MCQ 11 Mark
If $f(x) = x^{100} + x^{99} ....... + x + 1,$ then $f(1)$ is equal to:
- ✓
$5050$
- B
$5049$
- C
$5051$
- D
$50051$
AnswerCorrect option: A. $5050$
Given $f(x) = x^{100} + x^{99} + .... + x + 1$
$f(x) = 100.x^{100}+ 99.x^{98} + .... + 1$
S0$, f(1) = 100 + 99 + 98 + ..... + 1$
$=\frac{100}{2}[2\times100+(100-1)(-1)]$
$=50[200-99]=50\times101=5050$
$=5050$
View full question & answer→MCQ 21 Mark
$ \lim\limits_{\text{x} \rightarrow 0}\frac{\tan2\text{x}-\text{x}}{3\text{x}-\sin\text{x}}$ is equal to:
- A
$2$
- ✓
$\frac{1}{2}$
- C
$-\frac{1}{2}$
- D
$\frac{1}{4}$
AnswerCorrect option: B. $\frac{1}{2}$
Given $ \lim\limits_{\text{x} \rightarrow 0}\frac{\tan2\text{x}-\text{x}}{3\text{x}-\sin\text{x}} =\lim\limits_{\text{x} \rightarrow 0}\frac{\text{x}\big[\frac{\tan2\text{x}}{\text{x}}-1\big]}{\text{x}\big[3-\frac{\sin\text{x}}{\text{x}}\big]}$
$ =\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\tan2\text{x}}{2\text{x}}\times2-1}{3-\frac{\sin\text{x}}{\text{x}}}$
$=\frac{1.2-1}{3-1}$
$=\frac{2-1}{2}$
$=\frac{1}{2}$
View full question & answer→MCQ 31 Mark
$\lim\limits_{\text{x} \rightarrow0}\frac{(1+\text{x})^{\text{n}}-1}{\text{x}}$ is equal to:
- ✓
$\text{n}$
- B
$1 $
- C
$-\text{n}$
- D
$0$
AnswerCorrect option: A. $\text{n}$
Given $\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{\text{n}}-1}{\text{x}}=\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{\text{n}}-(1)^{\text{n}}}{(1+\text{x})-(1)}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{\text{n}}-(1)^{\text{n}}}{1+\text{x}-(2)}$
$=\text{n}(1)^{\text{n}-1}$
$=\text{n}$
View full question & answer→MCQ 41 Mark
$\lim\limits_{\text{x} \rightarrow0}\frac{\text{cosec}-\cot\text{x}}{\text{x}}$ is equal to:
- A
$-\frac{1}{2}$
- B
$1$
- ✓
$\frac{1}{2}$
- D
$-1$
AnswerCorrect option: C. $\frac{1}{2}$
Given $\lim\limits_{\text{x} \rightarrow 0}\frac{\text{cosec}\text{x}-\cot\text{x}}{\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}}{\text{x}}$
$ =\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{x}}{\text{x}\sin\text{x}}$
$=\frac{2\sin^{2}\frac{\text{x}}{2}}{\text{x}\cdot\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}$
$=\lim\limits_{\text{x} \rightarrow1}\frac{\sin\frac{\text{x}}{2}}{\text{x}\cos\frac{\text{x}}{2}} $
$=\frac{\tan\frac{\text{x}}{2}}{\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\tan\frac{\text{x}}{2}}{2\times\frac{\text{x}}{2}}$
$=\frac{1}{2}\times1$
$=\frac{1}{2}$
View full question & answer→MCQ 51 Mark
If $\text{y}=\frac{1+\frac{1}{\text{x}^{2}}}{1-\frac{1}{\text{x}^{2}}}$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:
- ✓
$\frac{-4\text{x}}{(\text{x}^{2}-1)^{2}}$
