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Maths 5 Marks Questions

Limits and Derivatives · CBSE STD 11 Science (CBSE - English Medium). 15 questions.

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Question 15 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin\text{x}-\sin2\text{x}}{\text{x}^{3}}$ 
Answer
Given that $\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin\text{x}-\sin2\text{x}}{\text{x}^{3}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin\text{x}-2\sin\text{x}\cos\text{x}}{\text{x}^{3}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin\text{x}-(1-\cos\text{x})}{\text{x}^{3}}$
 $=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin\text{x}}{\text{x}}\Big(\frac{1-\cos\text{x}}{\text{x}^{2}}\Big)$
$=\lim\limits_{\text{x} \rightarrow 0}2\Big(\frac{\sin\text{x}}{\text{x}}\Big)\bigg(\frac{\frac{2\sin^{2}\text{x}}{2}}{\text{x}^{2}}\bigg)$
$=\lim\limits_{\text{x} \rightarrow 0}2\Big(\frac{\sin\text{x}}{\text{x}}\Big)\Bigg(2\frac{\sin^{2}\frac{\text{x}}{2}}{\frac{\text{x}^{4}}{4}}\times\frac{1}{4}\Bigg)$
 $=\lim\limits_{\text{x} \rightarrow 0}2\Big(\frac{\sin\text{x}}{\text{x}}\Big)\Bigg(2\frac{\sin^{2}\frac{\text{x}}{2}}{\frac{\text{x}^{4}}{4}}\Bigg)^{2}.\frac{1}{4}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{4}{4}\Big(\frac{\sin\text{x}}{\text{x}}\Big)\lim\limits_{\frac{\text{x}}{2} \rightarrow 0}\Bigg(\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\Bigg)$
$1.1.(1)^{2}=1$
Hence, the required answer is 1.
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Question 25 Marks
Differentiate the functions with respect to 'x'.
$\text{x}\cos\text{x}$
Answer
Let $\text{y}=\text{x}\cos\text{x}\ ...(\text{i})$
$\text{y}+\Delta\text{y}=(\text{x}+\Delta\text{x})\cos(\text{x}+\Delta\text{x})\ ...(\text{ii})$
Subtracting eq. (i) from eq. (ii)
$\text{y}+\Delta\text{y}-\text{y}=(\text{x}+\Delta\text{x})\cos(\text{x}+\Delta\text{x})-\text{x}\cos\text{x}$ 
$\Delta\text{y}=\text{x}\cos(\text{x}+\Delta\text{x})+\Delta\text{x}\cos(\text{x}+\Delta\text{x})-\text{x}\cos\text{x}$
Dividing both sides by taken the limit, we get
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\Delta\text{y}}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\text{x}\cos(\text{x}+\Delta\text{x})-\text{x}\cos\text{x}+\Delta\text{x}\cos(\text{x}+\Delta\text{x})}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\text{x}\big[\cos(\text{x}+\Delta\text{x})-\cos\text{x}\big]}{\Delta\text{x}}$ 
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\text{x}\bigg[-2\sin\frac{(\text{x}+\Delta\text{x}+\text{x})}{2}\cdot\sin\frac{(\text{x}+\Delta\text{x}-\text{x})}{2}\bigg]}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\text{x}\bigg[-2\sin\Big(\text{x}+\frac{\Delta\text{x}}{2}\Big)\cdot\sin\frac{\Delta\text{x}}{2}\bigg]}{2\times\frac{\Delta\text{x}}{2}}$
$\frac{\Delta\text{x}}{2}\rightarrow0$ Taking the limits, we have
$=\text{x}[-\sin\text{x}]+\cos\text{x}$
$=-\text{x}\sin\text{x}+\cos\text{x}$
Hence, the required answer is $-\text{x}\sin\text{x}+\cos\text{x}.$
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Question 35 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow0}\frac{\sin2\text{x}+3\text{x}}{2\text{x}+\tan3\text{x}}$
Answer
Given that $\lim\limits_{\text{x} \rightarrow0}\frac{\sin2\text{x}+3\text{x}}{2\text{x}+\tan3\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\big(\frac{\sin2\text{x}+3\text{x}}{2\text{x}}\big)\times2\text{x}}{\big(\frac{2\text{x}+\tan3\text{x}}{3\text{x}}\big)\times3\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\Big(\frac{\sin2\text{x}}{2\text{x}}+\frac{3\text{x}}{2\text{x}}\Big)\times2\text{x}}{\Big(\frac{2\text{x}\tan3\text{x}}{3\text{x}}\Big)\times3\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\Big(\frac{\sin2\text{x}}{2\text{x}}+\frac{3\text{x}}{2\text{x}}\Big)\times2\text{x}}{\Big(\frac{2\text{x}}{3\text{x}}+\frac{\tan3\text{x}}{3\text{x}}\Big)\times3\text{x}}$
$=\frac{\Big(\lim\limits_{2\text{x} \rightarrow 0}\frac{\sin2\text{x}}{2\text{x}}+\frac{3}{2}\Big)}{\Big(\frac{2}{3}+\lim\limits_{3\text{x} \rightarrow 0}\frac{\tan3\text{x}}{3\text{x}}\Big)}\times\frac{2}{3}$
$=\Big(\frac{1+\frac{3}{2}}{\frac{2}{3}+1}\Big)\times\frac{2}{3}$
$=\frac{\frac{5}{2}}{\frac{5}{3}}\times\frac{2}{3}=\frac{3}{2}\times\frac{2}{3}=1$
Hence, the required answer is 1.
