Mathematics Test-32

we know, when a line passing through a print \(\vec{a}\) and perallel to the vector \(\vec{b}\), the equation of line,

\(\vec{r}=a+\lambda \vec{b}\)

Given line, \(\frac{x-2}{2}=\frac{y+1}{3}=\frac{z-3}{-2}\)

we see that this line passing through the point \((2,-1,3)\) and parallel to the vector \((2 \hat{i}+3 \hat{j}-2 \hat{k})\)

Therefor the vector equation of this line

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

Given,

The sequence 2, 6, 18, 54, ..... is a GP.

Here, we have to find which term of the given sequence is 4374.

As we know that, the the general term of a GP is given by:

\(a_{n}=a r^{n-1}\)

Here, \(a=2, r=3\) and let \(a_{n}=4374\)

\(\Rightarrow 4374=(2) \cdot(3)^{n-1}\)

\(\Rightarrow 3^{n-1}=2187\)

\(\Rightarrow 3^{n-1}=3^{7}\)

\(\Rightarrow n-1=7\)

\(\Rightarrow n = 8\)

Thus, 4374 is the 8th term of the given sequence.

Given, 

\(\mathrm{X}=8^{\mathrm{n}}-7 \mathrm{n}-1\)

\(=(1+7)^{\mathrm{n}}-7 n-1\)

\(=1+7 \mathrm{n}+\frac{\mathrm{n}(\mathrm{n}-1)}{2} 7^{2}+\ldots+7^{\mathrm{n}}-7 \mathrm{n}-1\)

\(=\frac{\mathrm{n}(\mathrm{n}-1)}{2} 7^{2}+\ldots+7^{\mathrm{n}}\)

\(=49\left[\frac{\mathrm{n}(\mathrm{n}-1)}{2}+\ldots+7^{\mathrm{n}-2}\right]\)

So, the set \(\mathrm{X}\) will be some specific multiples of \(49\).

\(\Rightarrow\)\(\mathrm{Y}=49(\mathrm{n}-1)\)

Therefore, the set \(Y\) will be all multiples of \(49 .\) So, it will contain the elements of \(\mathrm{X}\) too.

So, \(\mathrm{X} \subset \mathrm{Y}\)

We know that, the general form of a quadratic equation is

\(x^2-(M+N) x-M N=0\)

Here, \(M\) and \(N\) are the roots of the equation.

Let \(1^{\text {st}}\) root \(M=7+\sqrt{3}\)

And \(2^{\text {nd }}\) is \(N=7-\sqrt{3}\)

By puting the value of \(M\) and \(N\) and solving them, we get,

\(x^2-(7+\sqrt{3}+7-\sqrt{3}) x+[(7+\sqrt{3}) \times(7-\sqrt{3})]=0\)

We know that,

\(\left[(a+b)(a-b)=\left(a^2-b^2\right)\right]\)

Then, \(x^2-14 x+(49-3)=0\)

So the quadratic equation is \(x^2-14 x+46=0\)

\(\sec ^2 \theta=3 \Rightarrow \sec \theta=\frac{\sqrt{3}}{1}=\frac{\text { Hypotenuse }}{\text { Base }}\)

By Pythagoras Theorem,

\((\text {Hypotenuse})^2=(\text {Base})^2+(\text {Altitude})^2\)

\((\sqrt{3})^2=(1)^2+(\text {Altitude})^2 \)

\(\Rightarrow 3=1+(\text {Altitude})^2 \Rightarrow(\text {Altitude})^2=3-1=2 \)

\(\therefore \text { Altitude }=\sqrt{2}\)

\(\therefore \tan \theta=\frac{\text { Altitude }}{\text { Base }}=\frac{\sqrt{2}}{1}=\sqrt{2} \)

\(\operatorname{cosec} \theta=\frac{\text { Hypotenuse }}{\text { Altitude }}=\frac{\sqrt{3}}{\sqrt{2}}=\sqrt{\frac{3}{2}}\)

Now, \(\frac{\tan ^2 \theta-\operatorname{cosec}^2 \theta}{\tan ^2 \theta+\operatorname{cosec}^2 \theta}\)

\(=\frac{(\sqrt{2})^2-\left(\sqrt{\frac{3}{2}}\right)^2}{(\sqrt{2})^2+\left(\sqrt{\frac{3}{2}}\right)^2}=\frac{2-\frac{3}{2}}{2+\frac{3}{2}}\)

\(=\frac{\frac{1}{2}}{\frac{7}{2}}=\frac{1}{2} \times \frac{2}{7}=\frac{1}{7}\)

Given equation is \(y ^2+( x - b )^2= c\)

There are two constants \(b\) and \(c\) so differentiate two times

Differentiating w.r.t \(x\)

\(2 y \frac{d y}{d x}+2(x-b)=0 \)

\(y \frac{d y}{d x}=b-x\)

Differentiating again w.r.t \(x\)

\(\left(\frac{d y}{d x}\right)^2+y \frac{d^2 y}{d x^2}=-1 \)

\(y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2+1=0\)

Given equation is:

\(a^2 x^2-3 a b x+2 b^2=0\)

\(\Rightarrow a^2 x^2-2 a b x-abx+2 b^2=0\)

\(\Rightarrow a x(a x-2 b)-b(a x-2 b)=0\)

\(\Rightarrow (a x-b)(a x-2 b)=0\)

If \(a x-b=0\)

\(\therefore x=\frac{b}{a}\)

If \(a x-2 b=0\)

\(\therefore x=\frac{2 b}{a}\)

So the roots are \(\frac{b}{a}\) and \(\frac{2 b}{a}\).

It is given that \(f: R-\left\{-\frac{4}{3}\right\} \rightarrow R\) be a function as \(f(x)=\frac{4 x}{3 x+4}\)

Let \(y\) be an arbitrary element of Range \(f\).

Then, there exists \(x \in R-\left\{-\frac{4}{3}\right\}\) such that \(y=f(x)\)

\(\Rightarrow y=\frac{4 x}{3 x+4}\)

\(\Rightarrow 3 x y+4 y=4 x\)

\(\Rightarrow x(4-3 y)=4 y\)

\(\Rightarrow x=\frac{4 y}{4-3 y}\)

Let us define g: Range \(\mathrm{f} \rightarrow \mathrm{R}-\left\{-\frac{4}{3}\right\}\) as \({g}({y})=\frac{4 {y}}{4-3 {y}}\)

Now, \(g o f(x)=g(f(x))=g\left(\frac{4 x}{3 x+4}\right)=\frac{4\left(\frac{4 x}{3 x+4}\right)}{4-3\left(\frac{4 x}{3 x+4}\right)}=\frac{16 x}{12 x+16-12 x}=\frac{16 x}{16}=x\)

and \(f o g(y)=f(g(y))=f\left(\frac{4 y}{4-3 y}\right)=\frac{4\left(\frac{4 y}{4-3 y}\right)}{3\left(\frac{4 y}{4-3 y}\right)+4}=\frac{16 y}{12 y+16-12 y}=\frac{16 y}{16}=y\)

\(\therefore g o f=I_{R-\left\{-\frac{4}{3}\right\}}\) and fog \(=I_{\text {Range } f}\)

Thus, \(g\) is the inverse of \(f\) i.e. \(f^{-1}=g\).

Thus, the inverse of \(f\) is the map \(g:\) Range \(f \rightarrow R-\left\{-\frac{4}{3}\right\},\) which is given by \(g(y)=\frac{4 y}{4-3 y}\).