- B
$\frac{-4\text{x}}{(\text{x}^{2}-1)^{2}}$
- C
$\frac{1-\text{x}^{2}}{4\text{x}}$
- D
$\frac{4\text{x}}{\text{x}^{2}-1}$
AnswerCorrect option: A. $\frac{-4\text{x}}{(\text{x}^{2}-1)^{2}}$
Given, $\text{y}=\frac{1+\frac{1}{\text{x}^{2}}}{1-\frac{1}{\text{x}^{2}}}$
$\Rightarrow\text{y}=\frac{\text{x}^{2}+1}{\text{x}^{2}-1}$
$\therefore \frac{\text{dy}}{\text{dx}}=\frac{(\text{x}^{2}-1).2\text{x}-(\text{x}^{2}+1).2\text{x}}{(\text{x}^{2}-1)^{2}}$
$=\frac{2\text{x}(\text{x}^{2}-1-\text{x}^{2}-1)}{(\text{x}^{2}-1)^{2}}$
$=\frac{2\text{x}(-2)}{(\text{x}^{2}-1)^{2}}$
$=\frac{-4\text{x}}{(\text{x}^{2}-1)^2{}}$
View full question & answer→MCQ 61 Mark
If $\text{y}=\frac{\sin(\text{x}+9)}{\cos\text{x}}$ then $\frac{\text{dy}}{\text{dx}}$ at $x = 0$ is equal to:
- ✓
$\cos9$
- B
$\sin9$
- C
$0$
- D
$1$
AnswerCorrect option: A. $\cos9$
Given $\text{y}=\frac{\sin(\text{x}+9)}{\cos\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}.\cos(\text{x}+9)-\sin(\text{x}+9)(-\sin\text{x})}{\cos^{2}\text{x}}$
$=\frac{\cos\text{x}\cos(\text{x}+9)+\sin\text{x}\sin(\text{x}+9)}{\cos^{2}\text{x}}$
$=\frac{\cos(\text{x}+9-\text{x})}{\cos^{2}\text{x}}=\frac{\cos9}{\cos^{2}\text{x}}$
$=\frac{\cos9}{(1)^{2}}$
$=\cos9$
View full question & answer→MCQ 71 Mark
$\lim\limits_{\text{x} \rightarrow 0}\frac{|\sin\text{x}|}{\text{x}}$ is equal to:
AnswerGiven $\lim\limits_{\text{x} \rightarrow 0}\frac{|\sin\text{x}|}{\text{x}}$
$\text{L}.\text{H}.\text{L}=\lim\limits_{\text{x} \rightarrow 0}\frac{-\sin\text{x}}{\text{x}}=-1$
$\text{R}.\text{H}.\text{L}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\text{x}}=1$
$\text{L}.\text{H}.\text{L}\neq\text{R}.\text{H}.\text{L}$
So, the limit does not exist.
View full question & answer→MCQ 81 Mark
If $\text{f}(\text{x})=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$ then $\frac{\text{dy}}{\text{dx}}$ at $x = 1$ is equal to:
- A
$1$
- B
$\frac{1}{2}$
- C
$\frac{1}{\sqrt{2}}$
- ✓
$0$
AnswerGiven that $\text{f}(\text{x})=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{x}}}-\frac{1}{2\text{x}^{\frac{3}{2}}}$
$\big(\frac{\text{dy}}{\text{dx}}\big)=\frac{1}{2}-\frac{1}{2}=0$
View full question & answer→MCQ 91 Mark
If $\text{f}(\text{x})=1+\text{x}+\frac{\text{x}^{2}}{2}+.....+\frac{\text{x}^{100}}{100}$ then $f'(1)$ is equal to:
- A
$\frac{1}{100}$
- ✓
$100$
- C
- D
$0$
AnswerGiven $\text{f}(\text{x})=1+\text{x}+\frac{\text{x}^{2}}{2}+.....+\frac{\text{x}^{100}}{100}$
$\text{f}(\text{x})=1+\frac{2\text{x}}{2}+......+\frac{100\text{x}}{100}$
$\therefore \text{f'}(1)=1+11+....+1=100$
View full question & answer→MCQ 101 Mark
