Question 13 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow 0}\frac{\sin3\text{x}}{\sin7\text{x}}$
AnswerGiven that $\lim\limits_{\text{x} \rightarrow 0}\frac{\sin3\text{x}}{\sin7\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\sin3\text{x}}{3\text{x}}\times3\text{x}}{\frac{\sin7\text{x}}{7\text{x}}\times7\text{x}}$
$=\frac{\lim\limits_{3\text{x} \rightarrow 0}\Big(\frac{\sin3\text{x}}{3\text{x}}\Big)}{\lim\limits_{7\text{x} \rightarrow 0}\Big(\frac{\sin7\text{x}}{7\text{x}}\Big)}\times\frac{3}{7}$
$=\frac{1}{1}\times\frac{3}{7}=\frac{3}{7}$
Hence, the required answer is $\frac{3}{7}.$
View full question & answer→Question 23 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow{\text{a}}}\frac{(2+\text{x})^\frac{5}{2}-(\text{a}+2)^\frac{5}{2}}{\text{x}-\text{a}}$
AnswerGiven that $\lim\limits_{\text{x} \rightarrow{\text{a}}}\frac{(2+\text{x})^\frac{5}{2}-(\text{a}+2)^\frac{5}{2}}{\text{x}-\text{a}}$
$=\lim\limits_{\text{x} \rightarrow{\text{a}}}\frac{(2+\text{x})^\frac{5}{2}-(\text{a}+2)^\frac{5}{2}}{(2+\text{x})-(\text{a}+2)}$
$=\lim\limits_{2+\text{x} \rightarrow \text{a}+2}\frac{(2+\text{x})^\frac{5}{2}-(\text{a}+2)^\frac{5}{2}}{(2+\text{x})-(\text{a}+2)}$
$=\frac{5}{2}(\text{a}+2)^{\frac{5}{2}-1}$
$=\frac{5}{2}(\text{a}+2)^{\frac{3}{2}}$
Hence, the required answre is $\frac{5}{2}(\text{a}+2)^{\frac{3}{2}}.$
View full question & answer→Question 33 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow \sqrt{2}}\frac{\text{x}^{2}-4}{\text{x}^{2}+3\sqrt{2\text{x}}-8}$
AnswerGiven that $\lim\limits_{\text{x} \rightarrow \sqrt{2}}\frac{\text{x}^{2}-4}{\text{x}^{2}+3\sqrt{2\text{x}}-8}$
$=\lim\limits_{\text{x} \rightarrow \sqrt{2}}\frac{(\text{x}^{2}-2)(\text{x}^{2}+2)}{\text{x}^{2}+4\sqrt{2}\text{x}-\sqrt{2}\text{x}-8}$
$=\lim\limits_{\text{x} \rightarrow \sqrt{2}}\frac{(\text{x}+\sqrt{2})(\text{x}-\sqrt{2})(\text{x}^{2}+2)}{\text{x}(\text{x}+4\sqrt{2})-\sqrt{2}(\text{x}+4\sqrt{2})}$
$=\lim\limits_{\text{x} \rightarrow \sqrt{2}}\frac{(\text{x}+\sqrt{2})(\text{x}-\sqrt{2})(\text{x}^{2}+2)}{(\text{x}+4\sqrt{2})(\text{x}-\sqrt{2})}$
$=\lim\limits_{\text{x} \rightarrow \sqrt{2}}\frac{(\text{x}+\sqrt{2})(\text{x}^{2}+2)}{\text{x}+4\sqrt{2}}$
Taking limits we have
$=\frac{(\sqrt{2}+\sqrt{2})(2+2)}{\sqrt{2}+4\sqrt{2}}$
$=\frac{2\sqrt{2}\times4}{5\sqrt{2}}=\frac{8}{5}$
Hence, required answre is $\frac{8}{5}.$
View full question & answer→Question 43 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow 0}\frac{\sqrt{1+\text{x}^{3}}-\sqrt{1-\text{x}^{3}}}{\text{x}^{2}}$
AnswerGiven that $\lim\limits_{\text{x} \rightarrow 0}\frac{\sqrt{1+\text{x}^{3}}-\sqrt{1-\text{x}^{3}}}{\text{x}^{2}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\Big[\sqrt{1+\text{x}^{3}}-\sqrt{1-\text{x}^{3}}\Big]\Big[\sqrt{1+\text{x}^{3}}+\sqrt{1-3^{3}}\Big]}{\text{x}^{2}\Big[\sqrt{1+\text{x}^{3}}+\sqrt{1-\text{x}^{3}}\Big]} $
$=\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x}^{3})-(1-\text{x}^{3})}{\text{x}^{2}\Big[\sqrt{1+\text{x}^{3}}+\sqrt{1-\text{x}^{3}}\Big]}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{1+\text{x}^{3}-1-\text{x}^{3}}{\text{x}^{2}\Big[\sqrt{1+\text{x}^{3}}+\sqrt{1-\text{x}^{3}}\Big]}$
$=\lim\limits_{\text{x} \rightarrow 1}\text{f}(\text{x})$
Hence, the required answer is 0.
