Let \(m\) be the slope of required line.

\(\therefore\left|\frac{m-(-1)}{1+m(-1)}\right|=1\)

\(\Rightarrow \frac{m+1}{1-m}=\pm 1\)

\(\Rightarrow m+1=1-m\)

and \(m+1=-1+m\)

\(\Rightarrow m=0\) and \(m=\infty\)

\(\therefore\) Equation of line through \((1,1)\) is \(y-1=0, x-1=0\)

Thus, option \(x-1=0, y-1=0\) is correct.

Let

\(f(x)=e^{3 x+b}\)

\(\Rightarrow f(x+h)=e^{a(x+h)+b}\)

\(\frac{d}{d x}(f(x))=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

\(=\lim _{h \rightarrow 0} \frac{e^{3(x+h)+b}-e^{(2 x+b)}}{h}\)

\(=\lim _{h \rightarrow 0} \frac{e^{a x+b} e^{a x}-e^{a x+b}}{h}\)

\(=\lim _{h \rightarrow 0} e^{a x+b}\left\{\frac{\left(e^{a h}-1\right)}{a h}\right\} \times a\)

\(=a e^{a x+b}\)

\(\left[\right.\) Since, \(\left.\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1\right]\)

So,

\(\frac{d}{d x}\left(e^{a x+b}\right)=a e^{a x+b}\)

The two normal vectors are \(m=2 \hat{i}+3 \hat{j}+\hat{k}\) and \(n=\hat{i}+3 \hat{j}+2 \hat{k}\)

The line \(\mathrm{L}\) is along, \(m \times n=\left[\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 3 & 2\end{array}\right]\)

\(=\hat{i}(6-3)-\hat{j}(4-1)+\hat{k}(6-3)\)

\(=3 \hat{i}-3 \hat{j}+3 \hat{k}=3(\hat{i}-\hat{j}+\hat{k})\)

Now, the direction cosines of X-axis are (1, 0, 0).

\(\therefore \cos \alpha=\frac{3(i-j+k) \cdot i}{\sqrt{3^{2}\left(1^{2}+1^{2}+1^{2}\right) \sqrt{1}}}\)

\(=\frac{3}{3 \sqrt{3}}=\frac{1}{\sqrt{3}}\)

\(\Rightarrow \cos \alpha=\frac{1}{\sqrt{3}}\)

Let the set of examinees who solved Question 8 be A,

The set of candidates who solve Question 9 is B,

And the set of candidates who solve Question 10 is C, then

$\begin{array}{l}n\left(\mathrm{A}\right)=67\\ n\left(\mathrm{B}\right)=46\\ n\left(\mathrm{C}\right)=40\\ n(\mathrm{A}\cap \mathrm{B})=28\\ n(\mathrm{B}\cap \mathrm{C})=8\\ n(\mathrm{A}\cap \mathrm{C})=26\\ n(A\cap B\cap C)=2\end{array}$

Number of examinees who solved question 8 but not questions 9 and 10 =$=67-(26+24+2)$

$$\begin{gathered} =67-52 \\ =15 \end{gathered}$$

The given equation is,

$a{x}^2+2bx+c=0$

Here, addition of roots$=\alpha +\beta =\frac{-2b}{a}$

Multiplication of roots$=\alpha \beta =\frac{c}{a}$

$\Rightarrow \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$

$=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }$

$=\frac{\frac{4b^2}{a^2}-2\frac{c}{a}}{{\displaystyle \frac{c}{a}}}$

$=\frac{4b^2-2ac}{ac}$

Clearly, we have

\(a=1, h=-3, b=9, g=\frac{3}{2}, f=\frac{-9}{2}\) and \(c=-4\)

Required distance \(=\left|2 \sqrt{\frac{t^{2}-b c}{b(a+b)}}\right|\)

\(=\left|2 \sqrt{\frac{\left(\frac{-9}{2}\right)^{2}+9 \times 4}{9(9+1)}}\right|\)

\(=\left|2 \sqrt{\frac{225}{4 \times 90}}\right|=\left|\frac{2 \sqrt{5}}{2 \sqrt{2}}\right|=\sqrt{\frac{5}{2}}\)

Hence, the correct option is (C).

Given:

\(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)

\(\therefore e=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{16-9}{16}}\)

\(=\sqrt{\frac{7}{16}}=\frac{\sqrt{17}}{4}\)

\(\therefore\) Coordinates of foci are \((\pm \sqrt{7}, 0)\).

Since, centre of circle is \((0,3)\) and passing through foci \((\pm \sqrt{7}, 0)\)

\(\therefore\) Radius of the circle \(=\sqrt{(0 \pm \sqrt{7})^{2}+(3-0)^{2}}\)

\(=\sqrt{7+9}=4\)

\(A=\left[\begin{array}{cc}x & -7 \\ 7 & y\end{array}\right]\)

\(\operatorname{Now} A^T=\left[\begin{array}{cc}x & 7 \\ -7 & y\end{array}\right]\)

Now for \(A\) to be skew symmetric matrix, \(A+A^T=0\)

\(\Rightarrow\left[\begin{array}{cc} x & -7 \\ 7 & y \end{array}\right]+\left[\begin{array}{cc} x & 7 \\ -7 & y \end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\)

\(\Rightarrow\left[\begin{array}{cc}2 x & 0 \\ 0 & 2 y \end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\)

Now equating the corresponding elements we get, \(( x , y )=(0,0)\)

Equality of complex numbers.

Two complex numbers \(z_{1}=x_{1}+i y_{1}\) and \(z_{2}=x_{2}+i y_{2}\) are equal if and only if \(x_{1}=x_{2}\) and \(y_{1}=y_{2}\)

Or \(\operatorname{Re}\left(z_{1}\right)=\operatorname{Re}\left(z_{2}\right)\) and \(\operatorname{Im}\left(z_{1}\right)=\operatorname{Im}\left(z_{2}\right)\)

\(\Rightarrow z=\frac{1-i}{i}\)

Multiplying numerator and denominator by \({i}\)

\(\Rightarrow z=\frac{1-i}{i} \times \frac{i}{i}\)

\(=\frac{i-i^{2}}{i^{2}}\)

\(=\frac{i+1}{-1}\)

\(=-1-i\)

\(\operatorname{Re}(z)=-1\)

\(\operatorname{lm}(z)=-1\)

Given, \(e^{\theta \phi}=c+4 \theta \phi\), where \(c\) is an arbitrary constant and \(\phi\) is a function of \(\theta\)

Differentiate w.r.to \(\theta,\) we get

\(\Rightarrow \frac{\mathrm{d}}{\mathrm{d} \theta}\left(\mathrm{e}^{\theta \phi}\right)=\frac{\mathrm{d}}{\mathrm{d} \theta}(\mathrm{c}+4 \theta \phi)\)

\(\Rightarrow \mathrm{e}^{\theta \phi}\left(\theta \frac{\mathrm{d} \phi}{\mathrm{d} \theta}+\phi\right)=4\left(\theta \frac{\mathrm{d} \phi}{\mathrm{d} \theta}+\phi\right)\)

\(\Rightarrow\left(e^{\theta \phi}-4\right)\left(\theta \frac{\mathrm{d} \phi}{\mathrm{d} \theta}+\phi\right)=0\)

\(\Rightarrow\left(\theta \frac{\mathrm{d} \phi}{\mathrm{d} \theta}+\phi\right)=0\)

\(\Rightarrow \theta \mathrm{d} \phi+\phi \mathrm{d} \theta=0\)

\(\therefore \phi d \theta=-\theta \mathrm{d} \phi\)