(Each question 1 marks)

Question 11 Mark

Write the value of $\frac{\text{d}}{\text{dx}}\big\{(\text{x}+|\text{x}|)|\text{x}|\big\}$

Answer

We have, $\frac{\text{d}}{\text{dx}}\big\{(\text{x}+|\text{x}|)|\text{x}|\big\}=\begin{cases}\frac{\text{d}}{\text{dx}}\text{2x}^2&\big[\text{if }\text{x}>0\big]\\\frac{\text{d}}{\text{dx}}(-\text{x}+\text{x})\text{x}&\big[\text{if }\text{x}<0\big]\end{cases}$
$=\begin{cases}\text{4x}&[\text{if }\text{x}>0]\\\text{0}&[\text{if }\text{x}<0]\end{cases}$

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Question 21 Mark

If $\text{f}(\text{x})=\frac{\text{x}^2}{|\text{x}|},$ write $\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)$

Answer

We have, $\frac{\text{d}}{\text{dx}}\frac{\text{x}^2}{|\text{x}|}=\begin{cases}\frac{\text{d}}{\text{dx}}\text{x}&[\text{if }\text{x}>0]\\\frac{\text{d}}{\text{dx}}(-\text{x})&[\text{if }\text{x}<0]\end{cases}$
$=\begin{cases}1&[\text{if }\text{x}>0]\\-1&[\text{if }\text{x}<0]\end{cases}$

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Question 31 Mark

If $|\text{x}|<1$ and $\text{y}=1+\text{x}+\text{x}^2+\text{x}^3+\dots,$ then write the value of $\frac{\text{dy}}{\text{dx}}.$

Answer

We have,$\text{y}=1+\text{x}+\text{x}^2+\text{x}^3+\dots$
$=(1-\text{x})^{-1}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(1-\text{x})^{-1}$
$={-1}(1-\text{x})^{-2}\times(-1)$
$=(1-\text{x})^{-2}$
$=\frac{1}{(1-\text{x})^{2}}$

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Question 41 Mark

Write the value of $\frac{\text{d}}{\text{dx}}(\log|\text{x}|)$

Answer

$\frac{\text{d}}{\text{dx}}(\log|\text{x}|)=\frac{1}{|\text{x}|},\text{x}\neq0$

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Question 51 Mark

Write the value of $\frac{\text{d}}{\text{dx}}(\text{x}|\text{x}|)$

Answer

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

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Question 61 Mark

If x<2, then write the value of $\frac{\text{d}}{\text{dx}}(\sqrt{\text{x}^2-\text{4x}+4})$

Answer

$\frac{\text{d}}{\text{dx}}(\sqrt{\text{x}^2-\text{4x}+4})=\frac{\text{d}}{\text{dx}}\Big(\sqrt{(\text{x}-2)^2}\Big)$$=\frac{\text{d}}{\text{dx}}(|\text{x}-2|)$
$=\frac{\text{d}}{\text{dx}}(-(\text{x}-2))\ (\because\text{x}<2)$
$=-1$

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Question 71 Mark

Write the value of the derivative of f(x) = |x + 1| + |x - 3| at x = 2

Answer

For x = 2 |x - 1| = (x - 1) and |x - 3| = -(x - 3)$\frac{\text{d}}{\text{dx}}(|\text{x}-1|+|\text{x}-3|)$
$\frac{\text{d}}{\text{dx}}(|\text{x}-1|)+\frac{\text{d}}{\text{dx}}(|\text{x}-3|)$
$\frac{\text{d}}{\text{dx}}(\text{x}-1)+\frac{\text{d}}{\text{dx}}(-(\text{x}-3))$
$\frac{\text{d}}{\text{dx}}(\text{x}-1)+\frac{\text{d}}{\text{dx}}(3-\text{x})$
$=1-1$
$=0$