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Question 45 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow0}\frac{\sqrt{2}-\sqrt{1+\cos\text{x}}}{\sin^{2}\text{x}}$
Answer
Given that $\lim\limits_{\text{x} \rightarrow0}\frac{\sqrt{2}-\sqrt{1+\cos\text{x}}}{\sin^{2}\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\sqrt{2}-\sqrt{1+\cos\text{x}}}{\sin^{2}\text{x}}\times\frac{\sqrt{2}+\sqrt{1+\cos\text{x}}}{\sqrt{2}+\sqrt{1+\cos\text{x}}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{2-(1+\cos\text{x})}{\sin^{2}\text{x}\big[\sqrt{2}+\sqrt{1+\cos\text{x}}\big]}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{1+\cos\text{x}}{\sin^{2}\text{x}\Big[\sqrt{2}+\sqrt{1+\cos\text{x}}\Big]}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin^{2}\frac{\text{x}}{2}}{(2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2})^{2}}\times\frac{1}{\sqrt{2}+\sqrt{1+\cos\text{x}}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin^{2}\frac{\text{x}}{2}}{4\sin^{2}\frac{\text{x}}{2}\cos^{2}\frac{\text{x}}{2}}\times\frac{1}{\sqrt{2}+\sqrt{1+\cos\text{x}}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{2}{4\cos^{2}\frac{\text{x}}{2}}\times\frac{1}{\sqrt{2}+\sqrt{1+\cos\text{x}}}$
Taking limit, we get
$=\frac{2}{4\cos^{2}0}\times\frac{1}{(\sqrt{2}+\sqrt{2})}$
$=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2\sqrt{2}}=\frac{1}{4\sqrt{2}}$
Hence, the required answer is $\frac{1}{4\sqrt{2}}.$
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Question 55 Marks
Evaluate the following limits.
Let $\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-{2}\text{x}}, \\3,\text{x}=\frac{\pi}{2}\text{and } \text{f}(\text{x})=\text{f}\big(\frac{\pi}{2}\big)\end{cases}$ Find the value of k.
Answer
Given $\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-{2}\text{x}}, \\3,\text{x}=\frac{\pi}{2} \end{cases}$
$\text{L}.\text{H}.\text{L}.\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{2}}}\frac{\text{k}\cos\text{x}}{\pi-{2}\text{x}}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{k}\cos\big(\frac{\pi}{2}+\text{h}\big)}{\pi-{2}\big(\frac{\pi}{2}-\text{h}\big)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{k}\sin\text{h}}{\pi-\pi+2\text{h}}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{k}\sin\text{h}}{2\text{h}}$
$=\frac{\text{k}}{2}.1=\frac{\text{k}}{2}$
$\text{R}.\text{H}.\text{L}.\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{2}}}\frac{\text{k}\cos\text{x}}{\pi-{2}\text{x}}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{k}\cos\big(\frac{\pi}{2}+\text{h}\big)}{\pi-{2}\big(\frac{\pi}{2}-\text{h}\big)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{k}\sin\text{h}}{\pi-\pi-2\text{h}}=\lim\limits_{\text{h} \rightarrow 0}\frac{-\text{k}\sin\text{h}}{-2\text{h}}$
$=\frac{\text{k}}{2}.1=\frac{\text{k}}{2}$
we are given that $\lim\limits_{\text{h} \rightarrow\frac{\pi}{2}}\text{f}(\text{x})=3$ 
Hence, the required answer is 6.