If $\text{f}(\text{x})=\frac{\text{x}-4}{2\sqrt{\text{x}}}$ then $f'(1)$ is equal to:
- ✓
$\frac{5}{4}$
- B
$\frac{4}{5}$
- C
$1$
- D
$0$
AnswerCorrect option: A. $\frac{5}{4}$
Given that $\text{f}(\text{x})=\frac{\text{x}-4}{2\sqrt{\text{x}}}$
$\therefore\text{f}(\text{x})=\frac{1}{2}\bigg[\frac{\sqrt{\text{x}.}1-(\text{x}-4).\frac{1}{2\sqrt{\text{x}}}}{\text{x}}\bigg]$
$=\frac{1}{2}\Big[\frac{2\text{x}-\text{x}+4}{2\sqrt{\text{x}.\text{x}}}\Big]$
$=\frac{1}{2}\Bigg[\frac{\text{x}+4}{2(\text{x})^{\frac{3}{2}}}\Bigg]$
$\therefore\text{f}(\text{x})\ \text{x}=1=\frac{1}{2}\Big[\frac{1+4}{2\times1}\Big]$
$=\frac{5}{4}$
View full question & answer→MCQ 111 Mark
If $\text{f}(\text{x})=\frac{\text{x}^{\text{n}}-\text{a}^{\text{n}}}{\text{x}-\text{a}}$ for some constant, $a,$ then $f'(a)$ is equal to:
- A
$1$
- B
$0$
- ✓
- D
$\frac{1}{2}$
AnswerGiven $\text{f}(\text{x})=\frac{\text{x}^{\text{n}}-\text{a}^{\text{n}}}{\text{x}-\text{a}}$
$\text{f'}(\text{x})=\frac{(\text{x}-\text{a})(\text{n.}\text{x}^{\text{n-1}}-(\text{x}^{\text{n}-\text{a}^{\text{n}}})1}{(\text{x}-\text{a})^{2}}$
So, $\text{f}(\text{a})=\frac{0}{0} =$ Does not exist.
View full question & answer→MCQ 121 Mark
If $f(x) = 1 - x + x^2 - x^3 + ....-x^{99} + x^{100},$ then $f(1)$ is equal to:
AnswerGiven that $f(x) = 1 - x + x^2 - x^3 + .......-x^{99} + x^{100}$
$f(x) = -1 + 2x - 3x^2 + ... -99.x^{98} + 100.x^{99}$
$f(x) = -1 + 2 - 3 + ....-99 + 100$
$=(- 1 - 3 - 5 ....-99) + (2 + 4 + 6 + ....100$)
$=\frac{50}{2}\big[2\times1+(50-1)(-2)\big]+\frac{50}{2}\big[2\times2(50-1)2\big] $
$=25\big[-11+102\big]=25\times2=50$
View full question & answer→MCQ 131 Mark
If $\text{f}(\text{x})=\text{x}-[\text{x}],\in\text{R}$ then $\text{f}'\big(\frac{1}{2}\big)$ is equal to:
- A
$\frac{3}{2}$
- ✓
$1$
- C
$0$
- D
$-1$
AnswerGiven $f(x) = x - [x]$
we have ti first check for differentiability of $f(x)$ at $\text{x}=\frac{1}{2}$
$\therefore \text{Lf}'\Big(\frac{1}{2}\Big)=\text{LHD}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{f}\big[\frac{1}{2}-\text{h}\big]-\text{f}\big[\frac{1}{2}\big]}{-\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\big(\frac{1}{2}-\text{h}\big)-\big[\frac{1}{2}-\text{h}\big]-\frac{1}{2}+\big[\frac{1}{2}\big]}{-\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{1}{2}-\text{h}-0-\frac{1}{2}+0}{-\text{h}}=\frac{-\text{h}}{-\text{h}}=1$
$\therefore \text{Rf}'\Big(\frac{1}{2}\Big)=\text{RHD}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{f}\big[\frac{1}{2}-\text{h}\big]-\text{f}\big[\frac{1}{2}\big]}{\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\big(\frac{1}{2}+\text{h}\big)-\big[\frac{1}{2}+\text{h}\big]-\frac{1}{2}+\big[\frac{1}{2}\big]}{-\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{1}{2}+\text{h}-1-\frac{1}{2}+1}{\text{h}}$