View full question & answer→Question 53 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow 1}\frac{\text{x}^{7}-2\text{x}^{5}+1}{\text{x}^{3}-3\text{x}^{2}+2}$
AnswerGiven that $\lim\limits_{\text{x} \rightarrow 1}\frac{\text{x}^{7}-2\text{x}^{5}+1}{\text{x}^{3}-3\text{x}^{2}+2}$
$=\lim\limits_{\text{x} \rightarrow 1}\frac{\text{x}^{7}-\text{x}^{5}-\text{x}^{5}+1}{\text{x}^{3}-\text{x}^{2}-2\text{x}^{2}+2}$
$=\lim\limits_{\text{x} \rightarrow 1}\frac{\text{x}^{5}(\text{x}^{2}-1)-1(\text{x}^{5}-1)}{\text{x}^{2}(\text{x}-1)-2(\text{x}^{2}-1)}$
Dividing the numerator and denominator by (x - 1) we get
$=\lim\limits_{\text{x} \rightarrow 1}\frac{\text{x}^{5}\Big(\frac{\text{x}^{2}-1}{\text{x}-1}\Big)-1\Big(\frac{\text{x}^{5}-1}{\text{x}-1}\Big)}{\text{x}^{2}\Big(\frac{\text{x}-1}{\text{x}-1}\Big)-2\Big(\frac{\text{x}^{2}-(1)^{5}}{\text{x}-1}\Big)}$
$=\frac{\lim\limits_{\text{x} \rightarrow 1}\text{x}^{5}(\text{x}+1)-\lim\limits_{\text{x} \rightarrow 1}\Big(\frac{\text{x}^{5}-(1)^{5}}{\text{x}-1}\Big)}{\lim\limits_{\text{x} \rightarrow 1}\text{x}^{2}-2\lim\limits_{\text{x} \rightarrow 1}(\text{x}+1)}$
$=\frac{1(2)-5.(1)^{5-1}}{1-2(2)}$
$=\frac{2-5}{1-4}=\frac{-3}{-3}=1$
Hence, the required answer is 1.
View full question & answer→Question 63 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow \frac{\pi}{3}}\frac{\sqrt{1-\cos6}}{\sqrt{2}\bigg(\frac{\pi}{3}-\text{x}\bigg)} $
AnswerGiven that $\lim\limits_{\text{x} \rightarrow \frac{\pi}{3}}\frac{\sqrt{1-\cos6}}{\sqrt{2}\bigg(\frac{\pi}{3}-\text{x}\bigg)} $
$=\lim\limits_{\text{x} \rightarrow\frac{\pi}{3}}\frac{\sqrt{2\sin^{2}3\text{x}}}{\sqrt{2}\bigg(\frac{\pi}{3}-\text{x}\bigg)} $
$=\lim\limits_{\text{x} \rightarrow\frac{\pi}{3}}\frac{\sqrt{2}\sin3\text{x}}{\sqrt{2}\bigg(\frac{\pi-3\text{x}}{3}\bigg)}$
$=\lim\limits_{\text{x} \rightarrow\frac{\pi}{3}}\frac{3.\sin(\pi-3\text{x})}{\pi-3\text{x}} $
$=3\Big[\therefore\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
Hence, the required answer is 3.
View full question & answer→Question 73 Marks
Differentiate the following functions.