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Question 65 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow 2}\frac{\text{x}^{2}-4}{\sqrt{3\text{x}-2}-\sqrt{\text{x}+2}}$
Answer
Given that $\lim\limits_{\text{x} \rightarrow 2}\frac{\text{x}^{2}-4}{\sqrt{3\text{x}-2}-\sqrt{\text{x}+2}}$
Rationalizing the denomonator, we get
$=\lim\limits_{\text{x} \rightarrow 2}\frac{(\text{x}+2)(\text{x}+2)\big[\sqrt{3\text{x}-2}+\sqrt{\text{x}+2}\big]}{\big[\sqrt{3\text{x}-2}-\sqrt{\text{x}+2}\big]\big[\sqrt{3\text{x}-2}+\sqrt{\text{x}+2}\big]}$
$=\lim\limits_{\text{x} \rightarrow 2}\frac{(\text{x}+2)(\text{x}+2)\big[\sqrt{3\text{x}-2}+\sqrt{\text{x}+2}\big]}{3\text{x}-2-\text{x}-2}$
$=\lim\limits_{\text{x} \rightarrow 2}\frac{(\text{x}+2)(\text{x}+2)\big[\sqrt{3\text{x}-2}+\sqrt{\text{x}+2}\big]}{2\text{x}-4}$
$=\lim\limits_{\text{x} \rightarrow 2}\frac{(\text{x}+2)(\text{x}+2)\big[\sqrt{3\text{x}-2}+\sqrt{\text{x}+2}\big]}{2(\text{x}-2)}$
$=\lim\limits_{\text{x} \rightarrow 2}\frac{(\text{x}+2)(\text{x}+2)\big[\sqrt{3\text{x}-2}+\sqrt{\text{x}+2}\big]}{2}$
Taking limits, we have
$=\frac{(2+2)\big[\sqrt{6-2}+\sqrt{2+2}\big]}{2}=\frac{4\big[2+2\big]}{2}$
$=\frac{4\times4}{2}=8$
Hence, the required answer is 8.
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Question 75 Marks
Evaluate the following limits.
$\lim\limits_{\text{y} \rightarrow 0}\frac{(\text{x}+\text{y})\sec(\text{x}+\text{y})-\text{x}\sec\text{x}}{\text{y}}$
Answer
$\lim\limits_{\text{y} \rightarrow 0}\frac{(\text{x}+\text{y})\sec(\text{x}+\text{y})-\text{x}\sec\text{x}}{\text{y}}$
$=\lim\limits_{\text{y} \rightarrow 0}\frac{\text{x}\sec(\text{x}+\text{y})+\text{y}\sec(\text{x}+\text{y})-\text{x}\sec\text{x}}{\text{y}}$
$=\lim\limits_{\text{y} \rightarrow 0}\frac{\Big[\text{x}\sec(\text{x}+\text{y})-\text{x}\sec\text{x}\big]}{\text{y}}+\lim\limits_{\text{y} \rightarrow 0}\frac{\text{y}\sec(\text{x}+\text{y})}{\text{y}}$
$=\lim\limits_{\text{y} \rightarrow 0}\frac{\text{x}\big[\sec(\text{x}+\text{y})-\sec\text{x}\big]}{\text{y}}+\lim\limits_{\text{y} \rightarrow 0}\sec(\text{x}+\text{y})$
$=\lim\limits_{\text{y} \rightarrow 0}\frac{\text{x}\big[\frac{1}{\cos(\text{x}+\text{y})}-\frac{1}{\cos\text{x}}\big]}{\text{y}}+\lim\limits_{\text{y} \rightarrow 0}\sec(\text{x}+\text{y})$
$=\lim\limits_{\text{y} \rightarrow 0}\text{x}\bigg[\frac{\cos\text{x}-\cos(\text{x}+\text{y})}{\text{y}\cdot\cos(\text{x}+\text{y})\cdot\cos\text{x}}\bigg]+\lim\limits_{\text{y} \rightarrow 0}\sec(\text{x}+\text{y})$
