Mathematics Test - 23

The vector equation of a line passing through a point with position vector \(\overrightarrow{r_{1}}\) and parallel to \(\vec{m}\) is given by:

\(\vec{r}=\overrightarrow{r_{1}}+\lambda \vec{m}\) where, \(\lambda\) is a scalar

Given here:

Equation of plane is \(\vec{r} \cdot(2 \hat{i}-3 \hat{j}-\hat{k})=0\) and the line is perpendicular to the given plane.

It is given that, the required line is perpendicular to the plane \(\vec{r} \cdot(2 \hat{i}-3 \hat{j}-\hat{k})=0\)

i.e., The required line is parallel to the vector \(\vec{m}=2 \hat{i}-3 \hat{j}-\hat{k}\)

The line passes through the point whose position vector is \(\hat{i}-2 \hat{j}+5 \hat{k}\) i.e., \(\overrightarrow{r_{1}}=\hat{i}-2 \hat{j}+5 \hat{k}\)

As we know that, the vector equation of a line passing through a point with position vector \(\overrightarrow{r_{1}}\) and parallel to \(\vec{m}\) is given by:

\(\vec{r}=\overrightarrow{r_{1}}+\lambda \vec{m}\)

So, the required equation of line is:

\(\vec{r}=(\hat{i}-2 \hat{j}+5 \hat{k})+\lambda \cdot(2 \hat{i}-3 \hat{j}-\hat{k})\)

\(f(x)=\frac{x^2-5 x-6}{x^2+5 x-6}\)

Iif any function is not continuous at \(x=a\) then \(\lim _{x \rightarrow a} f(x)=l \neq f(a)\)

So, for the function \(f(x)\) if denominator is 0 at \(x=\) a then we can say that \(f(a)\) is infinite and limit cannot exist.

Let's find the value of \(x\) for which the denominator of \(f(x)\) is 0.

\( x^2+5 x-6=0 \)

\(\Rightarrow(x+6)(x-1)=0 \)

\(\Rightarrow x=-6,1\)

The given equation is

\( (2 y+x) \frac{d y}{d x}=1 \)

\( \Rightarrow(2 y+x)=\frac{d x}{d y}\)

\( \Rightarrow \frac{d x}{d y}-x=2 y\)

\(\therefore\) It is linear differential equation is of first order

\( \text { IF }=e^{\int-1 d y}\)

\(\Rightarrow I F=e^{-y}\)

Now, \(x \times(I F)=\int Q(I F) d y\)

\( \Rightarrow x \times e^{-y}=\int 2 y \times e^{-y} d y \)

\( \Rightarrow x e^{-y}=2\left[y \int e^{-y} d y-\int\left\{\frac{d y}{d y} \times \int e^{-y} d y\right\} d y\right] \)

\( \Rightarrow x e^{-y}=2\left[-y e^{-y}+\int e^{-y} d y\right]+c \)

\(\Rightarrow x e^{-y}=2\left[-y e^{-y}-e^{-y}\right]+c \)

\( \Rightarrow x+2 y+2=c e^y\)

Clearly \(f(A)=I+A+A^2+\ldots \ldots+A^{16}\)

\(A^2=A A=\left[\begin{array}{ll}0 & 5 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}0 & 5 \\ 0 & 0\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=O\)

\(A^3=0, A^4=0, \ldots \ldots A^{16}=0\)

\(\therefore f(A)=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]+\left[\begin{array}{ll}0 & 5 \\ 0 & 0\end{array}\right]+0+0+\ldots \ldots+0\)

\(=\left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]\)

Let \({r}\) be the common ratio of the given GP

Now,

\(\left|\begin{array}{lll}\ln a_{1} & \ln a_{2} & \ln a_{3} \\ \ln a_{4} & \ln a_{5} & \ln a_{6} \\ \ln a_{7} & \ln a_{8} & \ln a_{9}\end{array}\right|\)

Apply, \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}\) and \(\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}\)

\(\begin{aligned}&=\left|\begin{array}{lll}\ln a_{1} & \ln a_{2}-\ln a_{1} & \ln a_{3}-\ln a_{2} \\ \ln a_{4} & \ln a_{5}-\ln a_{4} & \ln a_{6}-\ln a_{5} \\ \ln a_{7} & \ln a_{8}-\ln a_{7} & \ln a_{9}-\ln a_{8}\end{array}\right| \\&=\left|\begin{array}{lll}\ln a_{1} & \ln r & \ln r \\ \ln a_{4} & \ln r & \ln r \\ \ln a_{7} & \ln r & \ln r\end{array}\right| \quad\left[\text { Since, } \frac{a_{p+1}}{a_{p}}=r \text { and } \log a_{p+1}-\log a_{p}=\log \frac{a_{p+1}}{a_{p}}=\log \right]\end{aligned}\)

\(=0\qquad  \qquad{\left[\text { Since, } \text{C}_{3}=\text{C}_{2}\right]}\)

Let the three numbers in A.P. be \(a-d, a\), and \(a+d\).

Given: the sum of three numbers in A.P. is 24

\( \Rightarrow( a - d )+ a +( a + d )=24 \)

\( \Rightarrow 3 a =24 \)

\( \Rightarrow a =8\)

Also, given: the product of three numbers in A.P. is 440 .

\(\Rightarrow(a-d)(a)(a+d)=440\)

Put \(a =8\)

\( \Rightarrow(8-d)(8)(8+d)=440 \)

\( \Rightarrow(8-d)(8+d)=55 \)

\( \Rightarrow 64-d^2=55 \)

\( \Rightarrow d^2=9 \)

\( \Rightarrow d= \pm 3\)

Therefore, when d = 3, the numbers are 5, 8, and 11 and when d = –3, the numbers are 11, 8, and 5.
So, the three numbers are 5, 8, and 11.

As we know,

\(\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}\)

Given,

\(\tan ^{-1} x +\cot ^{-1} x =\frac{\pi}{2}\)

\(\Rightarrow \tan ^{-1}( x )+\tan ^{-1}\left(\frac{1}{ x }\right)=\frac{\pi}{2}\)

\(\Rightarrow \tan ^{-1} \frac{ x +\frac{1}{x}}{1- x \times \frac{1}{x}}=\frac{\pi}{2}\)

\(\Rightarrow \tan ^{-1} \frac{ x +\frac{1}{x}}{0}=\frac{\pi}{2}\)

\(\Rightarrow \tan ^{-1} \frac{ x +\frac{1}{x}}{0}=\frac{\pi}{2}\)

\(\Rightarrow \tan ^{-1}(\infty)=\frac{\pi}{2}\)

This is true for all\(x \in R-(-1,1)\)