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And \( R f^{\prime}(0)=\lim _{h \rightarrow 0.0} \frac{f(0+h)-f(0)}{h} \) where \( f(x)=x \) \[ =\lim _{h \rightarrow 0+0} \frac{0+h-0}{h}=1 \rightarrow(2) \] Hence from (1) and (2) \( L f^{\prime}(0) \neq R f^{\prime}(0) \) \( \therefore f^{\prime}(0) \) does not exists. Q.6. Let \( f(x)= \) at +1 , If \( x \geq 1 \) \[ =x^{2}+a \text { if } x<1 \] Find ' \( a \) ' if \( f(x) \) is derivable at \( x=1 \) Sol: \( L f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \rightarrow(1) \) where \( f(x), x \) a \[ \begin{array}{l} f(1)=1 \cdot a \text { and } f(1+h)=(1+h)^{2}+a \\ \begin{array}{r} \therefore(1) \Rightarrow L f^{\prime}(1)=\lim _{h \rightarrow 0-0} \frac{f(1+h)^{2}+a-(1+a)}{h} \\ =\lim _{h \rightarrow 0-0} \frac{1+h^{2}+2 h+a-1-a}{h} \\ =\lim _{h \rightarrow 0-0} \frac{h(h+2)}{h}=2 \rightarrow(A) \end{array} \\ \text { Rf } \begin{array}{r} \text { vhere } f(x)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \rightarrow(2) \\ f(1)=a+1 \text { and } f(1+h)=a(1+h)+1 \\ (2) \rightarrow R f^{\prime}(1)=\lim _{h \rightarrow 0 \rightarrow 0} \frac{a+a h+1-a-1}{h} \end{array} \end{array} \]

          And \( R f^{\prime}(0)=\lim _{h \rightarrow 0.0} \frac{f(0+h)-f(0)}{h} \)
where \( f(x)=x \)
\[
=\lim _{h \rightarrow 0+0} \frac{0+h-0}{h}=1 \rightarrow(2)
\]

Hence from (1) and (2) \( L f^{\prime}(0) \neq R f^{\prime}(0) \)
\( \therefore f^{\prime}(0) \) does not exists.
Q.6. Let \( f(x)= \) at +1 , If \( x \geq 1 \)
\[
=x^{2}+a \text { if } x<1
\]

Find ' \( a \) ' if \( f(x) \) is derivable at \( x=1 \)
Sol: \( L f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \rightarrow(1) \)
where \( f(x), x \) a
\[
\begin{array}{l}
f(1)=1 \cdot a \text { and } f(1+h)=(1+h)^{2}+a \\
\begin{array}{r}
\therefore(1) \Rightarrow L f^{\prime}(1)=\lim _{h \rightarrow 0-0} \frac{f(1+h)^{2}+a-(1+a)}{h} \\
=\lim _{h \rightarrow 0-0} \frac{1+h^{2}+2 h+a-1-a}{h} \\
=\lim _{h \rightarrow 0-0} \frac{h(h+2)}{h}=2 \rightarrow(A)
\end{array} \\
\text { Rf } \begin{array}{r}
\text { vhere } f(x)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \rightarrow(2) \\
f(1)=a+1 \text { and } f(1+h)=a(1+h)+1 \\
(2) \rightarrow R f^{\prime}(1)=\lim _{h \rightarrow 0 \rightarrow 0} \frac{a+a h+1-a-1}{h}
\end{array}
\end{array}
\]

And R f^'(0)=limh → 0.0(f(0+h)-f(0))/(h)
where f(x)=x

=limh → 0+0(0+h-0)/(h)=1 →(2)

Hence from (1) and (2) L f^'(0) ≠ R f^'(0)
∴ f^'(0) does not exists.
Q.6. Let f(x)= at +1 , If x ≥ 1

=x^2+a if x<1

Find ' a ' if f(x) is derivable at x=1
Sol: L f^'(1)=limh → 0(f(1+h)-f(1))/(h)→(1)
where f(x), x a

f(1)=1 · a and f(1+h)=(1+h)^2+a

∴(1) ⇒ L f^'(1)=limh → 0-0(f(1+h)^2+a-(1+a))/(h)

=limh → 0-0(1+h^2+2 h+a-1-a)/(h)

=limh → 0-0(h(h+2))/(h)=2 →(A)

Rf
vhere f(x)=limh → 0(f(1+h)-f(1))/(h)→(2)

f(1)=a+1 and f(1+h)=a(1+h)+1

(2) → R f^'(1)=limh → 0 → 0(a+a h+1-a-1)/(h)

Added by Kaleem S.

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