Download App

Inverse Trigonometric Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions1 Mark
MCQ
Solve for $x : \sin^{-1}2\text{x}+\sin^{-1}3\text{x}=\frac{\pi}{3}$
  • A
    $\sqrt{\frac{76}{3}}$
  • ✓
    $\sqrt{\frac{3}{76}}$
  • C
    $\frac{3}{\sqrt{76}}$
  • D
    $\frac{\sqrt{3}}{76}$
✓

Answer

Correct option: B.
$\sqrt{\frac{3}{76}}$

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

If $A + B = \left[ {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right]$and $A - 2B = \left[ {\begin{array}{*{20}{c}}{ - 1}&1\\0&{ - 1}\end{array}} \right]\,,$ then $A=$
→
The value of $p$ for which the function $f(x) = \left\{ \begin{array}{l}\frac{{{{({4^x} - 1)}^3}}}{{\sin \frac{x}{p}\log \left[ {1 + \frac{{{x^2}}}{3}} \right]}},\,x \ne 0\\\,\,\,\,\,\,\,\,\,\,\,12{(\log 4)^3},\,\,x = 0\end{array} \right.$ may be continuous at $x = 0$, is
→
The equation of the plane through the intersection of the planes $x + 2y + 3z = 4$ and $2x + y - z = -5$ and perpendicular to the plane $5x + 3y + 6z + 8 = 0$ is :
→
Number of solution(s) satisfying the equation, $3x^2 - 2x^3 = log_2 (x^2 + 1) -log_2 x$ is
→
The general solution of the differential equation $x d y-\left(1+x^2\right) d x=d x$ is :
→
If $OP = 8$  and $\overrightarrow {OP} $ makes angles ${45^o}$ and ${60^o}$ with $OX-$ axis and $OY-$axis respectively, then $\overrightarrow {OP} = $
→
If  $ [x] $ denotes the greatest integer less than or equal to $ x$ , then the value of $\int_{\,1}^{\,5} {\,\,[|x - 3|]\,dx} $ is
→
If $A$ is a $m \times n$ matrix such that $A B$ and $B A$ are both defined, then $B$ is an
→
The point at the curve $y = 12x - x^2$ where the slope of the tangent is zero will be :
→
ABCD is a parallelogram with AC and BD as diagonals. Then, $\overrightarrow{\text{AC}}-\overrightarrow{\text{BD}}=$
→