Question 1 |
Given that the determinant of the matrix
\begin{bmatrix} 1 & 3 & 0\\ 2 & 6 & 4\\ -1 & 0 & 2 \end{bmatrix}
is -12, the determinant of the matrix
\begin{bmatrix} 2 & 6 & 0\\ 4 & 12 & 8\\ -2 & 0 & 4 \end{bmatrix}
\begin{bmatrix} 1 & 3 & 0\\ 2 & 6 & 4\\ -1 & 0 & 2 \end{bmatrix}
is -12, the determinant of the matrix
\begin{bmatrix} 2 & 6 & 0\\ 4 & 12 & 8\\ -2 & 0 & 4 \end{bmatrix}
-96 | |
-24 | |
24 | |
96 |
Question 1 Explanation:
Let D=-12 for the given matrix
|A|=\left|\begin{array}{ccc}2 & 6 & 0 \\4 & 12 & 8 \\-2 & 0 & 4\end{array}\right|=(2)^{3}\left|\begin{array}{ccc}1 & 3 & 0 \\2 & 6 & 4 \\-1 & 0 & 2\end{array}\right|
(Taking 2 common from each row)
\begin{aligned}\therefore \text{Det}(A) &=(2)^{3} \times D \\=& 8 x-12=-96\end{aligned}
|A|=\left|\begin{array}{ccc}2 & 6 & 0 \\4 & 12 & 8 \\-2 & 0 & 4\end{array}\right|=(2)^{3}\left|\begin{array}{ccc}1 & 3 & 0 \\2 & 6 & 4 \\-1 & 0 & 2\end{array}\right|
(Taking 2 common from each row)
\begin{aligned}\therefore \text{Det}(A) &=(2)^{3} \times D \\=& 8 x-12=-96\end{aligned}
Question 2 |
\lim_{x\rightarrow 0} \frac{x-sin x}{1-cos x} is
0 | |
1 | |
3 | |
not define |
Question 2 Explanation:
\lim_{x \rightarrow 0} \frac{x-\sin x}{1-\cos x}
Applying L ' Hospital's rule
\lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x-sinx)}{\frac{d}{dx}(1-cosx)}=\lim_{x \rightarrow 0} \frac{(1-\cos x)}{\sin x}
(It is still of \frac{0}{0} form)
Again applying L' Hospital's rule
\lim_{x \rightarrow 0} \frac{\frac{d}{d x}(1-\cos x)}{\frac{d}{d x}(\sin x)}=\lim_{x \rightarrow 0} \frac{\sin x}{\cos x}=\frac{0}{1}=0
Applying L ' Hospital's rule
\lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x-sinx)}{\frac{d}{dx}(1-cosx)}=\lim_{x \rightarrow 0} \frac{(1-\cos x)}{\sin x}
(It is still of \frac{0}{0} form)
Again applying L' Hospital's rule
\lim_{x \rightarrow 0} \frac{\frac{d}{d x}(1-\cos x)}{\frac{d}{d x}(\sin x)}=\lim_{x \rightarrow 0} \frac{\sin x}{\cos x}=\frac{0}{1}=0
Question 3 |
The argument of the complex number \frac{1+i}{1-i},where i=\sqrt{-1}, is
-\pi | |
-\pi/2 | |
\pi/2 | |
\pi |
Question 3 Explanation:
\begin{aligned} Let z&=\frac{1+i}{1-i} \\ or z&=\frac{(1+i)(1+i)}{(1-i)(1+i)}=\frac{1+i^{2}+2 i}{1-i^{2}}=\frac{2 i}{2}=i \\ z&=x+i y=i \\ SO, x&=0 \\ y &=1 \end{aligned}
\begin{aligned} Arg(z) &=\tan ^{-1}\left(\frac{y}{x}\right) \\ &=\tan ^{-1}\left(\frac{1}{0}\right)=\tan ^{-1} \infty=\frac{\pi}{2} \end{aligned}
\begin{aligned} Arg(z) &=\tan ^{-1}\left(\frac{y}{x}\right) \\ &=\tan ^{-1}\left(\frac{1}{0}\right)=\tan ^{-1} \infty=\frac{\pi}{2} \end{aligned}
Question 4 |
The matrix form of the linear system \frac{dx}{dt}=3x-5y and \frac{dy}{dt}=4x+8y is
\frac{d}{dt} \begin{Bmatrix} x\\ y \end{Bmatrix}=\begin{bmatrix} 3 & -5\\ 4& 8 \end{bmatrix}\begin{Bmatrix} x\\ y \end{Bmatrix} | |
\frac{d}{dt} \begin{Bmatrix} x\\ y \end{Bmatrix}=\begin{bmatrix} 3 & 8\\ 4& -5 \end{bmatrix}\begin{Bmatrix} x\\ y \end{Bmatrix} | |
\frac{d}{dt} \begin{Bmatrix} x\\ y \end{Bmatrix}=\begin{bmatrix} 4 & -5\\ 3& 8 \end{bmatrix}\begin{Bmatrix} x\\ y \end{Bmatrix} | |
\frac{d}{dt} \begin{Bmatrix} x\\ y \end{Bmatrix}=\begin{bmatrix} 4 & 8\\ 3& -5 \end{bmatrix}\begin{Bmatrix} x\\ y \end{Bmatrix} |
Question 4 Explanation:
\begin{aligned} \frac{d x}{d t} &=3 x-5 y \\ \frac{d y}{d t} &=4 x+8 y \\ \frac{d}{d t}\left\{\begin{array}{l}x \\y\end{array}\right\}&=\left[\begin{array}{cc}3 & -5 \\4 & 8\end{array}\right]\left\{\begin{array}{l}x \\y\end{array}\right\}=\left[\begin{array}{cc} 3 x & -5 y \\4 x & +8 y\end{array}\right]\end{aligned}
Question 5 |
Which one of the following describes the relationship among the three vectors \hat{i}+\hat{j}+\hat{k}, 2\hat{i}+3\hat{j}+\hat{k} and 5\hat{i}+6\hat{j}+4\hat{k}
The vectors are mutually perpendicular | |
The vectors are linearly dependent | |
The vectors are linearly independent | |
The vectors are unit vectors |
Question 5 Explanation:
For linear dependency, det \left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & 3 & 1 \\ 5 & 6 & 4\end{array}\right| must be zero.
\therefore \Delta=1(12-6)-1(8-5)+1(12-15)
=6-3-3=0
\therefore There three vectors are linearly dependent.
\therefore \Delta=1(12-6)-1(8-5)+1(12-15)
=6-3-3=0
\therefore There three vectors are linearly dependent.
There are 5 questions to complete.