MATHS 1 Marks Question

$\frac{\text{d}}{\text{dx}}\frac{\text{x}^2}{|\text{x}|}=\begin{cases}\frac{\text{d}}{\text{dx}}\text{x}&[\text{if }\text{x}>0]\\\frac{\text{d}}{\text{dx}}(-\text{x})&[\text{if }\text{x}<0]\end{cases}$

$=\begin{cases}1&[\text{if }\text{x}>0]\\-1&[\text{if }\text{x}<0]\end{cases}$

$\frac{\text{d}}{\text{dx}}(|\text{x}-1|+|\text{x}-3|)$

$\frac{\text{d}}{\text{dx}}(|\text{x}-1|)+\frac{\text{d}}{\text{dx}}(|\text{x}-3|)$

$\frac{\text{d}}{\text{dx}}(\text{x}-1)+\frac{\text{d}}{\text{dx}}(-(\text{x}-3))$

$\frac{\text{d}}{\text{dx}}(\text{x}-1)+\frac{\text{d}}{\text{dx}}(3-\text{x})$

$=1-1$

$=0$

$\text{y}=1+\text{x}+\text{x}^2+\text{x}^3+\dots$

$=(1-\text{x})^{-1}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(1-\text{x})^{-1}$

$={-1}(1-\text{x})^{-2}\times(-1)$

$=(1-\text{x})^{-2}$

$=\frac{1}{(1-\text{x})^{2}}$

$=\frac{\text{d}}{\text{dx}}\cos\text{x}$

$=-\sin\text{x}$

$=\sin\text{x}\ \Big[\because\frac{\pi}{2}<\text{x}<\pi\Big]$

When x > 1

$\text{f}(\text{x})=\text{x}+\text{x}-1=\text{2x}-1$

$\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=2$

When 0 < x < 1

$\text{f}(\text{x})=\text{x}-\text{x}+1=1$

$\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=0$

When x < 0

$\text{f}(\text{x})=-\text{x}-\text{x}+1=-\text{2x}+1$

$\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=-2$

$\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\begin{cases}2,&\text{x}>1\\0,&0<\text{x}<1\\-2,&\text{x}<0\end{cases}$

$\frac{\text{d}}{\text{dx}}(\text{f}(\text{x}))=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h}-\text{f}(\text{x}))}{\text{h}}$

Let,

$\text{h}=\text{x}-\text{c}$ and $\text{f}(\text{x})=\text{c}$

If $\text{h}\rightarrow0$ then $\text{x}\rightarrow\text{c}$

Therefore,

$\frac{\text{d}}{\text{dx}}(\text{c})=\lim_\limits{\text{x}\rightarrow\text{c}}\frac{\text{f}(\text{c}+\text{x}-\text{c})-\text{f}(\text{c})}{\text{x}-\text{c}}$

$=\lim_\limits{\text{x}\rightarrow\text{c}}\frac{\text{f}(\text{x})-\text{f}(\text{c})}{\text{x}-\text{c}}$

$=\text{f}'(\text{c})$

$\lim_\limits{\text{x}\rightarrow\text{a}}\frac{\text{f}(\text{a})(\text{x}-\text{a})-\text{a}(\text{f}(\text{x})-\text{f}(\text{a}))}{\text{x}-\text{a}}$

$=\lim_\limits{\text{x}\rightarrow\text{a}}\frac{\text{f}(\text{a})(\text{x}-\text{a})}{(\text{x}-\text{a})}-\text{a}\lim_\limits{\text{x}\rightarrow\text{a}}\frac{\text{f}(\text{x})-\text{f}(\text{a})}{\text{x}-\text{a}}$

$=\text{f}(\text{a})-\text{a}\text{f}'(\text{a})$

$=\frac{\text{d}}{\text{dx}}(|\text{x}-2|)$

$=\frac{\text{d}}{\text{dx}}(-(\text{x}-2))\ (\because\text{x}<2)$

$=-1$

f(1) = 1

f'(1) = 2

Now,

$\lim_\limits{\text{x}\rightarrow1}\frac{\sqrt{\text{f}(\text{x})}-1}{\sqrt{\text{x}}-1}=\lim_\limits{\text{x}\rightarrow1}\frac{\frac{\text{f}'(\text{x})}{2\sqrt{\text{f}(\text{x})}}}{\frac{1}{2\sqrt{\text{x}}}}$

$=\lim_\limits{\text{x}\rightarrow1}\frac{\text{f}'(\text{x})\sqrt{\text{x}}}{\sqrt{\text{f}(\text{x})}}$

$=\frac{\text{f}'(1)\sqrt1}{\sqrt{\text{f}(1)}}$

$=2$

$\text{f}(\text{x})=\log_{\text{x}^2}\text{x}^3$

$=\frac{\log\text{x}^3}{\log\text{x}^2}$

$=\frac{3\log\text{x}}{2\log\text{x}}$

$=\frac{3}{2}$

$\text{f}'(\text{x})=0$

$\text{f}(\text{x})=3|2+\text{x}|$

$\text{f}(\text{x})=\begin{cases}3(2+\text{x})&\text{at }\text{x}>-2\\-3(2+\text{x})&\text{at }\text{x}<-2\end{cases}$

Since x = -3, therefore f(x) = -3(2 + x)

f'(x) = -3

$\frac{\text{d}}{\text{dx}}\big\{(\text{x}+|\text{x}|)|\text{x}|\big\}=\begin{cases}\frac{\text{d}}{\text{dx}}\text{2x}^2&\big[\text{if }\text{x}>0\big]\\\frac{\text{d}}{\text{dx}}(-\text{x}+\text{x})\text{x}&\big[\text{if }\text{x}<0\big]\end{cases}$

$=\begin{cases}\text{4x}&[\text{if }\text{x}>0]\\\text{0}&[\text{if }\text{x}<0]\end{cases}$

$\frac{\text{d}}{\text{dx}}(\text{x}|\text{x}|)=\begin{cases}\frac{\text{d}}{\text{dx}}\text{x}^2&\big[\text{if }\text{x}>0\big]\\\frac{\text{d}}{\text{dx}}(-\text{x})\text{x}&\big[\text{if }\text{x}<0\big]\end{cases}$

$=\begin{cases}\text{2x}&[\text{if }\text{x}>0]\\-\text{2x}&[\text{if }\text{x}<0]\end{cases}$