$=\frac{\text{h}}{\text{h}}$
$=1$
Since, $\text{LHD = RHD}$
$\text{f}'\big(\frac{1}{2}\big)=1$
View full question & answer→MCQ 141 Mark
$\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}^{\text{m}}-1}{\text{x}^{\text{n}}-1}$ is equal to:
- A
$1$
- ✓
$\frac{\text{m}}{\text{n}}$
- C
$\frac{-\text{m}}{\text{n}}$
- D
$\text{m}^{2}\text{n}^{2}$
AnswerCorrect option: B. $\frac{\text{m}}{\text{n}}$
Given $\lim\limits_{\text{x} \rightarrow 1}\frac{\text{x}^{\text{m}}-1}{\text{x}^{\text{n}}-1}=\lim\limits_{\text{x} \rightarrow 1}\frac{\frac{\text{x}^{\text{m}-(1)^\text{m}}}{\text{x}-1}}{\frac{\text{x}^{\text{n}-(1)^{\text{n}}}}{\text{x}-1}}$
$=\frac{\text{m}(1)^{\text{m}-1}}{\text{n}(1)^{\text{n}-1}}$
$=\frac{\text{m}}{\text{n}}$
View full question & answer→MCQ 151 Mark
$\lim\limits_{\text{x} \rightarrow0}\frac{1-\cos4\theta}{1-\cos6\theta}$ is equal to:
- ✓
$\frac{4}{9}$
- B
$\frac{1}{2}$
- C
$\frac{-1}{2}$
- D
$-1$
AnswerCorrect option: A. $\frac{4}{9}$
Given $\lim\limits_{\theta \rightarrow 0}\frac{1-\cos4\theta}{1-\cos6\theta}=\lim\limits_{\theta \rightarrow 0}\frac{2\sin^{2}2\theta}{2\sin^{2}3\theta}$
$=\lim\limits_{\theta \rightarrow 0}\frac{\sin^{2}2\theta}{\sin^{2}3\theta}$
$=\lim\limits_{\theta \rightarrow 0}\Big[\frac{\sin2\theta}{\sin3\theta}\Big]^{2}$
$=\lim\limits_{\theta \rightarrow 0}\bigg[\frac{\frac{\sin2\theta}{2\theta}\times2\theta}{\frac{\sin3\theta}{2\theta}\times3\theta}\bigg]$
$=\Big[\frac{2\theta}{2\theta}\Big]^{2}$
$=\Big(\frac{2}{3}\Big)^{2}$
$=\frac{4}{9}$
View full question & answer→MCQ 161 Mark
If $\begin{cases}\frac{\sin[\text{x}]}{[\text{x}]} & \text{x}\neq0\\0, & [\text{x}]=0\end{cases}$ where $[.]$ denotes the greatest integer function. then $\lim\limits_{\text{x} \rightarrow 0}\text{f}(\text{x})$ is equal to :
AnswerGiven $\begin{cases}\frac{\sin[\text{x}]}{[\text{x}]} & \text{x}\neq0\\0, & [\text{x}]=0\end{cases}$
$\text{L}.\text{H}.\text{H}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin[\text{x}]}{[\text{x}]}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[0-\text{h}]}{[0-\text{h}]} $
$=\lim\limits_{\text{h} \rightarrow 0}\frac{-\sin[\text{-h}]}{[-\text{h}]}$
$=-1$
$\text{R}.\text{H}.\text{L}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin[\text{x}]}{[\text{x}]}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[0+\text{h}]}{[0+\text{h}]}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[\text{h}]}{[\text{h}]}$
$=1$
$\text{L}.\text{H}.\text{L}\neq\text{R}.\text{H}.\text{L}$
So, the limit does not exist.
