Download App

Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants1 Mark
MCQ
If $\text{x},\text{ y}\in\text{R},$ then the determinant $\triangle=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&0\end{vmatrix}$ lies in the interval:
  • ✓
    $\Big[-\sqrt{2},\sqrt{2}\Big]$
  • B
    $[-1,1]$
  • C
    $\Big[-\sqrt{2},1\Big]$
  • D
    $\Big[-1,-\sqrt{2}\Big]$
✓

Answer

Correct option: A.
$\Big[-\sqrt{2},\sqrt{2}\Big]$
$\triangle=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&0\end{vmatrix}$
$=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\0&0&\sin\text{y}-\cos\text{y}\end{vmatrix} [$Applying $R_3 \rightarrow R_3 - \cos y\ R_1 + \sin y\ R_2]$
$=(\sin\text{y}-\cos\text{y})(\cos^2\text{x}+\sin^2\text{x})$
$=\sin\text{y}-\cos\text{y}$
$=\sqrt{2}\Big(\frac{1}{\sqrt{2}}\sin\text{y}-\frac{1}{\sqrt{2}}\cos\text{y}\Big)$
$=\sqrt{2}\Big(\cos\frac{\pi}{4}\sin\text{y}-\sin\frac{\pi}{4}\cos\text{y}\Big)$
$=\sqrt{2}\sin\Big(\text{y}-\frac{\pi}{4}\Big)$
Therefore, $-\sqrt{2}\leq\triangle\leq\sqrt{2}$
Hence, the correct option is $(a)$

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 Determinants→Determinants — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Choose the correct answer from the given four options.
The vector in the direction of the vector $\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$ that has magnitude 9 is:
→
If $\sin(\text{x}+\text{y})=\log(\text{x}+\text{y}),$ then $\frac{\text{dy}}{\text{dx}}=$
→
$\int\text{x}\sec\text{x}^2\text{ dx}$ is equal to:
→
The probabilities of a student getting $\text{I, II}$ and $\text{III}$ division in an examination are $\frac{1}{10},\frac{3}{5}$ and $\frac{1}{4}$ respectively. The probability that the student fails in the examination is.
→
$\int\limits^\frac{\pi}{2}_0\frac{1}{1+\tan\text{x}}\text{dx}$ is equal to:
→
Let $\vec a\, = \,\hat i\, + \,\hat j\, + \,\sqrt 2 \hat k,\,\,\vec b\, = \,{b_1}\hat i\, + \,{b_2}\hat j\, + \sqrt 2 \hat k$ and $\vec c\, = \,5\hat i\, + \,\hat j + \sqrt 2 \hat k$ be three vectors such that the projection vector of $\vec b$ on $\vec a$ is $\vec a$. If $\vec a\, + \vec b$ is perpendicular to $\vec c$ , then $\left| {\vec b} \right|$ is equal to
→
If $A^3=O$, then $A^2+A+I=$
→
The function$f(x) = [x]\cos \left[ {\frac{{2x - 1}}{2}} \right]\pi ,\,$ where$[.]$ denotes the greatest integer function, is discontinuous at
→
The differential equation ${\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2} - {\left( {\frac{{dy}}{{dx}}} \right)^{1/2}} = {y^3}$ has the degree
→
The value of the integral $\int_0^{\frac{1}{2}} \frac{1+\sqrt{3}}{\left((x+1)^2(1-x)^6\right)^{\frac{1}{4}}} d x$  is . . . . . . . .
→