$(2\text{x}-7)^{2}(3\text{x}+5)^{3}$
Answer$\frac{\text{d}}{\text{dx}}(2\text{x}-7)^{2}(3\text{x}+5)^{3}$
$=(2\text{x}-7^{2}).\frac{\text{d}}{\text{dx}}(3\text{x}+5)^{3}+(3\text{x}+5)^{3}.\frac{\text{d}}{\text{dx}}(2\text{x}-7)^{2}$
[Using product Rule]
⇒ (2x - 7)2.3(3x + 5)2.3 + (3x + 5)3.2(2x - 7).2
⇒ 9(2x - 7)2(3x + 5)2[9(2x) - 7) + 4(3x + 5)]
⇒ (2x - 7)(3x - 5)2(18 - 63 + 12x + 20)
⇒ (2x - 7)(3x + 5)2(30x - 43)
Hence, the required answer is (2x - 7)(30x - 43)(3x + 5)2.
View full question & answer→Question 83 Marks
Differentiate the following functions.
$(\sec\text{x}-1)(\sec\text{x}+1)$
Answer$\frac{\text{d}}{\text{dx}}(\sec\text{x}-1)(\sec\text{x}+1)$
$=(\sec\text{x}-1).\frac{\text{d}}{\text{dx}}(\sec\text{x}+1)+(\sec\text{x}+1)\frac{\text{d}}{\text{dx}}(\sec\text{x}-1)$
$=(\sec\text{x}-1)(\sec\text{x}\tan\text{x})+(\sec\text{x}+1)(\sec\text{x}\tan\text{x})$
$=\sec\text{x}\tan\text{x}(\sec\text{x}-1+\sec\text{x}+1)$
$=\sec\text{x}\tan\text{x}\ 2\sec\text{x}=2\sec\text{x}^{2}.\tan\text{x}$
Hence, the required answer is $2\sec^{2}\text{x}.\tan\text{x}.$
View full question & answer→Question 93 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow 0}\frac{(\text{x}+2)^\frac{1}{3}-2^\frac{1}{3}}{\text{x}}$
AnswerGiven that $\lim\limits_{\text{x} \rightarrow 0}\frac{(\text{x}+2)^\frac{1}{3}-2^\frac{1}{3}}{\text{x}}$
Put x + 2 = y
⇒ x= y - 2
$=\lim\limits_{\text{y-2} \rightarrow 0}\frac{\text{y}^\frac{1}{3}-2^\frac{1}{3}}{\text{y}-2}$
$=\lim\limits_{\text{y} \rightarrow 2}\frac{\text{y}^\frac{1}{3}-2^\frac{1}{3}}{\text{y}-2}$
$=\frac{1}{3}.(2)^{\frac{1}{3}-1}=\frac{1}{3}.2^{\frac{-2}{3}}$
Hence, the answer is $\frac{1}{3}.(2)^{\frac{-2}{3}}.$
View full question & answer→Question 103 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{mx}}{1-\cos\text{nx}}$
AnswerGiven that $\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{mx}}{1-\cos\text{nx}}$
$=\lim\limits_{\text{x} \rightarrow 0}\bigg(\frac{2\sin^{2}\frac{\text{m}}{2}\text{x}}{2\sin^{2}\frac{\text{n}}{2}\text{x}}\bigg)$
$=\lim\limits_{\text{x} \rightarrow 0}\bigg(\frac{\sin\frac{\text{m}}{2}\text{x}}{\sin\frac{\text{n}}{2}\text{x}}\bigg)^{2}$