$=\lim\limits_{\text{y} \rightarrow 0}\frac{\text{x}\bigg[-2\sin\Big(\frac{\text{x}+\text{x}+\text{y}}{2}\Big)\cdot\sin\Big(\frac{\text{x}-\text{x}-\text{y}}{2}\Big)\bigg]}{\text{y}\cos(\text{x}+\text{y})\cdot\text{x}}+\lim\limits_{\text{y} \rightarrow 0}\sec(\text{x}+\text{y})$
$=\lim\limits_{\text{y} \rightarrow 0}\frac{\text{x}\Big[-2\sin\big(\text{x}+\frac{\text{y}}{2}\Big)\cdot\Big(\frac{-\text{y}}{2}\Big)\Big]}{\cos(\text{x}+\text{y})\cdot\cos\text{x}\cdot\text{y}}+\lim\limits_{\text{y} \rightarrow 0}\sec(\text{x}+\text{y})$
Taking the limits we have
$=\text{x}\Big[\sin\text{x}\cdot\frac{1}{\cos\text{x}\cdot\cos\text{x}}\Big]+\sec\text{x}$
$=\text{x}\sec\text{x}\tan\text{x}+\sec\text{x}$
$=\sec\text{x}(\text{x}\tan\text{x}+1)$
Hence, the required answer is $\sec\text{x}(\text{x}\tan\text{x}+1).$
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Question 85 Marks
Evaluate the following limits.
$\lim\limits_{\text{x} \rightarrow\frac{\pi}{4}}\frac{\tan^{3}\text{x}-\tan\text{x}}{\cos\Big(\text{x}+\frac{\pi}{4}\Big)}$ 
Answer
Given$\lim\limits_{\text{x} \rightarrow\frac{\pi}{4}}\frac{\tan^{3}\text{x}-\tan\text{x}}{\cos\Big(\text{x}+\frac{\pi}{4}\Big)}$
$=\lim\limits_{\text{x} \rightarrow\frac{\pi}{4}}\frac{\tan\text{x}(\tan^{2}\text{x}-1)}{\cos\Big(\text{x}+\frac{\pi}{4}\Big)}$
$=\lim\limits_{\text{x} \rightarrow\frac{\pi}{4}}\tan\text{x}\cdot\lim\limits_{\text{x} \rightarrow\frac{\pi}{4}}\Bigg[\frac{(1-\tan^{2}\text{x})}{\cos\Big(\text{x}+\frac{\pi}{4}\Big)}\Bigg]$
$=-1\times\lim\limits_{\text{x} \rightarrow\frac{\pi}{4}}\frac{(1-\tan\text{x})(1+\tan\text{x})}{\cos\Big(\text{x}+\frac{\pi}{4}\Big)}$
$=-(1+1)\times\lim\limits_{\text{x} \rightarrow\frac{\pi}{4}}\frac{(\cos\text{x}-\sin\text{x})}{\cos\text{x}\cdot\cos\Big(\text{x}+\frac{\pi}{4}\Big)}$
$=-2\times\lim\limits_{\text{x} \rightarrow\frac{\pi}{4}}\frac{\sqrt{2}\Big(\frac{1}{\sqrt{2}}\cos\text{x}-\frac{1}{\sqrt{2}}\sin\text{x}\Big)}{\cos\text{x}\cdot\cos\Big(\text{x}+\frac{\pi}{4}\Big)}$
$=-2\sqrt{2}\lim\limits_{\text{x} \rightarrow\frac{\pi}{4}}\frac{\Big[\cos\frac{\pi}{4}\cdot\cos\text{x}-\sin\frac{\pi}{4}\sin\text{x}\Big]}{\cos\text{x}\cdot\cos\Big(\text{x}+\frac{\pi}{4}\Big)}$
$=\lim\limits_{\text{x} \rightarrow\frac{\pi}{4}}\frac{-2\sqrt{2}\cdot\cos\Big(\text{x}+\frac{\pi}{4}\Big)}{\cos\text{x}\cdot\cos\Big(\text{x}+\frac{\pi}{4}\Big)}$
$=\frac{-2\sqrt{2}}{\cos\frac{\pi}{4}}$
$=\frac{-2\sqrt{2}}{\frac{1}{\sqrt{2}}}=-2\times2=-4$
Hence, the rquired answer is -4.
