2 Marks Questions

Question 1012 Marks

Evaluate the following:
$\tan\Big(\cos^{-1}\frac{8}{17}\Big)$

Answer

$\tan\Big(\cos^{-1}\frac{8}{17}\Big)$
$=\begin{Bmatrix}\tan^{-1}\frac{\sqrt{1-\Big(\frac{8}{17}\Big)^2}}{\frac{8}{17}}\end{Bmatrix}$ $\bigg[\because\ \cos^{-1}\text{x}=\tan^{-1}\bigg(\frac{\sqrt{1-\text{x}^2}}{\text{x}}\bigg)\bigg]$
$=\tan\Bigg(\tan^{-1}\frac{\frac{15}{17}}{\frac{8}{17}}\Bigg)$
$=\frac{15}{8}$

View full question & answer→

Question 1022 Marks

Find the domain of the following function:$\text{f(x)}=\sin^{-1}\text{x}+\sin^{-1}2\text{x}$

Answer

Let f(x) = g(x) + h(x), where
Therefore the domain of f(x) is given by intersection of the domain of g(x) and h(x)
The domain of g(x) is [-1, 1]
The domain of h(x) is $\Big[-\frac{1}{2},\frac{1}{2}\Big]$
Therefore, the intersection of g(x) and h(x) is $\Big[-\frac{1}{2},\frac{1}{2}\Big]$
Hence, the domain is $\Big[-\frac{1}{2},\frac{1}{2}\Big]$

View full question & answer→

Question 1032 Marks

Write the value of $\cos\Big(2\sin^{-1}\frac{1}{2}\Big).$

Answer

$\cos\Big(2\sin^{-1}\frac{1}{2}\Big)$
$=\cos\Big(2\times\frac{\pi}{6}\Big)$ $\Big\{\text{Since},\sin^{-1}\text{x}=\text{An angle in }\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\text{ whose sine is x}\Big\}$
$=\cos\Big(\frac{\pi}{3}\Big)$
$=\frac{1}{2}$
Hence,
$\cos\Big(2\sin^{-1}\frac{1}{2}\Big)=\frac{1}{2}$

View full question & answer→

Question 1042 Marks

Write the value of $2\sin^{-1}\frac{1}{2}+\cos^{-1}\Big(-\frac{1}{2}\Big).$

Answer

$2\sin^{-1}\frac{1}{2}+\cos^{-1}\Big(-\frac{1}{2}\Big)$
$=\sin^{-1}2\times\frac{1}{2}\sqrt{1-\Big(\frac{1}{2}\Big)^2}+\cos^{-1}\Big(-\frac{1}{2}\Big)$
$=\sin^{-1}\frac{\sqrt3}{2}+\cos^{-1}\Big(-\frac{1}{2}\Big)$
$=\sin^{-1}\Big(\sin\frac{\pi}{3}\Big)+\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)$
$=\frac{\pi}{3}+\frac{2\pi}{3}$
$=\pi$

View full question & answer→

Question 1052 Marks

Evaluate:
$\cos\Big(\sec^{-1}\text{x}+\text{cosec}^{-1}\text{x}\Big)$

Answer

$\cos\Big(\sec^{-1}\text{x}+\text{cosec}^{-1}\text{x}\Big)$
$=\cos\Big(\frac{\pi}{2}\Big)$
$=0$

View full question & answer→

Question 1062 Marks

Evaluate the following:
$\cot^{-1}\frac{1}{\sqrt3}-\text{cosec}^{-1}(-2)+\sec^{-1}\Big(\frac{2}{\sqrt3}\Big)$

Answer

$\cot^{-1}\frac{1}{\sqrt3}-\text{cosec}^{-1}(-2)+\sec^{-1}\Big(\frac{2}{\sqrt3}\Big)$$=\frac{\pi}{6}-\Big(-\frac{\pi}{6}\Big)+\frac{\pi}{3}$
$=\frac{\pi}{6}+\frac{\pi}{6}+\frac{\pi}{3}$
$=\frac{4\pi}{6}$
$=\frac{2\pi}{3}$

