Choose the correct answer from the given four options. If 3 ^ -1 … - Vidyadip

MCQ

Choose the correct answer from the given four options. If $3\tan^{-1}\text{x}+\cot^{-1}\text{x}=\pi,$ then $x$ equals:

A
$0$
✓
$1$
C
$-1$
D
$\frac{1}{2}$

✓

Answer

Correct option: B.

$1$

We have, $3\tan^{-1}\text{x}+\cot^{-1}\text{x}=\pi$
$\Rightarrow 2\tan^{-1}\text{x}+(\tan^{-1}\text{x}+\cot^{-1}\text{x})=\pi$
$\Rightarrow 2\tan^{-1}\text{x}+\frac{\pi}{2}=\pi$
$\Big(\because \tan^{-1}\text{x}+\cot^{-1}\text{x}=\frac{\pi}{2}\Big)$
$\Rightarrow 2\tan^{-1}\text{x}=\frac{\pi}{2}$
$\Rightarrow \tan^{-1}\text{x}=\frac{\pi}{4}$
$\Rightarrow \text{x}=1$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Inverse Trigonometric Functions→Inverse Trigonometric Functions — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Assume X, Y, Z, W and P are matrices of order $2 \times n, 3 \times k, 2 \times p, n \times 3$ and $p \times k$ respectively. Then the restriction on n, k and p so that PY + WY will be defined are:

$\int_{}^{} {\sqrt {\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right)} } \;dx = $

The order of the differential equation of all circles of radius $r$, having centre on $y$-axis and passing through the origin is

The function $f(x) = {x^{ - x}},\,(x\, \in \,R)$ attains a maximum value at $x =$

The lines $\frac{1-x}{2}=\frac{y-1}{3}=\frac{z}{1}$ and $\frac{2 x-3}{2 p}=\frac{y}{-1}=\frac{z-4}{7}$ are perpendicular to each other for $p$ equal to :

If $\int {f(x)\,\,dx = g(x),} $ then $\int {{f^{ - 1}}(x)} \,\,dx$ is equal to

The area bounded by the curve $x^2 = 4y$ and straight line $x = 4y - 2$ is:

A curve is such that the area of the region bounded by the co-ordinate axes, the curve $\&$ the ordinate of any point on it is equal to the cube of that ordinate. The curve represents

Let $f(x) = \left\{ \begin{array}{l}{x^p}\sin \frac{1}{x},x \ne 0\\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,,x = 0\end{array} \right.$ then $f(x)$ is continuous but not differential at $x = 0$ if

Let $f (x) =$ $\left[ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\frac{{\begin{array}{*{20}{c}} {{2^x}}&{ + {2^{3 - x}}}&{ - 6} \end{array}}}{{\begin{array}{*{20}{c}} {\sqrt {{2^{ - x}}} }&{ - {2^{1 - x}}} \end{array}}}}&{if}&{x > 2} \end{array}} \\ {} \\ {} \\ {} \\ {\begin{array}{*{20}{c}} {\frac{{\begin{array}{*{20}{c}} {{x^2}}&{ - 4} \end{array}}}{{x - \sqrt {3x - 2} }}}&{if}&{x < 2} \end{array}} \end{array}} \right.$ then