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Need solution for rd sharma maths class 12 chapter 9 Differentiability exercise Multiple choice question, question 15

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HINTS: As x=1 put LHD=RHD

GIVEN: f(x)=x-[x]


f(x)= \begin{cases}a x^{2}+1 & x>1 \\ x+\frac{1}{2} & x \leq 1\end{cases}

As f(x) is derivable at x=1

LHD=RHD   at x=1

\begin{aligned} \lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} &=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} \\ \end{aligned}

\begin{aligned} \lim _{h \rightarrow 0} \frac{f(1-h)-f(x)}{-h} &=\lim _{x \rightarrow 0} \frac{a(1+h)^{2}+1-(h)}{h} \\ \end{aligned}

\begin{aligned}\lim _{x \rightarrow 0} \frac{-h}{h} &=\lim _{h \rightarrow 0} \frac{a\left[1+h^{2}+2 h\right]-1-\frac{3}{2}}{h} \\ \end{aligned}

\begin{aligned} &1 =\lim _{h \rightarrow 0} \frac{a\left[1+h^{2}+2 h\right]-\frac{1}{2}}{h} \\ &1-2 =\lim _{h \rightarrow 0} \frac{a-\frac{1}{2}}{h} \\ &-1=\lim _{h \rightarrow 0} \frac{a-\frac{1}{2}}{h} \end{aligned}

\begin{aligned} &\lim _{x \rightarrow 1} \frac{x+\frac{1}{2}-\frac{3}{2}}{x-1}=\lim _{x \rightarrow 1} \frac{a x^{2}+1-\frac{3}{2}}{x-1} \\ \end{aligned}

\begin{aligned}&\lim _{x \rightarrow 1} \frac{x-1}{x-1}=\lim _{x \rightarrow 1} \frac{a x^{2}-\frac{1}{2}}{x-1} \\ \end{aligned}

\begin{aligned} &\pm=\lim _{x \rightarrow 1} \frac{a x^{2}-\frac{1}{2}}{x-1}\end{aligned}

As , at x=1 denominator becomes 0.So f limits exist so it must be \frac{0}{0} form

\begin{aligned} &1=\lim _{x \rightarrow 1} \frac{2 a x}{1} \\ &1=2 a \times 1 \\ &2 a=1 \\ &a=\frac{1}{2} \end{aligned}

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