View full question & answer→MCQ 171 Mark
If $\text{y}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}}$ then $\frac{\text{dy}}{\text{dx}}$ at $x = 0$ is equal to:
- ✓
$2$
- B
$0$
- C
$\frac{1}{2}$
- D
AnswerGiven $\text{y}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\frac{-(\sin\text{x}+\cos\text{x})(\cos\text{x}+\sin\text{x})}{(\sin\text{x}-\cos\text{x})^{2}}$
$=\frac{-(\sin\text{x}+\cos\text{x})^{2}(\sin\text{x}+\cos\text{x})}{(\sin\text{x}-\cos\text{x})^{2}}$
$=\frac{\sin^{2}\text{x}+\cos^{2}\text{x}+2\sin\text{x}\cos\text{x}}{(\sin\text{x}-\cos\text{x})^{2}}$
$=\frac{-2}{(\sin\text{x}-\cos\text{x})^{2}}$
$\therefore \Big(\frac{\text{dy}}{\text{dx}}\Big)$
$=\frac{-2}{(\sin\text{0}-\cos0)^{2}}$
$=\frac{-2}{(-1)^{2}}$
$=2$
View full question & answer→MCQ 181 Mark
$\lim\limits_{\text{x} \rightarrow0}\frac{\sec^{2}\text{x}-2}{\tan\text{x}-1}$ is:
AnswerGiven $\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{\sec^{2}\text{x}-2}{\tan\text{x}-1}$
$=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{1+\tan^{2}\text{x}-2}{\tan\text{x}-1}$
$=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{\tan\text{x}-1}{\tan\text{x}-1}$
$=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{(\tan\text{x}+1)(\tan\text{x}-1)}{(\tan\text{x}-1)}$
$=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}(\tan\text{x}+1)$
$=\tan\frac{\pi}{4}+1$
$=1+1=2$
$=2$
View full question & answer→MCQ 191 Mark
$\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\sqrt{\text{x}+1}-\sqrt{1-\text{x}}}$ is:
AnswerGiven $\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\sqrt{\text{x}+1}-\sqrt{1-\text{x}}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}\big[\sqrt{\text{x}+1}\sqrt{1-\text{x}}\big] }{\big(\sqrt{\text{x}+1}-\sqrt{1-\text{x}}\big)\big(\sqrt{\text{x}+1}+\sqrt{1-\text{x}}\big)}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}\big[\sqrt{\text{x}+1}\sqrt{1-\text{x}}\big] }{\text{x}+1-1+\text{x}}$
$=\frac{1}{2}\cdot\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\text{x}}\big[\sqrt{\text{x}+1}+\sqrt{1-\text{x}}\big]$
Taking limit, we get
$=\frac{1}{2}\times1\times\big[\sqrt{0+1}+\sqrt{0-1}\big]$
$=\frac{1}{2}\times1\times2$
$=1$
View full question & answer→MCQ 201 Mark
$\lim\limits_{\text{x} \rightarrow \pi}\frac{\sin\text{x}}{\text{x}-\pi}$ is:
AnswerGiven, $\lim\limits_{\text{x} \rightarrow \pi}\frac{\sin\text{x}}{\text{x}-\pi}$
$=\lim\limits_{\text{x} \rightarrow\pi}\frac{\sin(\pi)-\text{x}}{-(\pi-\text{x})}$
$=-1$
View full question & answer→MCQ 211 Mark
$\lim\limits_{\text{x} \rightarrow0}\frac{\big(\sqrt{\text{x}}-1\big)\big(2\text{x}-3\big)}{2\text{x}^{2}+\text{x}-3}$ is:
- A
$\frac{1}{10}$
- ✓
$-\frac{1}{10}$
- C
$1$
- D