$=\frac{\lim\limits_{\text{x} \rightarrow 0}\bigg(\frac{\sin\frac{\text{m}}{2}\text{x}}{\frac{\text{m}}{2}\text{x}}\times\frac{\text{m}}{2}\text{x}\bigg)^{2}}{\lim\limits_{\text{x} \rightarrow 0}\bigg(\frac{\sin\frac{\text{n}}{2}\text{x}}{\frac{\text{n}}{2}\text{x}}\times\frac{\text{n}}{2}\text{x}\bigg)^{2}}$
$=\frac{1.\frac{\text{m}^{2}}{4}\text{x}^{2}}{1.\frac{\text{n}^{2}}{4}\text{x}^{2}}$
$=\frac{\text{m}^{2}}{\text{n}^{2}}$
Hence, the required answer is $\frac{\text{m}^{2}}{\text{n}^{2}}.$
View full question & answer→Question 113 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow\frac{\pi}{6}}\frac{\sqrt{3}\sin\text{x}-\cos\text{x}}{\text{x}-\frac{\pi}{6}}$
AnswerGiven that $\lim\limits_{\text{x} \rightarrow\frac{\pi}{6}}\frac{\sqrt{3}\sin\text{x}-\cos\text{x}}{\text{x}-\frac{\pi}{6}}$
$=\lim\limits_{\text{x} \rightarrow\frac{\pi}{6}}\frac{2\Big[\frac{\sqrt{3}}{2}\sin\text{x}-\frac{1}{2}\cos\text{x}\Big]}{\text{x}-\frac{\pi}{6}}$
$=\lim\limits_{\text{x} \rightarrow\frac{\pi}{6}}\frac{2\big[\cos\frac{\pi}{6}\sin\text{x}-\sin\frac{\pi}{6}\cos\text{x}\big]}{\text{x}-\frac{\pi}{6}}$
$=\lim\limits_{\text{x} \rightarrow\frac{\pi}{6}}\frac{2\sin\big(\text{x}-\frac{\pi}{6}\big)}{\big(\text{x}-\frac{\pi}{6}\big)}$ $\bigg[ \therefore\lim\limits_{\text{x} \rightarrow\frac{\pi}{6}}\frac{\sin\text{x}}{\text{x}}=1\bigg] $
$=2.1=2$
Hence, the required answer is 2.
View full question & answer→Question 123 Marks
Differentiate the following functions.
$\sin^{3}\text{x}\cos^{3}\text{x}$
Answer$\frac{\text{d}}{\text{dx}}(\sin^{3}\text{x}\cos^{3}\text{x})$
$=\sin^{3}\text{x}\cdot\frac{\text{d}}{\text{dx}}\cos^{3}\text{x}\cdot\frac{\text{d}}{\text{dx}}(\sin^{3}\text{x})$
$=\sin^{3}\text{x}.3\cos^{3}\text{x}(-\sin\text{x})+\cos^{3}\text{x}\cdot3\sin^{2}\text{x}\cdot\cos\text{x}$
$=-3\sin^{4}\text{x}\cos^{2}+3\cos^{4}\text{x}\sin^{2}\text{x}$
$=3\sin^{2}\text{x}\cos^{2}\text{x}(-\sin^{2}\text{x}+\cos^{2}\text{x})$
$=3\sin^{2}\text{x}\cos^{2}\text{x}\cdot\cos2\text{x}$
$=\frac{3}{4}\cdot4\sin^{2}\text{x}\cos^{2}\text{x}\cdot\cos2\text{x}$
$=\frac{3}{4}(2\sin\text{x}\cos\text{x})^{2}\cos2\text{x}$
$=\frac{3}{4}\sin^{2}2\text{x}\cdot\cos2\text{x}$
Hence, the required answer is $\frac{3}{4}\sin^{2}2\text{x}\cdot\cos2\text{x}.$
View full question & answer→Question 133 Marks
Differentiate the following functions.
