \frac{g (e^{x}) - g (1)}{x} \\ = \lim_{x \to 0^{+}} \frac{g (e^{x}) - g (1)}{e^{x} - 1} \frac{e^{x} - 1}{x} = g^{1} (1) \to (1) \\ A l s o \lim_{x \to 0^{-}} \frac{f o h (x) - f o h (0)}{x} = \lim_{x \to 0^{-}} \frac{f (e^{- x}) - f (1)}{x} = \lim_{x \to 0^{-}} \frac{g (e^{- x}) - g (1)}{x} \\ = \lim_{x \to 0^{-}} \frac{g (e^{- x}) - g (1)}{e^{- x} - 1} \frac{e^{- x} - 1}{x} = - g^{1} (1) \to (2) \\ B y (1) a n d (2), f o h (x) i s n o t d i f f e r e n t i a b l e a t x = 0 \end{array}$

$\begin{array}{l} 4) h o f (x) = \{\begin{cases} h [- g (x)], x < 0 \\ h [g (x)], x > 0 \\ 1, x = 0 \end{cases} \\ \Rightarrow h o f (x) = \{\begin{cases} e^{|g (x)|}, x < 0 \\ e^{|g (x)|}, x > 0 \\ 1, x = 0 \end{cases} \\ ∴ (hof)^{1} (x) = e^{|g (x)|} (g^{1} (x)) \\ = 0, a t x = 0 (∵ g^{1} (0) = 0) \\ ∴ h o f (x) i s differentiable at x = 0 \end{array}$