AnswerCorrect option: B. $-\frac{1}{10}$
Given $\lim\limits_{\text{x} \rightarrow0}\frac{\big(\sqrt{\text{x}}-1\big)\big(2\text{x}-3\big)}{2\text{x}^{2}+\text{x}-3}$
$=\lim\limits_{\text{x} \rightarrow1}\frac{\big(\sqrt{\text{x}}-1\big)\big(2\text{x}-3\big)}{\text{x}\big(2\text{x}+3\big)-1\big(2\text{x}+3\big)}$
$=\lim\limits_{\text{x} \rightarrow 1}\frac{\big(\sqrt{\text{x}}-1\big)(\big(2\text{x}-3\big)}{\big(\text{x}-1\big)\big(2\text{x}+3\big)}$
$=\lim\limits_{\text{x} \rightarrow 1}\frac{2\text{x}-3}{\big(\sqrt{\text{x}}+1\big)\big(2\text{x}+3\big)}$
Taking limit, we get
$=\frac{2(1)-3}{\big(\sqrt{\text{x}}+1\big)\big(2\times1+3\big)}$
$=\frac{-1}{2\times5}$
$=\frac{-1}{10}$
View full question & answer→MCQ 221 Mark
If $\text{f}(\text{x})=\begin{cases}\text{x}^{2}-1 & 0<\text{x}<2\\2\text{x}+3, & 2\geq\text{3}<3\end{cases}$ then the quadeatic equation whose roots are $\lim\limits_{\text{x} \rightarrow 2^{-}}\text{f}(\text{x})$ and $\lim\limits_{\text{x} \rightarrow 2^{+}}\text{f}(\text{x})$ is:
- A
$\text{x}^{2}-6\text{x}+9=0$
- B
$\text{x}^{2}-7\text{x}+8=0$
- C
$\text{x}^{2}+14\text{x}+49=0$
- ✓
$\text{x}^{2}-10\text{x}+21=0$
AnswerCorrect option: D. $\text{x}^{2}-10\text{x}+21=0$
Given $\text{f}(\text{x})=\begin{cases}\text{x}^{2}-1 & 0<\text{x}<2\\2\text{x}+3, & 2\geq\text{3}<3\end{cases}$
$\therefore\ \lim\limits_{\text{x} \rightarrow 2}\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow 2^{-}}(\text{x}^{2}-1)$
$ \lim\limits_{\text{h} \rightarrow 0}[(2-\text{x})^{2}-1]=\lim\limits_{\text{h} \rightarrow 0}(4+\text{h}^{2}-4\text{h}-1)$
$ =\lim\limits_{\text{h} \rightarrow 0}(\text{h}^{2}-4\text{h}+3)=3$
$\therefore\ \lim\limits_{\text{x} \rightarrow 2}\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow 2^{+}}(2\text{x}+3)$
$ =\lim\limits_{\text{h} \rightarrow 0}[2(2+\text{h})+3]$
$=7$
Therefore, the quadratic equation whose roots are $3$ and $7$ is $\text{x}^{2}-10\text{x}+21=0$
View full question & answer→MCQ 231 Mark
$\lim\limits_{\text{x} \rightarrow \pi}\frac{\text{x}^{2}\cos\text{x}}{1-\cos\text{x}}$ is equal to:
- ✓
$2$
- B
$\frac{3}{2}$
- C
$-\frac{3}{2}$
- D
$1$
AnswerGiven $\lim\limits_{\text{x} \rightarrow \pi}\frac{\text{x}^{2}\cos\text{x}}{1-\cos\text{x}}=\lim\limits_{\text{x} \rightarrow 0}\frac{\text{x}^{2}\cos\text{x}}{2\sin^{2}\frac{\text{x}}{2}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\text{x}^{2}}{4}\times4\cos\text{x}}{2\sin^{2}\frac{\text{x}}{2}}=\lim\limits_{\text{x} \rightarrow 0}\frac{\Big(\frac{\text{x}}{2}\Big)\cdot2\cos\text{x}}{\sin^{2}\frac{\text{x}}{2}}$
$=\lim\limits_{\text{x} \rightarrow 0}\bigg(\frac{\frac{\text{x}}{2}}{\sin\frac{\text{x}}{2}}\bigg)\cdot2\cos\text{x}$
$=2\cos\text{x}0=2\times1=2$
View full question & answer→