$\frac{\text{x}^{5}-\cos\text{x}}{\sin\text{x}}$
Answer$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^{5}-\cos\text{x}}{\sin\text{x}}\Big)$
$=\frac{\sin\text{x}\frac{\text{d}}{\text{dx}}-(\text{x}^{5}-\cos\text{x}).\frac{\text{d}}{\text{dx}}(\sin\text{x})}{\sin^{2}\text{x}}$
$=\frac{\sin\text{x}(5\text{x}^{4}+\sin\text{x}-(\text{x}^{5}-\cos\text{x})(\cos\text{x})}{\sin^{2}\text{x}}$
$=\frac{5\text{x}^{4}.\sin\text{x}+\sin^{2}\text{x}-\text{x}^{5}\cos\text{x}+\cos^{2}\text{x}}{\sin^{2}\text{x}}$
$=\frac{5\text{x}^{4}.\sin\text{x}-\text{x}^{5}\cos\text{x}+(\sin^{2}\text{x}+\cos^{2}\text{x})}{\sin^{2}\text{x}}$
$=\frac{5\text{x}^{4}\sin\text{x}-\text{x}^{5}\cos\text{x}+1}{\sin^{2}\text{x}}$
Hence, the required answer is $\frac{5\text{x}^{4}\sin\text{x}-\text{x}^{5}\cos\text{x}+1}{\sin^{2}\text{x}}.$
View full question & answer→Question 143 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow\frac{\pi}{6}}\frac{\cot^{2}\text{x}-3}{\text{cosec}\text{x}-2}$
AnswerGiven that $\lim\limits_{\text{x} \rightarrow\frac{\pi}{6}}\frac{\cot^{2}\text{x}-3}{\text{cosec}\text{x}-2}$
$=\lim\limits_{\text{x} \rightarrow\frac{\pi}{6}}\frac{\text{cosec}^{2}\text{x}-1-3}{\text{cosec}\text{x}-2}$
$=\lim\limits_{\text{x} \rightarrow\frac{\pi}{6}}\frac{\text{cosec}^{2}\text{x}-4}{\text{cosec}\text{x}-2}$
$=\lim\limits_{\text{x} \rightarrow\frac{\pi}{6}}\frac{(\text{cosec}\text{x}+2)(\text{cosec}\text{x}-2)}{(\text{cosec}\text{x}-2)}$
$=\lim\limits_{\text{x} \rightarrow\frac{\pi}{6}}({\text{cosec}\text{x}+2})$
Taking limit we have
$=\text{cosec}\frac{\pi}{6}+2=2+2=4$
Hence, the required answer is 4.
View full question & answer→Question 153 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow \text{1}}\frac{\text{x}^{4}-\sqrt{\text{x}}}{\sqrt{\text{x}}-1}$
AnswerGiven that $\lim\limits_{\text{x} \rightarrow \text{1}}\frac{\text{x}^{4}-\sqrt{\text{x}}}{\sqrt{\text{x}}-1}$
$=\lim\limits_{\text{x} \rightarrow 1}\frac{\sqrt{\text{x}}\Big[ (\text{x})^\frac{7}{2}-1\Big]}{\sqrt{\text{x}-1}}$
$=\lim\limits_{\text{x} \rightarrow 1}\frac{\sqrt{\text{x}}\frac{\bigg[\text{x}^\frac{7}{2}-(1)^{\frac{7}{2}}\bigg]}{\text{x}-1}}{\frac{(\text{x})^\frac{1}{2}-(1)^\frac{1}{2}}{\text{x}-1}}$
Dividing the numerator and denominator of x - 1
$=\lim\limits_{\text{x} \rightarrow1}\frac{\sqrt{\text{x}}\frac{\bigg[\text{x}^\frac{7}{2}-(1)^{\frac{7}{2}}\bigg]}{\text{x}-1}}{\frac{(\text{x})^\frac{1}{2}-(1)^\frac{1}{2}}{\text{x}-1}}\times\lim\limits_{\text{x} \rightarrow 1}\sqrt{\text{x}}$
$=\frac{\frac{7}{2}(1)^{\frac{7}{2}-1}}{\frac{1}{2}(1)^{\frac{1}{2}-1}}\times\sqrt{1}$
$=\frac{\frac{7}{2}}{\frac{1}{2}}=7$
Hence, the required answer is 7.
View full question & answer→Question 163 Marks
Differentiate the following functions.
