Mathematics Test - 21

\(\begin{aligned} & 2 y=\left[\cot ^{-1}\left(\frac{\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x}{\frac{1}{2} \cos x-\frac{\sqrt{3}}{2} \sin x}\right)\right]^2 \\ & \Rightarrow 2 y=\left[\cot ^{-1}\left(\frac{\cos \left(\frac{\pi}{6}-x\right)}{\sin \left(\frac{\pi}{6}-x\right)}\right)\right]^2 \\ & \Rightarrow 2 \mathrm{y}=\left[\cot ^{-1}\left(\cot \left(\frac{\pi}{6}-\mathrm{x}\right)\right)\right]^2 \because \frac{\pi}{6}-\mathrm{x} \in\left(-\frac{\pi}{3}, \frac{\pi}{6}\right) \\ & \Rightarrow 2 y= \begin{cases}\left(\frac{7 \pi}{6}-x\right)^2, & \text { if } \frac{\pi}{6}-x \in\left(\frac{-\pi}{3}, 0\right) \\ \left(\frac{\pi}{6}-x\right)^2, & \text { if } \frac{\pi}{6}-x \in\left(0, \frac{\pi}{0}\right)\end{cases} \\ & \Rightarrow \frac{d y}{d x}=\left\{\begin{array}{c}x-\frac{7 \pi}{6} \text { if } x \in\left(\frac{\pi}{6}, \frac{\pi}{2}\right) \\ x-\frac{\pi}{6} \text { if } x \in\left(0, \frac{\pi}{6}\right)\end{array}\right. \\ & \end{aligned}\)

We know that some quadratic equations exist with no real solutions. So, we need to extend the real number system to a larger system so that we can find the solution of a quadratic equation. In fact. the main objective is to solve the equation \(a x^{2}+b x+c=0\), where \(D=b^{2}-4 a c<0\), which is not possible in the system of real numbers. Therefore, the complex numbers came into the picture.

For the complex number \(z=a+i b, a\) is called the real part, denoted by \(\operatorname{Re} z\) and \(b\) is called the imaginary part denoted by \(I {~m} {z}\) of the complex number \({z}\).

\(z=\frac{5+2 i}{2-5 i}-\frac{3-4 i}{4+3 i}-\frac{1}{i}\)

\(z=\frac{5+2 i}{2-5 i} \times \frac{2+5 i}{2+5 i}-\frac{3-4 i}{4+3 i} \times \frac{4-3 i}{4-3 i}-\frac{1}{i} \times \frac{i}{i}\)

\(z=\frac{10+29 i+10(i)^{2}}{4-25 i^{2}}-\frac{12-25 i+12 i^{2}}{16-9 i^{2}}-\frac{i}{i^{2}}\)

\(z=\frac{10+29 i-10}{4+25}-\frac{12-25 i-12}{16+9}-\frac{i}{(-1)}\)

\(z=\frac{29 i}{29}-\frac{(-25 i)}{25}-\frac{i}{(-1)}\)

\(z=i+i+i\)

\(z=3 i\)

Compare this equation with \(z=a+b i\)

\(a=0, b=3\)

\(\therefore\) Real part \({a}=0\)

Given,

\(3 \sin \alpha=5 \sin \beta\)

\(\Rightarrow \frac{\sin \alpha}{\sin \beta}=\frac{5}{3}\)

As we know that componendo and dividendo rule is given as,

If \(\frac{a}{b}=\frac{c}{d}\)

Then \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)

Then applying this rule, we get

\(\Rightarrow \frac{\sin \alpha+\sin \beta}{\sin \alpha-\sin \beta}=\frac{5+3}{5-3}=4\)

As we know,

\(\sin C+\sin D=2 \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)\)

\(\sin C-\sin D=2 \sin \left(\frac{C-D}{2}\right) \cos \left(\frac{C+D}{2}\right)\)

\(\Rightarrow \frac{2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}}{2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}}=4\)

\(\Rightarrow \frac{\tan \frac{\alpha+\beta}{2}}{\tan \frac{\alpha-\beta}{2}}=4\)

Let \(\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=\vec{b}, \overrightarrow{O C}=\vec{c}\) and \(\overrightarrow{O D}=\vec{d}\).

Therefore,

\(\overrightarrow{ OA }+\overrightarrow{ OB }+\overrightarrow{ OC }+\overrightarrow{ OD }=\overrightarrow{ a }+\overrightarrow{ b }+\overrightarrow{ c }+\overrightarrow{ d }\)

The midpoint \(P\) of \(A B\), is \(\frac{\vec{a}+\vec{b}}{2}\).

The position vector of the midpoint \(Q\) of \(C D\), is \(\frac{\vec{c}+\vec{d}}{2}\).