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Question 95 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow{\text{a}}}\frac{\sin\text{x}-\sin\text{a}}{\sqrt{\text{x}}-\sqrt{\text{a}}}$
Answer
Given that $\lim\limits_{\text{x} \rightarrow{\text{a}}}\frac{\sin\text{x}-\sin\text{a}}{\sqrt{\text{x}}-\sqrt{\text{a}}}$
$=\lim\limits_{\text{x} \rightarrow{\text{a}}}\frac{\sin\text{x}-\sin\text{a}}{\sqrt{\text{x}}-\sqrt{\text{a}}}\times\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{\sqrt{\text{x}}+\sqrt{\text{a}}}$
$=\lim\limits_{\text{x} \rightarrow{\text{a}}}\frac{\big(2\cos\frac{\text{x}+\text{a}}{2}.\sin\frac{\text{x}-\text{a}}{2}\big)\sqrt{\text{x}}+\sqrt{\text{a}}}{\text{x}-\text{a}}$
$=\lim\limits_{\text{x} \rightarrow{\text{a}}}\cos\big(\frac{\text{x}+\text{a}}{2}\big)\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)$
Taking limits we have
$=\cos\big(\frac{\text{a}+\text{a}}{2}\big)\big(\sqrt{\text{a}}+\sqrt{\text{a}}\big)$
$=\cos\text{x}\times2\sqrt{\text{a}}=2\sqrt{\text{a}}.\cos\text{a}$
Hence, the required answer is $2\sqrt{\text{a}}.\cos\text{a}.$
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Question 105 Marks
Evaluate the following limits.
$\lim\limits_{\text{x} \rightarrow\pi}\frac{1-\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}\Big(\cos\frac{\text{x}}{4}-\sin\frac{\text{x}}{4}\Big)}$ 
Answer
Given $\lim\limits_{\text{x} \rightarrow\pi}\frac{1-\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}\Big(\cos\frac{\text{x}}{4}-\sin\frac{\text{x}}{4}\Big)}$
$=\lim\limits_{\text{x} \rightarrow\pi}\frac{\cos^{2}\frac{\text{x}}{4}+\sin^{2}\frac{\text{x}}{4}-2\sin\frac{\text{x}}{4}\cdot\cos\frac{\text{x}}{4}}{\cos\frac{\text{x}}{2}\Big(\cos\frac{\text{x}}{4}-\sin\frac{\text{x}}{4}\Big)\Big(\cos\frac{\text{x}}{4}-\sin\frac{\text{x}}{4}\Big)}$
$=\lim\limits_{\text{x} \rightarrow\pi}\frac{\Big(\cos\frac{\text{x}}{4}-\sin\frac{\text{x}}{4}\Big)^{2}}{\cos\frac{\text{x}}{2}\Big(\cos\frac{\text{x}}{4}-\sin\frac{\text{x}}{4}\Big)\Big(\cos\frac{\text{x}}{4}-\sin\frac{\text{x}}{4}\Big)\Big(\cos\frac{\pi}{4}-\sin\frac{\text{x}}{4}\big)}$
$=\lim\limits_{\text{x} \rightarrow\pi}\frac{1}{\Big(\cos\frac{\pi}{4}+\sin\frac{\pi}{4}\Big)}$
$=\frac{1}{\cos\frac{\pi}{4}+\sin\frac{\pi}{4}}$
$=\frac{1}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}=\frac{1}{\frac{2}{\sqrt{2}}}$
$=\frac{1}{\sqrt{2}}$
Hence. the required answer is $\frac{1}{\sqrt{2}}.$
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Question 115 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow \frac{1}{2}}\Big(\frac{8\text{x}-3}{2\text{x}-1}-\frac{4\text{x}^{2}+1}{4\text{x}^{2}-1}\Big)$