$\frac{\text{x}^{5}-\cos\text{x}}{\sin\text{x}}$
Answer$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^{5}-\cos\text{x}}{\sin\text{x}}\Big)$
$=\frac{\sin\text{x}\frac{\text{d}}{\text{dx}}-(\text{x}^{5}-\cos\text{x}).\frac{\text{d}}{\text{dx}}(\sin\text{x})}{\sin^{2}\text{x}}$
$=\frac{\sin\text{x}(5\text{x}^{4}+\sin\text{x})-(\text{x}^{5}-\cos\text{x})(\cos\text{x})}{\sin^{2}\text{x}}$
$=\frac{5\text{x}^{4}.\sin\text{x}+\sin^{2}\text{x}-\text{x}^{5}\cos\text{x}+\cos^{2}\text{x}}{\sin^{2}\text{x}}$
$=\frac{5\text{x}^{4}.\sin\text{x}-\text{x}^{5}\cos\text{x}+(\sin^{2}\text{x}+\cos^{2}\text{x})}{\sin^{2}\text{x}}$
$=\frac{5\text{x}^{4}\sin\text{x}-\text{x}^{5}\cos\text{x}+1}{\sin^{2}\text{x}}$
Hence, the required answer is $\frac{5\text{x}^{4}\sin\text{x}-\text{x}^{5}\cos\text{x}+1}{\sin^{2}\text{x}}.$
View full question & answer→Question 173 Marks
Evaluate:
$\lim\limits_{\text{h} \rightarrow 0}\frac{\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{\text{h}}$
AnswerGiven that $\lim\limits_{\text{h} \rightarrow 0}\frac{\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{\text{h}\big[\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}\big]}\times\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{x}+\text{h}-\text{x}}{\text{h}\big[\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}\big] }$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}}{\text{h}\big[\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}\big] }$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{1}}{\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}} $
Talking the limits, We have
$\frac{1}{\sqrt{\text{x}}+\sqrt{\text{x}}}=\frac{1}{2\sqrt{\text{x}}}$
Hence, the answer is $\frac{1}{2\sqrt{\text{x}}}.$
View full question & answer→Question 183 Marks
If $\lim\limits_{\text{x} \rightarrow 1}\frac{\text{x}^{4}-1}{\text{x}-1}=\lim\limits_{\text{x} \rightarrow{\text{k}}}\frac{\text{x}^{3}-\text{k}^{3}}{\text{x}^{2}-\text{k}^{2}}$ then find the value of K.
AnswerGiven that $\lim\limits_{\text{x} \rightarrow 1}\frac{\text{x}^{4}-1}{\text{x}-1}=\lim\limits_{\text{x} \rightarrow{\text{k}}}\frac{\text{x}^{3}-\text{k}^{3}}{\text{x}^{2}-\text{k}^{2}}$
$\Rightarrow 4(1)^{4-1}=\lim\limits_{\text{x} \rightarrow{\text{k}}}\frac{(\text{x}-\text{k})(\text{x}^{2}+\text{k}^{2}+\text{kx})}{(\text{x}-\text{k})(\text{x}+\text{k})}$
$\Rightarrow 4=\lim\limits_{\text{x} \rightarrow{\text{k}}}\frac{(\text{x}-\text{k})(\text{x}^{2}+\text{k}^{2}+\text{kx})}{(\text{x}-\text{k})(\text{x}+\text{k})}$
$\Rightarrow 4=\frac{\text{k}^{2}+\text{k}^{2}+\text{k}^{2}}{2\text{k}}$
$\Rightarrow4=\frac{3\text{k}^{2}}{2\text{k}}$
$\Rightarrow 4 =\frac{3}{2}\text{k}$
$\Rightarrow \text{k}=\frac{8}{3}$
Hence, the required value of k is $\frac{8}{3} .$
View full question & answer→Question 193 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow 3}\frac{\text{x}^{3}+27}{\text{x}^{5}+243}$
AnswerGiven that $\lim\limits_{\text{x} \rightarrow 3}\frac{\text{x}^{3}+27}{\text{x}^{5}+243}$
$=\lim\limits_{\text{x} \rightarrow 3}\frac{\frac{\text{x}^{3}+(3)^{3}}{\text{x}-3}}{\frac{\text{x}^{5}+(3)^{3}}{\text{x}-3}}$ [Dividing the Nr and Den. By x - 3]
$=\lim\limits_{\text{x} \rightarrow 3}\frac{\frac{\text{x}^{3}-(3)^{3}}{\text{x}+3}}{\frac{\text{x}^{5}-(3)^{5}}{\text{x}+3}}$
$=\frac{3(-3)^{3-1}}{5(-3)^{5-1}}$
$=\frac{3\times(-3)^{2}}{5\times(-3)^{4}}$
$=\frac{1}{5\times3}=\frac{1}{15}$
Hence, the required answer is $\frac{1}{15}.$
View full question & answer→Question 203 Marks
Differentiate the following functions.
