2 Marks Questions

Question 12 Marks

Evaluate:
$\lim\limits _{\text{x} \rightarrow 3}\frac{\text{x}^{2}-9}{\text{x}-3}$

Answer

Given that $\lim\limits_{\text{x} \rightarrow 3}\frac{\text{x}^{2}-9}{\text{x}-3}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 3}\frac{(\text{x}+3)(\text{x}-3)}{(\text{x}-3)}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 3}\text{x}+3$
Taking limit, We have
3 + 3 = 6
Hence, the answer is 6.

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Question 22 Marks

Evaluate:
$\lim\limits_{\text{x} \rightarrow 0}\frac{\sin^{2}2\text{x}}{\sin^{2}4\text{x}}$

Answer

Given that $\lim\limits_{\text{x} \rightarrow 0}\frac{\sin^{2}2\text{x}}{\sin^{2}4\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin^{2}2\text{x}}{\sin^{2}2(2\text{x})}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin^{2}2\text{x}}{4\sin^{2}2\text{x}.\cos^{2}2\text{x}}$
$=\frac{1}{4.\cos^{2}2\text{x}}$
Taking limits we have
$=\frac{1}{4.\cos^{2}0}=\frac{1}{4}$
Hence, the required is $\frac{1}{4.}$

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Question 32 Marks

Differentiate the following functions.
$(\text{ax}^{2}+\cot\text{x})(\text{p}+\text{q}\cos\text{x})$

Answer

$\frac{\text{d}}{\text{dx}}(\text{ax}^{2}+\cot\text{x})(\text{p}+\text{q}\cos\text{x})$
$=(\text{ax}^{2}+\cot\text{x})\frac{\text{d}}{\text{dx}}(\text{p}+\text{q}\cot\text{x})+(\text{p}+\text{q}\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{ax}^{2}+\cot\text{x})$
$=(\text{ax}^{2}+\cot\text{x})(-\text{q}\sin\text{x})+(\text{p}\sin\text{x})+(\text{p}+\text{q}\cos\text{x})(2\text{ax}-\text{cosec}^{2}\text{x}\big)$
Hence, the required answer is
$(\text{ax}^{2}+\cot\text{x})(-\text{q}\sin\text{x})+(\text{p}\sin\text{x})+(\text{p}+\text{q}\cos\text{x})(2\text{ax}-\text{cosec}^{2}\text{x}\big)$

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Question 42 Marks

Differentiate the following functions.
$\frac{1}{\text{ax}^{2}+\text{bx}+\text{c}}$

Answer

$\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\text{ax}^{2}+\text{bx}+\text{c}}\Big)$
$=\frac{(\text{ax}^{2}+\text{bx}+\text{c})\frac{\text{d}}{\text{dx}}(1)-1\cdot\frac{\text{d}}{\text{dx}}(\text{ax}^{2}+\text{bx}+\text{c})}{(\text{ax}^{2}+\text{bx}+\text{c})^{2}}$
$=\frac{(\text{ax}^{2}+\text{bx}+\text{c})\times0-(2\text{ax}+\text{b})}{(\text{ax}^{2}+\text{bx}+\text{c})^{2}}$
$\frac{-(2\text{ax}+\text{b})}{(\text{ax}^{2}+\text{bx}+\text{c})^{2}}$
Hence, the required answer is $\frac{-(2\text{ax}+\text{b})}{(\text{ax}^{2}+\text{bx}+\text{c})^{2}}.$

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Question 52 Marks

Differentiate the following functions.
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^{3}$

Answer

$\frac{\text{d}}{\text{dx}}\Big(\text{x}+\frac{1}{\text{x}}\Big)^{3}$
$=\frac{\text{d}}{\text{dx}}\Big(\text{x}^{3}+\frac{1}{\text{x}^{3}}+3\text{x}+\frac{3}{\text{x}}\Big)$
$=\frac{\text{d}}{\text{dx}}(\text{x}^{3}+\text{x}^{-3}+3.\text{x}^{-1})$
$=3\text{x}^{2}-3\text{x}^{-4}{+3-3.\text{x}^{-2}}$
$=3\text{x}^{2}-\frac{3}{\text{x}^{4}}+3-\frac{3}{\text{x}^{2}}$
Hence, the required answer is $3\text{x}^{2}-\frac{3}{\text{x}^{4}}+3-\frac{3}{\text{x}^{2}}.$

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Question 62 Marks

Differentiate the following functions.
$(3\text{x}+5)(1+\tan\text{x})$

Answer

$\frac{\text{d}}{\text{dx}}(3\text{x}+5)(1+\tan\text{x})$
$=(3\text{x}+5)\frac{\text{d}}{\text{dx}}(1+\tan\text{x})+(1+\tan\text{x})\frac{\text{d}}{\text{dx}}(3\text{x}+5)$
$=(3\text{x+5)}(\sec^{2}\text{x})+(1+\tan\text{x})(3)$
$=3\text{x}\sec^{2}\text{x}+5\sec^{2}\text{x}+3+3\tan\text{x}$
Hence, the required answer is $3\text{x}\sec^{2}\text{x}+5\sec^{2}\text{x}+3+3\tan\text{x}.$

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Question 72 Marks

Evaluate the following limits.
Show that $\lim\limits_{\text{x} \rightarrow 4}\frac{|\text{x}+4|}{\text{x}-4}$ does not exists.