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Question 81 Mark

If $\frac{\pi}{2}<\text{x}<\pi,$ then find $\frac{\text{d}}{\text{dx}}\Bigg(\sqrt{\frac{1+\cos\text{2x}}{2}}\Bigg)$

Answer

$\frac{\text{d}}{\text{dx}}\sqrt{\frac{1+\cos\text{2x}}{2}}=\frac{\text{d}}{\text{dx}}\sqrt{\frac{\sin^2\text{x}+\cos^2\text{x}+\cos^2\text{x}-\sin^2\text{x}}{2}}$$=\frac{\text{d}}{\text{dx}}\cos\text{x}$
$=-\sin\text{x}$
$=\sin\text{x}\ \Big[\because\frac{\pi}{2}<\text{x}<\pi\Big]$

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Question 91 Mark

If f(1) = 1, f'(1) = 2, then write the value of $\lim_\limits{\text{x}\rightarrow1}\frac{\sqrt{\text{f}(\text{x})}-1}{\sqrt{\text{x}}-1}$

Answer

Givenf(1) = 1
f'(1) = 2
Now,
$\lim_\limits{\text{x}\rightarrow1}\frac{\sqrt{\text{f}(\text{x})}-1}{\sqrt{\text{x}}-1}=\lim_\limits{\text{x}\rightarrow1}\frac{\frac{\text{f}'(\text{x})}{2\sqrt{\text{f}(\text{x})}}}{\frac{1}{2\sqrt{\text{x}}}}$
$=\lim_\limits{\text{x}\rightarrow1}\frac{\text{f}'(\text{x})\sqrt{\text{x}}}{\sqrt{\text{f}(\text{x})}}$
$=\frac{\text{f}'(1)\sqrt1}{\sqrt{\text{f}(1)}}$
$=2$

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Question 101 Mark

If f(x) = |x| + |x - 1|, write the value of $\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)$

Answer

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

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Question 111 Mark

Write the value of $\lim_\limits{\text{x}\rightarrow\text{c}}\frac{\text{f}(\text{x})-\text{f}(\text{c})}{\text{x}-\text{c}}$

Answer

Since we know that,$\frac{\text{d}}{\text{dx}}(\text{f}(\text{x}))=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h}-\text{f}(\text{x}))}{\text{h}}$
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,
$\frac{\text{d}}{\text{dx}}(\text{c})=\lim_\limits{\text{x}\rightarrow\text{c}}\frac{\text{f}(\text{c}+\text{x}-\text{c})-\text{f}(\text{c})}{\text{x}-\text{c}}$
$=\lim_\limits{\text{x}\rightarrow\text{c}}\frac{\text{f}(\text{x})-\text{f}(\text{c})}{\text{x}-\text{c}}$
$=\text{f}'(\text{c})$

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Question 121 Mark

Write the value of $\lim_\limits{\text{x}\rightarrow\text{a}}\frac{\text{xf}(\text{a})-\text{af}(\text{x})}{\text{x}-\text{a}}$

Answer

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

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Question 131 Mark

Write the derivative of f(x) = 3|2 + x| at x = -3.

Answer

Here,$\text{f}(\text{x})=3|2+\text{x}|$
$\text{f}(\text{x})=\begin{cases}3(2+\text{x})&\text{at }\text{x}>-2\\-3(2+\text{x})&\text{at }\text{x}<-2\end{cases}$
Since x = -3, therefore f(x) = -3(2 + x)
f'(x) = -3

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Question 141 Mark

If $\text{f}(\text{x})=\log_{\text{x}^2}\text{x}^3$ write the value of f'(x).

Answer

Given,$\text{f}(\text{x})=\log_{\text{x}^2}\text{x}^3$
$=\frac{\log\text{x}^3}{\log\text{x}^2}$
$=\frac{3\log\text{x}}{2\log\text{x}}$
$=\frac{3}{2}$
$\text{f}'(\text{x})=0$

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MATHS (Each question 1 marks)

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