$(\sin\text{x}+\cos\text{x})^{2}$
Answer$\frac{\text{d}}{\text{dx}}(\sin\text{x}+\cos\text{x})^{2}$
$=\frac{\text{d}}{\text{dx}}(\sin\text{x}+\cos\text{x})(\sin\text{x}+\cos\text{x})$
$=(\sin\text{x}+\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}+\cos\text{x})$
$=(\sin\text{x}+\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}+\cos\text{x})\\+(\sin\text{x}+\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}+\cos\text{x})$
$=2(\sin\text{x}+\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}+\cos\text{x})$
$=2(\sin\text{x}+\cos\text{x})(\cos\text{x}-\sin\text{x})$
$=2(\cos^{2}\text{x}-\sin^{2}\text{x})$
$=2\cos2\text{x}$
Hence, the required answer is $2\cos2\text{x}.$
View full question & answer→Question 213 Marks
Differentiate the following functions.
$\frac{\text{x}^{2}\cos\frac{\pi}{4}}{\sin\text{x}}$
Answer$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^{2}\cos\frac{\pi}{4}}{\sin\text{x}}\Big)$
$=\cos\frac{\pi}{4}.\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^{2}}{\sin\text{x}}\Big)$
$=\frac{\frac{1}{\sqrt{2}}\Big[\sin\text{x}.\frac{\text{d}}{\text{dx}}(\text{x})^{2}-\text{x}^{2}.\frac{\text{d}}{\text{dx}}(\sin\text{x})\Big]}{\sin^{2}\text{x}}$
$=\frac{1}{\sqrt{2}}\Big[\frac{\sin\text{x}2\text{x}-\text{x}^{2}\cos\text{x}}{\sin^{2}\text{x}}\Big]$
$=\frac{1}{\sqrt{2}}\Big[\frac{2\text{x}}{\sin\text{x}}-\frac{\text{x}^{2}\cos\text{x}}{\sin^{2}\text{x}}\Big]$
$=\frac{1}{\sqrt{2}}\big[2\text{x}\ \text{cosec}\text{x}-\text{x}^{2}\ \text{cot}\text{x}\ \text{cosec}\text{x}\big]$
$=\frac{\text{x}}{\sqrt{2}}\text{cosec}\text{x}\big[2-\text{x}\text{cot}\text{x}\big]$
Hence, the required answer is $\frac{\text{x}}{\sqrt{2}}\text{cosec}\text{x}\big[2-\text{x}\text{cot}\text{x}\big].$
View full question & answer→Question 223 Marks
Evaluate:
$\lim\limits _{\text{x} \rightarrow \frac{1}{2}}\frac{4\text{x}^{2}-1}{2\text{x}-1}$
AnswerGiven that $\lim\limits _{\text{x} \rightarrow \frac{1}{2}}\frac{4\text{x}^{2}-1}{2\text{x}-1}$
$=\lim\limits_{\text{x} \rightarrow \frac{1}{2}}\frac{(2\text{x})^{2}-(1)^{2}}{2\text{x}-1}$
$=\lim\limits_{\text{x} \rightarrow \frac{1}{2}}\frac{(2\text{x}+1)(2\text{x}-1)}{2\text{x}-1}$
$=\lim\limits_{\text{x} \rightarrow \frac{1}{2}}(2\text{x}+1)$
Taking limit, We have
$=2 \times\frac{1}{2}+1$
$=1+1 $
$=2$
Hence, the answer is 2.
View full question & answer→Question 233 Marks
Evaluate:
$\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{6}-1}{(1+\text{x})^{2}-1}$
AnswerGiven that $\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{6}-1}{(1+\text{x})^{2}-1}$
Dividing the numerator and denominator by x, we get
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{(1+\text{x})^{6}-1}{\text{x}}}{\frac{(1+\text{x})^{2}-1}{\text{x}}}$
Putting 1 + x = y
⇒ x = y - 1
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{\text{y}^{6}-(1)^{6}}{\text{y}-1}}{\frac{\text{y}^{2}-(1)^{2}}{\text{y}-1}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{\text{y}^{6}-(1)^{6}}{\text{y}-1}}{\frac{\text{y}^{2}-(1)^{2}}{\text{y}-1}}$
$\frac{6.(1)^{6-1}}{2.(1)^{2-1}}=\frac{6}{2}$
$=3$
Hence, the required qnswer is 3.