Answer
Given that $\lim\limits_{\text{x} \rightarrow \frac{1}{2}}\Big(\frac{8\text{x}-3}{2\text{x}-1}-\frac{4\text{x}^{2}+1}{4\text{x}^{2}-1}\Big)$
$=\lim\limits_{\text{x} \rightarrow\frac{1}{2} }\Big[\frac{(8\text{x}-3)(2\text{x}-1)-(4\text{x}^{2}+1)}{(4\text{x}^{2}-1)}\Big]$
$=\lim\limits_{\text{x} \rightarrow \frac{1}{2}}\Big[\frac{16\text{x}^{2}-6\text{x}+8\text{x}-3-4\text{x}^{2}-1}{4\text{x}^{2}-1}\Big] $
$=\lim\limits_{\text{x} \rightarrow \frac{1}{2}}\frac{2(6\text{x}^{2}+\text{x}-2)}{4\text{x}^{2}-1}$
$=\lim\limits_{\text{x} \rightarrow\frac{1}{2}} \frac{2[6\text{x}^{2}+4\text{x}-3\text{x}-2]}{(2\text{x}+1)(2\text{x}-1)}$
$=\lim\limits_{\text{x} \rightarrow\frac{1}{2}} \frac{2[ 2\text{x}(3\text{x}+2)-1(3\text{x}+2)]}{(2\text{x}+1)(2\text{x}-1)}$
$=\lim\limits_{\text{x} \rightarrow\frac{1}{2}} \frac{ 2(3\text{x}+2)(2\text{x}-1)]}{(2\text{x}+1)(2\text{x}-1)}$
$=\lim\limits_{\text{x} \rightarrow\frac{1}{2}} \frac{ 2(3\text{x}+2)}{(2\text{x}+1)}$
Taking limit, we have
$=\frac{2\big(3\times\frac{1}{2}+2\big)}{2\times\frac{1}{2}+1}$
$=\frac{2\big(\frac{7}{2}\big)}{2}=\frac{7}{2}$
Hence, the required answer is $\frac{7}{2}.$
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Question 125 Marks
Evaluate the following limits.
$\lim\limits_{\text{x} \rightarrow 0}\frac{(\sin(\alpha+\beta)\text{x}+\sin(\alpha-\beta)\text{x}+\sin2\alpha\cdot\text{x}}{\cos2\beta\text{x}-\cos2\alpha\text{x}}$ 
Answer
Given$\lim\limits_{\text{x} \rightarrow 0}\frac{(\sin(\alpha+\beta)\text{x}+\sin(\alpha-\beta)\text{x}+\sin2\alpha\cdot\text{x}}{\cos2\beta\text{x}-\cos2\alpha\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\big[2\sin\alpha\text{x}\cdot\cos\beta\text{x}+\sin2\alpha\cdot\text{x}\big]\cdot\text{x}}{2\sin(\alpha+\beta)\text{x}\cdot\sin(\text{x}-\beta)\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\big[2\sin\alpha\text{x}\cdot\cos\beta\text{x}+2\sin\alpha\text{x}\cdot\cos\alpha\text{x}\big]\cdot\text{x}}{2\sin(\alpha+\beta)\text{x}\cdot\sin(\text{x}-\beta)\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin\alpha\text{x}(\cos\beta\text{x}+\cos\alpha\text{x})\cdot\text{x}}{2\sin(\alpha+\beta)\text{x}\cdot\sin(\alpha-\beta)\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\alpha\text{x}\Big[2\cos\Big(\frac{\alpha+\beta}{2}\Big)\text{x}\cdot\cos\Big(\frac{\alpha-\beta}{2}\Big)\text{x}\Big]\cdot\text{x}}{\sin(\alpha+\beta)\text{x}\cdot\sin(\alpha-\beta)\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\alpha\text{x}\Big[2\cos\Big(\frac{\alpha+\beta}{2}\Big)\text{x}\cdot\cos\Big(\frac{\alpha-\beta}{2}\Big)\text{x}\Big]\cdot\text{x}}{2\sin\Big(\frac{\alpha+\beta}{2}\Big)\text{x}\cdot\cos\Big