View full question & answer→

Question 1072 Marks

$\sin^{-1}\frac{1}{2}-2\sin^{-1}\frac{1}{\sqrt2}$

Answer

$\sin^{-1}\frac{1}{2}-2\sin^{-1}\frac{1}{\sqrt2}$
$=\sin^{-1}\frac{1}{2}-\sin^{-1}\Bigg(2\times\frac{1}{\sqrt2}\sqrt{1-\Big(\frac{1}{\sqrt2}\Big)^2}\Bigg)$
$=\sin^{-1}\frac{1}{2}-\sin^{-1}(1)$
$=\frac{\pi}{6}-\frac{\pi}{2}$
$=-\frac{\pi}{3}$

View full question & answer→

Question 1082 Marks

For the principal values, evaluate the following:
$\sin^{-1}\Big(-\frac{\sqrt3}{2}\Big)+\text{cosec}^{-1}\Big(-\frac{2}{\sqrt3}\Big)$

Answer

$\sin^{-1}\Big(-\frac{\sqrt3}{2}\Big)+\text{cosec}^{-1}\Big(-\frac{2}{\sqrt3}\Big)$
$=-\sin^{-1}\Big(\frac{\sqrt3}{2}\Big)+\text{cosec}^{-1}\Big(-\frac{2}{\sqrt3}\Big)$
$=-\sin^{-1}\Big(\sin\frac{\pi}{3}\Big)+\text{cosec}^{-1}\Big[\text{cosec}\Big(-\frac{\pi}{3}\Big)\Big]$
$=-\frac{\pi}{3}-\frac{\pi}{3}$
$=-\frac{2\pi}{3}$

View full question & answer→

Question 1092 Marks

For the principal values, evaluate the following:
$\tan^{-1}(-1)+\cos^{-1}\Big(-\frac{1}{2}\Big)$

Answer

$\tan^{-1}(-1)+\cos^{-1}\Big(-\frac{1}{2}\Big)$
$\tan^{-1}\Big\{\tan\Big(-\frac{\pi}{4}\Big)\Big\}+\cos^{-1}\Big(\cos\frac{3\pi}{4}\Big)$$\begin{bmatrix}\because\text{Range of tan is }\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big);-\frac{\pi}{4}\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\\\ \\\text{and range of cosine is }[0,\pi];\frac{3\pi}{4}\in[0,\pi]\end{bmatrix}$
$=-\frac{\pi}{4}+\frac{3\pi}{4}$
$=\frac{\pi}{2}$
$\therefore\tan^{-1}(-1)+\cos^{-1}\Big(-\frac{1}{2}\Big)=\frac{\pi}{2}$

View full question & answer→

Question 1102 Marks

Evaluate the following:
$\text{cosec}^{-1}\Big(\text{cosec}\frac{11\pi}{6}\Big)$

Answer

$\text{cosec}^{-1}\Big(\text{cosec}\frac{11\pi}{6}\Big)$
$=\text{cosec}^{-1}\Big(\text{cosec}\Big(2\pi-\frac{\pi}{6}\Big)\Big)$
$=\text{cosec}^{-1}\Big(\text{cosec}\Big(-\frac{\pi}{6}\Big)\Big)$
$=\frac{\pi}{6}$

View full question & answer→

Question 1112 Marks

Find the domain of the following function:$\text{f(x)}=\sin^{-1}\sqrt{\text{x}^2-1}$

Answer

To the domain of $sin^{-1}y$ which is [-1, 1] $\therefore x^2 -1 \in$ [0, 1] as square root can not be negative
$\Rightarrow x^2 \in [1, 2]$
$\Rightarrow\text{x}\in\big[-\sqrt2,-1\big]\cup\big[1,\sqrt2\big]$
Hence, the domain is $\big[-\sqrt2,-1\big]\cup\big[1,\sqrt2\big]$

View full question & answer→

Question 1122 Marks

Write the value of $\tan\Big(2\tan^{-1}\frac{1}{5}\Big)$

Answer

Let $\tan\theta=\frac{1}{5}$$\tan\Big(2\tan^{-1}\frac{1}{5}\Big)$
$=\tan2\theta$
$=\frac{2\tan\theta}{1-\tan^2\theta}$
$=\frac{2\times\frac{1}{5}}{1-\frac{1}{25}}$
$=\frac{\frac{2}{5}}{\frac{24}{25}}$
$=\frac{5}{12}$

View full question & answer→

MATHS 2 Marks Questions