View full question & answer→Question 243 Marks
Differentiate the following functions.
$\frac{\text{a}+\text{b}\sin\text{x}}{\text{c}+\text{d}\cos\text{x}}$
Answer$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{a}+\text{b}\sin\text{x}}{\text{c}+\text{d}\cos\text{x}}\Big)$
$=\frac{(\text{c}+\text{d}\cos\text{x}).\frac{\text{d}}{\text{dx}}(\text{a}+\text{b}\sin\text{x})-(\text{a}+\text{b}\sin\text{x})\frac{\text{d}}{\text{dx}}(\text{c}+\text{d}\cos\text{x})}{(\text{c}+\text{d}\cos\text{x})^{2}}$
$=\frac{\text{cb}\cos\text{x}+\text{bd}\cos^{2}\text{x}+\text{ad}\sin\text{x}+\text{bd}\sin^{2}\text{x}}{(\text{c}+\text{d}\cos\text{x})^{2}}$
$=\frac{\text{cb}\cos\text{x}+\text{ad}\sin\text{x}+\text{bd}(\cos^{2}\text{x}+\sin^{2}\text{x})}{(\text{c}+\text{d}\cos\text{x}^{2})}$
$=\frac{\text{cb}\cos\text{x}+\text{ad}\sin\text{x}+\text{bd}}{(\text{c}+\text{d}\cos\text{x})^{2}}$
View full question & answer→Question 253 Marks
Evaluate the following limits.
If $\text{f}(\text{x})=\begin{cases}\text{x}+2, & \text{x}\leq-1\\\text{cx}^{2}, & \text{x} > -1\end{cases}$ then find c when $\lim\limits_{\text{x} \rightarrow 1}\text{f}(\text{x})$ exists.
AnswerGiven $\text{f}(\text{x})=\begin{cases}\text{x}+2, & \text{x}\leq-1\\\text{cx}^{2}, & \text{x} > -1\end{cases}$
$\text{L}.\text{H}.\text{L}.\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow1}\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow 1}(\text{x}+2)$
$=\lim\limits_{\text{x} \rightarrow 1}(-1-\text{h}+2)=\lim\limits_{\text{h} \rightarrow 0}(1-\text{h})=1$
$=1$
$\text{R}.\text{H}.\text{L}.\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow{1}}\text{cx}^{2}=\lim\limits_{\text{h} \rightarrow 0}\text{c}(1+\text{h})^{2}=\text{c}$
Since, the limita exist.
Hence, the requried answer 1.
View full question & answer→Question 263 Marks
Differentiate the following functions.
$\frac{3\text{x}+4}{5\text{x}^{2}-7\text{x}+9}$
Answer$\frac{\text{d}}{\text{dx}}\Big(\frac{3\text{x}+4}{5\text{x}^{2}-7\text{x}+9}\Big)$
$=\frac{(5\text{x}^{2}-7\text{x}+9)\frac{\text{d}}{\text{dx}}(3\text{x}+4)-(3\text{x}+4).\frac{\text{d}}{\text{dx}}(5\text{x}^{2}-7\text{x}+9)}{(5\text{x}^{2}-7\text{x}+9)^{2}}$
$=\frac{(5\text{x}^{2}-7\text{x}+9)(3)-(3\text{x}+4)(10\text{x}-7)}{(5\text{x}^{2}-7\text{x}+9)^{2}}$
$=\frac{15\text{x}^{2}-21\text{x}+27-30\text{x}^{2}+21\text{x}-40\text{x}+28}{(5\text{x}^{2}-7\text{x}+9)^{2}}$
$=\frac{-15\text{x}^{2}-40\text{x}+55}{(5\text{x}^{2}-7\text{x}+9)^{2}}$
$=\frac{55-40-15\text{x}^{2}}{(5\text{x}^{2}-7\text{x}+9)^{2}}$
Hence, the required answer is $\frac{55-40-15\text{x}^{2}}{(5\text{x}^{2}-7\text{x}+9)^{2}}.$
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