(\frac{\alpha-\beta}{2}\Big)\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\alpha\text{x}\cdot\text{x}}{2\sin\Big(\frac{\alpha+\beta}{2}\Big)\text{x}\cdot\cos\Big(\frac{\alpha-\beta}{2}\Big)\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{1}{2}\frac{\frac{\sin\alpha\text{x}}{\alpha\text{x}}\cdot(\alpha\text{x})\cdot\text{x}}{\Bigg[\frac{\frac{\sin\alpha+\beta}{2}\text{x}}{\frac{\alpha+\beta}{2}\text{x}}\times\Big(\frac{\alpha+\beta}{2}\Big)\text{x}\Bigg]\Bigg[\frac{\sin\Big(\frac{\alpha-\beta}{2}\Big)\text{x}}{\Big(\frac{\alpha-\beta}{2}\Big)\text{x}}\times\frac{(\alpha-\beta)}{2}\text{x}\Bigg]}$
$=\frac{1}{2}\cdot\frac{\alpha\text{x}^{2}}{\Big(\frac{\alpha+\beta}{2}\Big)\text{x}\cdot\Big(\frac{\alpha-\beta}{2}\Big)\text{x}}$
$=\frac{1}{2}\begin{bmatrix}\frac{\alpha}{\Big(\frac{\alpha+\beta}{2}\Big)\Big(\frac{\alpha-\beta}{2}\Big)} \end{bmatrix}$
$=\frac{1}{2}\cdot\frac{4\alpha}{\alpha^{2}-\beta^{2}}$
$=\frac{2\alpha}{\alpha^{2}-\beta^{2}}$
Hence, the required amswer is $\frac{2\alpha}{\alpha^{2}-\beta^{2}}.$
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Question 135 Marks
Differentiate the functions with respect to 'x'.
$\text{x}^{\frac{2}{3}}$
Answer
Let $\text{f}(\text{x})=\text{x}^{\frac{2}{3}}\ ...(\text{i})$
$\Rightarrow\text{f}(\text{x}+\Delta\text{x})=(\text{x}+\Delta\text{x})^{\frac{2}{3}}\ ...(\text{ii})$
Subtracting eq. (i) from eq. (ii)
$\Rightarrow\text{f}(\text{x}+\Delta\text{x}-\text{f}(\text{x})=(\text{x}+\Delta\text{x})^{\frac{2}{3}}-\text{x}^{\frac{2}{3}}$ 
Dividing both sides by taken the limit, we get
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\text{f}(\text{x}+\Delta\text{x})-\text{f}(\text{x})}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{(\text{x}+\Delta\text{x})^{\frac{2}{3}}}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\text{x}^{\frac{2}{3}}\Big[1+\frac{\Delta\text{x}}{\text{x}}\Big]^{\frac{2}{3}}-\text{x}^{\frac{2}{3}}}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\text{x}^{\frac{2}{3}}\Bigg[\bigg(1+\frac{\Delta\text{x}}{\text{x}}\bigg)^{\frac{2}{3}}\Bigg]-1}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\text{x}^{\frac{2}{3}}\bigg[\bigg(1+\frac{2}{3}\cdot\frac{\Delta\text{x}+\ ....}{\text{x}}\bigg)-1\bigg]}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\text{x}^{\frac{2}{3}}\cdot\frac{2}{3}\cdot\frac{\Delta\text{x}}{\text{x}}}{\Delta\text{x}}$
$=\frac{2}{3}\text{x}^{\frac{2}{3}-1}=\frac{2}{3}\text{x}^{\frac{-1}{3}}$
Hence, the required answer is $\frac{2}{3}\text{x}^{\frac{-1}{3}}.$
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Question 145 Marks
Differentiate the functions with respect to 'x'.
