MCQ 2011 MarkWhat is the value of $\int\limits_{-1}^{1} \sin^3\text{x}\cos^2\text{xdx:}$✓$0$B$1$C$\frac{1}{2}$D$2$AnswerCorrect option: A. $0$View full question & answer→
MCQ 2021 MarkChoose the correct answer in Exercise: $\text{If}\ \text{f}(\text{x})=\int^{\text{x}}_{0}\text{t}\sin\text{t}\ \text{dt}$, then $\text{f}\text{(x)}$ isA$\cos\text{x}+\text{x}\sin\text{x}$✓$\text{x}\sin\text{x}$C$\text{x}\cos\text{x}$D$\sin\text{x}+\text{x}\cos\text{x}$AnswerCorrect option: B. $\text{x}\sin\text{x}$$\text{f}\text{(x)}=\int\limits_{0}^{\text{x}}\text{t}\sin\text{t}\ \text{dt}$ $\therefore\text{f}\text{'(x)}=\text{x}\sin\text{x}\ $ $\big[$$\therefore$ of first fundamental theorem$\big]$ $\text{f}\text{(x)}=\int\limits_{0}^{\text{x}}\text{t}\sin\text{t}\ \text{dt}$ $\therefore\text{f}\text{'(x)}=\text{x}\sin\text{x}\ \ $ $\big[$$\therefore$ of first fundamental theorem$\big]$View full question & answer→
MCQ 2031 MarkChoose the correct answer in Exercise: $\int^{\sqrt{3}}_{1}\frac{\text{dx}}{1+\text{x}^{2}}\text{equals}$A$\frac{\pi}{3}$✓$\frac{2\pi}{3}$C$\frac{\pi}{6}$D$\frac{\pi}{12}$AnswerCorrect option: B. $\frac{2\pi}{3}$$\int\frac{\text{dx}}{1+\text{x}^{2}}=\tan^{-1}\text{x}=\text{F}\text{(x)}$ By second fundamental theorem of calculus, we obtain $\int\limits_{1}^{\sqrt{3}}\frac{\text{dx}}{1+\text{x}^{2}}=\text{F}(\sqrt{3})-\text{F}(1)$ $=\tan^{-1}\sqrt{3}-\tan^{-1}1$ $=\frac{\pi}{3}-\frac{\pi}{4}$ $=\frac{\pi}{12}$View full question & answer→
MCQ 2041 MarkChoose the correct answer in Exercises : $\int\frac{\sin^2\text{x}-\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$ is equal to✓$\tan\text{x}+\cot\text{x}+\text{C}$B$\tan\text{x}+\text{cosec x}+\text{C}$C$-\tan\text{x}+\cot\text{x}+\text{C}$D$\tan\text{x}+\sec\text{x}+\text{C}$AnswerCorrect option: A. $\tan\text{x}+\cot\text{x}+\text{C}$$\int\frac{\sin^2\text{x}-\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$ $=\int\bigg(\frac{\sin^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}-\frac{\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\bigg)\text{dx}$ $=\int(\sec^2\text{x}-\text{cosec}^2\text{x})\text{ dx}$ $=\tan\text{x}+\cot\text{x}+\text{C}$ Hence, the correct answer is $A$.View full question & answer→
MCQ 2051 MarkThe derivative of $\text{f(x)}=\int\limits^{\text{x}^3}_{\text{x}^2}\frac{1}{\log_{\text{e}}\text{t}}\text{ dt},(\text{x}>0),$ is:A$\frac{1}{3\ln\text{x}}$B$\frac{1}{3\ln\text{x}}-\frac{1}{2\ln\text{x}}$✓$\big(\ln\text{x}\big)^{-1}\text{x}(\text{x}-1)$D$\frac{3\text{x}^2}{\ln\text{x}}$AnswerCorrect option: C. $\big(\ln\text{x}\big)^{-1}\text{x}(\text{x}-1)$$\text{f}'(\text{x})=\frac{1}{\log_\text{e}\text{x}^3}(3\text{x}^2)-\frac{1}{\log_\text{e}\text{x}^2}(2\text{x})$ $=\frac{3\text{x}^2}{3\ln\text{ x}}-\frac{2\text{x}}{2\ln\text{ x}}$ $=\frac{\text{x}^2}{\ln\text{ x}^{-1}}-\frac{\text{x}}{\ln\text{ x}}$ $=\frac{1}{\ln\text{ x}}\text{x}(\text{x}-1)$ $=\big(\ln\text{ x}\big)^{-1}\text{x}(\text{x}-1)$View full question & answer→