Answer

Given $\lim\limits_{\text{x} \rightarrow 4}\frac{|\text{x}+4|}{\text{x}-4}$
$\text{L}.\text{H}.\text{L}=\lim\limits_{\text{x} \rightarrow 4}\frac{-(\text{x}+4)}{\text{x}-4}=-1$
$\text{R}.\text{H}.\text{L}=\lim\limits_{\text{x} \rightarrow 4}\frac{\text{x}-4}{\text{x}-4}=1$
Since $\text{L}.\text{H}.\text{L}\neq\text{R}.\text{H}.\text{L}$
Hence, the limit does not exist.

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Question 82 Marks

Find 'n' if $\lim\limits_{\text{x} \rightarrow 2}\frac{\text{x}^{\text{n}}-2^{\text{n}}}{\text{x}-2}=80,\text{x}\in\text{N}$

Answer

Given that $\lim\limits_{\text{x} \rightarrow 2}\frac{\text{x}^{\text{n}}-2^{\text{n}}}{\text{x}-2}=80$
$=\text{n}.(2)^{\text{n}-1}=80$
$=\text{n}\times2^{\text{n}-1}=5\times(2)^{5-1}$
$\therefore\text{n}=5$
Hence, the required answer is n = 5.

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Question 92 Marks

Evaluate:
$\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}-2\sin3\text{x}+\sin5\text{x}}{\text{x}}$

Answer

Given that $\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}-2\sin3\text{x}+\sin5\text{x}}{\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\text{x}}-\frac{2\sin3\text{x}}{\text{x}}+\frac{\sin5\text{x}}{\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\text{x}}-\lim\limits_{3\text{x} \rightarrow 0}2\big(\frac{\sin3\text{x}}{3\text{x}}\big)\times3+\lim\limits_{5\text{x} \rightarrow 0}\big(\frac{\sin5\text{x}}{5\text{x}}\big)\times5$
$=1-6+5=0$
Hence, the required answer is 0.

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Question 102 Marks

Differentiate the following functions.
$\frac{\text{x}^{4}+\text{x}^{3}+\text{x}^{2}+1}{\text{x}}$

Answer

$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^{4}+\text{x}^{3}+\text{x}^{2}+1}{\text{x}}\Big)$
$=\frac{\text{d}}{\text{dx}}\Big(\text{x}^{3}+\text{x}^{2}+\text{x}+\frac{1}{\text{x}}\Big)$
$=3\text{x}^{2}+2\text{x}^{2}+1-\frac{1}{\text{x}^{2}}$
Hence, the required answer is $3\text{x}^{2}+2\text{x}^{2}+1-\frac{1}{\text{x}^{2}}.$

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Question 112 Marks

Differentiate the following functions.
$\text{x}^{2}\sin\text{x}+\cos2\text{x}$

Answer

$\frac{\text{d}}{\text{dx}}(\text{x}^{2}\sin\text{x}+\cos2\text{x})$
$=\frac{\text{d}}{\text{dx}}(\text{x}^{2}\sin\text{x})+\frac{\text{d}}{\text{dx}}(\cos2\text{x})$
$=(\text{x}^{2}\cos\text{x}+\sin\text{x}.2\text{x})+(-2\sin2\text{x})$
$=\text{x}^{2}\cos\text{x}+2\text{x}+2\text{x}\sin\text{x}-2\sin2\text{x}$
Hence, the required answer is $\text{x}^{2}\cos\text{x}+2\text{x}+2\text{x}\sin\text{x}-2\sin2\text{x}.$

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Question 122 Marks

Evaluate:
$\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos2\text{x}}{\text{x}^{2}}$

Answer

Given that $\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos2\text{x}}{\text{x}^{2}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin^{2}\text{x}}{\text{x}^{2}}$
$=\lim\limits_{\text{x} \rightarrow 0}2\big(\frac{\sin\text{x}}{\text{x}}\big)^{2}$
$=2\times1=2$
Hence, the required answer is 2.

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Question 132 Marks

Evaluate:
$\lim\limits_{\text{x} \rightarrow\frac{\pi}{4}}\frac{\sin\text{x}-\cos\text{x}}{\text{x}-\frac{\pi}{4}}$

Answer

Given that $\lim\limits_{\text{x} \rightarrow\frac{\pi}{4}}\frac{\sin\text{x}-\cos\text{x}}{\text{x}-\frac{\pi}{4}}$
$=\lim\limits_{\text{x} \rightarrow\frac{\pi}{4}}\frac{\sqrt{2}\Big(\frac{1}{\sqrt{2}}\sin\text{x}-\frac{1}{\sqrt{2}}\cos\text{x}\Big)}{\text{x}-\frac{\pi}{4}}$
$=\lim\limits_{\text{x} \rightarrow\frac{\pi}{4}}\frac{\sqrt{2}\Big(\cos\frac{\pi}{4}\sin\text{x}-\sin\frac{\pi}{4}\cos\text{x}\Big)}{\text{x}-\frac{\pi}{4}}$
$=\lim\limits_{\text{x} \rightarrow\frac{\pi}{4}}\frac{\sqrt{2}\sin\Big(\text{x}-\frac{\pi}{4}\Big)}{\text{x}-\frac{\pi}{4}}$
$=\sqrt{2}.1=\sqrt{2}$
Hence, the required answer is $\sqrt{2}.$

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