$\frac{\text{ax}+\text{b}}{\text{cx}+\text{d}}$
Answer
Let $\text{f}(\text{x})=\frac{\text{ax}+\text{b}}{\text{cx}+\text{d}}\ ...(\text{i})$
$\Rightarrow\text{f}(\text{x}+\Delta\text{x})=\frac{\text{a}(\text{x}+\Delta\text{x})+\text{b}}{\text{c}(\text{x}+\Delta\text{x})+\text{d}}\ ...(\text{ii})$
Subtracting eq. (i) from eq. (ii)
$\Rightarrow\text{f}(\text{x}+\Delta\text{x}-\text{f}(\text{x})=\frac{\text{a}(\text{x}+\Delta\text{x})+\text{b}}{\text{c}(\text{x}+\Delta\text{x})+\text{d}}-\frac{\text{ax}+\text{b}}{\text{cx}+\text{d}}$
Dividing both sides by taken the limit, we get
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\text{f}(\text{x}+\Delta\text{x})-\text{f}(\text{x})}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\frac{\text{a}(\text{x}+\Delta\text{x})+\text{b}}{\text{c}(\text{x}+\Delta\text{x})+\text{d}}-\frac{\text{ax}+\text{b}}{\text{cx}+\text{d}}}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{-\text{acx}^{2}-\text{ac}\Delta\text{x}\cdot-\text{adx}-\text{bcx}-\text{bc}\cdot\Delta\text{x}-\text{bd}}{(\text{cx}+\text{c}\Delta\text{x}+\text{d})(\text{cx}+\text{d}).\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{(\text{ad}-\text{bc})\Delta\text{x}}{(\text{cx}+\text{c}\Delta\text{x}+\text{d})(\text{cx}+\text{d})\cdot.\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{(\text{ad}-\text{bc})}{(\text{cx}+\text{c}\Delta\text{x}+\text{d})(\text{cx}+\text{d})}$
Taking limit, we get
$=\frac{(\text{ad}-\text{bc})}{(\text{cx}+\text{d})(\text{cx}+\text{d})}=\frac{\text{ad}-\text{bc}}{(\text{cx}+\text{d})^{2}}$
Hence, the required answer is $\frac{\text{ad}}{(\text{cx}+\text{d})^{2}}.$
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Question 155 Marks
Differentiate the functions with respect to 'x'.
$\cos(\text{x}^{2}+1)$
Answer
Let $\text{f}(\text{x})=\cos(\text{x}^{2}+1)\ ...(\text{i})$
$\Rightarrow \text{f}(\text{x}\Delta\text{x})=\cos\big[(\text{x}+\Delta\text{x}^{2})+1\big]\ ...(\text{ii})$ 
Subtracting eq. (i) from eq. (ii) we get
$\text{f}(\text{x}+\Delta\text{x}-\text{f}(\text{x})=\cos\big[(\text{x}+\Delta\text{x}^{2})+1\big]-\cos(\text{x}^{2}+1)$
Dividing both sides by
$\frac{\text{f}(\text{x}+\Delta\text{x})-\text{f}(\text{x})}{\Delta\text{x}}=\frac{\cos\big[(\text{x}+\Delta\text{x}^{2})+1\big] -\cos(\text{x}^{2}+1)} {\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\text{f}(\text{x}+\Delta\text{x})-\text{f}(\text{x})}{\Delta\text{x}}=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\cos\big[(\text{x}+\Delta\text{x}^{2}+1\big] -\cos(\text{x}^{2}+1)} {\Delta\text{x}}$
$\text{f}'(\text{x})=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\cos\big[(\text{x}+\Delta\text{x}^{2}+1\big] -\cos(\text{x}^{2}+1)} {\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\sin\Big[\frac{(\text{x}+\Delta\text{x})^{2}+1-\text{x}^{2}-1}{2}\Big]}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\sin\Big[\frac{\text{x}^{2}+\Delta\text{x}^{2}+2\text{x}-\Delta\text{x}-\text{x}^{2}}{2}\Big]}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{-2\sin\bigg[\text{x}^{2}+\frac{\Delta\text{x}^{2}}{2}+\text{x}\Delta\text{x}+1\bigg]\sin\bigg[\Delta\text{x}\frac{(\Delta\text{x}+2\text{x})}{2}\bigg]}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\sin\Big[\Delta\text{x}\frac{(\Delta\text{x}+2\text{x})}{2}\Big]}{\Delta\big[\frac{\Delta\text{x}+2\text{x}}{2}\big]}\times\Big(\frac{\Delta\text{x}+2\text{x}}{2}\Big)$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}-2\sin\bigg[\text{x}^{2}+\frac{\Delta\text{x}^{2}}{2}+\text{x}\Delta\text{x}+1\bigg]\cdot\frac{\sin\Big[\Delta\text{x}\frac{(\Delta\text{x}+2\text{x})}{2}\Big]}{\Delta\text{x}\Big[\frac{\Delta\text{x}+2\text{x}}{2}\Big]}\times\bigg[\frac{\Delta\text{x}+2\text{x}}{2}\bigg] $
Taking limit, we have
$=-2\sin(\text{x}^{2}+1)\cdot1\cdot(\text{x})$
$=-2\sin(\text{x}^{2}+1)$
Hence, the required answer is $-2\sin(\text{x}^{2}+1).$
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