MCQ 2061 Mark$\int\frac{\sin^2\text{x}-\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{dx}$ is equal to:A$\tan\text{x}+\cos\text{x}+\text{c}$B$\tan\text{x}+\text{cosec}\ \text{x}+\text{c}$✓$\tan\text{x}+\text{cot}\ \text{x}+\text{c}$D$\tan\text{x}+\sec\text{x}+\text{c}$AnswerCorrect option: C. $\tan\text{x}+\text{cot}\ \text{x}+\text{c}$View full question & answer→
MCQ 2071 Mark$\int\log_{10}\text{xdx}=$A$\log_{\text{e}}10.\text{x}\log_{\text{e}}\big(\frac{\text{x}}{\text{e}}\big)+\text{c}$✓$\log_{10}\text{e.x}\log_{\text{e}}\big(\frac{\text{x}}{\text{e}}\big)+\text{c}$C$\log_{10}\text{e.x}\log_{\text{e}}\big(\frac{\text{x}}{\text{e}}\big)+\text{n}$D$\log_{10}\text{e.x}\log_{\text{e}}\big(\frac{\text{x}}{\text{n}}\big)+\text{n}$AnswerCorrect option: B. $\log_{10}\text{e.x}\log_{\text{e}}\big(\frac{\text{x}}{\text{e}}\big)+\text{c}$View full question & answer→
MCQ 2081 Mark$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin|\text{x}|\text{dx}$ is equal to:A1✓2C-1D-2AnswerCorrect option: B. 2$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin|\text{x}|\text{dx}$ $=-\int\limits^0_{-\frac{\pi}{2}}\sin\text{x}\text{ dx}+\int\limits^\frac{\pi}{2}_0\sin\text{x}\text{ dx}$ $=-\big[-\cos\text{x}\big]^0_{-\frac{\pi}{2}}+\big[-\cos\text{x}\big]^\frac{\pi}{2}_0$ $=1-0-0+1$ $=2$View full question & answer→
MCQ 2091 Mark$\lim\limits_{\text{n}\rightarrow\infty}\Big\{\frac{1}{2\text{n}+1}+\frac{1}{2\text{n}+2}+\ .....+\frac{1}{2\text{n}+\text{n}}\Big\}$ is equal to:A$\ln\Big(\frac{1}{3}\Big)$B$\ln\Big(\frac{2}{3}\Big)$✓$\ln\Big(\frac{3}{2}\Big)$D$\ln\Big(\frac{4}{3}\Big)$AnswerCorrect option: C. $\ln\Big(\frac{3}{2}\Big)$$\lim\limits_{\text{n}\rightarrow\infty}\Big\{\frac{1}{2\text{n}+1}+\frac{1}{2\text{n}+2}+\ .....\ +\frac{1}{2\text{n}+\text{n}}\Big\}$ $=\lim\limits_{\text{n}\rightarrow\infty}\sum\limits^\text{n}_{\text{r}=1}\frac{1}{2\text{n}+\text{r}}$ $=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\sum\limits^\text{n}_{\text{r}=1}\frac{1}{2+\frac{\text{r}}{\text{n}}}$ let $\frac{\text{r}}{\text{n}}=\text{x}$ $=\int\limits^\infty_0\frac{1}{2+\text{x}}\text{dx}$ $=\Big[\log\big(2+\text{x}\big)\Big]^\infty_0$ $=\log3-\log2$ $=\log\frac{3}{2}$ $=\ln\Big(\frac{3}{2}\Big)$View full question & answer→
MCQ 2101 MarkWhat is the value of $\int_{0}^{\frac{ \pi}{2}}\frac{\sqrt{\tan\text{x}}}{\sqrt{\tan}\ \text{x}+\sqrt{\cot}\ \text{x}}\text{dx}?$A$\frac{\pi}{2}$✓$\frac{\pi}{4}$C$\frac{\pi}{8}$DNone of theseAnswerCorrect option: B. $\frac{\pi}{4}$View full question & answer→