Therefore, the position vector of the midpoint of \(P Q\) is \(\frac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}\)

Similarly, the position vector of the midpoint of RS is \(\frac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}\),

i.e., \(\overrightarrow{O E}=\frac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}\)

\(x=4\)

Given: \(p(x)=(4 e)^{2 x}\)

Let \(I=\int \mathrm{p}(\mathrm{x}) \mathrm{dx}\)

\(=\int(4 e)^{2 x} d x\)

Let \(2 x=t\)

Differentiating with respect to \(\mathrm{x},\) we get

\(\Rightarrow 2 \mathrm{dx}=\mathrm{dt}\)

\(\Rightarrow d x=\frac{d t}{2}\)

Now,

\(\int \mathrm{p}(\mathrm{x}) \mathrm{dx}\)

\(=\frac{1}{2} \int(4 \mathrm{e})^{\mathrm{t}} \mathrm{dt}\)

\(=\frac{1}{2} \frac{(4 e)^{t}}{\ln 4 \mathrm{e}}+\mathrm{c}\)

\(=\frac{1}{2} \frac{(4 \mathrm{e})^{t}}{(\ln 4+\ln \mathrm{e})}+\mathrm{c} \quad(\because \log \mathrm{mn}=\log \mathrm{m}+\log \mathrm{n})\)

\(=\frac{1}{2} \frac{(4 \mathrm{e})^{t}}{(1+2 \ln 2)}+\mathrm{c}\)

Given, \(\mathrm{e}^{\mathrm{y}}+\mathrm{xy}=\mathrm{e}\)........(i)

Putting \(\mathrm{x}=0\) in (i), \(\Rightarrow \mathrm{e}^{\mathrm{y}}=\mathrm{e} \Rightarrow \mathrm{y}=1\)

On differentiating (i) w. r. to \(x\)

\(e^y \frac{d y}{d x}+x \frac{d y}{d x}+y=0\)........(ii)

Putting \(y=1\) and \(x=0\) in (ii),

\(e \frac{d y}{d x}+0+1=0 \Rightarrow \frac{d y}{d x}=-\frac{1}{e}\)

On differentiating (ii) w. r. to \(\mathrm{x}\),

\(e^y \frac{d^2 y}{d x^2}+\frac{d y}{d x} \cdot e^y \cdot \frac{d y}{d x}+x \frac{d^2 y}{d x^2}+\frac{d y}{d x}+\frac{d y}{d x}=0\)

Putting \(y=1, x=0\) and \(\frac{d y}{d x}=-\frac{1}{e}\) in (iii),

\(e \frac{d^2 y}{d x^2}+\frac{1}{e}-\frac{2}{e}=0 \Rightarrow \frac{d^2 y}{d x^2}=\frac{1}{e^2}\)

Hence, \(\left(\frac{d y}{d x}=\frac{d^2 y}{d x^2}\right) \equiv\left(-\frac{1}{e}, \frac{1}{e^2}\right)\)

\(\cos ^{-1}\left(4 x^3-3 x\right)\)

Put \(x=\cos \theta \)

\(\Rightarrow \theta=\cos ^{-1} x\)

\(\cos ^{-1}\left(4 x^3-3 x\right)\)

\(=\cos ^{-1}\left(4 \cos ^3 \theta-3 \cos \theta\right)\)

\(=\cos ^{-1}(\cos 3 \theta) \quad\left(\because \cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta\right)\)

\(=3 \theta \quad\left(\because \cos ^{-1} \cos \theta=\theta\right)\)

\(=3 \cos ^{-1} x\)

Since f(x) is an odd periodic function with period 2.

\(\therefore f(-x)=-f(x)\) and \(f(x+2)=f(x)\)

\(\therefore f(2)=f(0+2)=f(0)\)

and \(f(-2)=f(-2+2)=f(0)\)

Now, \(f(0)=f(-2)=-f(2)=-f(0)\)

\(\Rightarrow 2 f(0)=0\), i.e.,\(f(0)=0\)

\(\therefore f(4)=f(2+2)=f(2)=f(0)=0\)

Thus, \(f(4)=0\)

Given:

The equation of a line is,

\(\frac{x-1}{3}=\frac{y+2}{4}=\frac{z-3}{-2}=k\)

Any point on the line is,

\((3 k+1,4 k-2,-2 k+3)\)

If the given line intersects the plane \(2 x-y+3 z-1=0,\) then any point on the line lies in the plane. \(\therefore 2(3 k+1)-(4 k-2)+3(-2 k+3)-1=0\)

\(\Rightarrow -4 k+12=0\)

\(\Rightarrow k=3\)

\(\therefore\) Point is \((9+1,12-2-6+3)\)

i.e., (10, 10, -3)

The \((\mathrm{r}+1)\) th term in the expansion of \(\left(\sqrt{\frac{\mathrm{x}}{3}}-\frac{\sqrt{3}}{\mathrm{x}^{2}}\right)^{10}\) is given by:

\(\mathrm{T}_{\mathrm{r}+1}=(-1)^{\mathrm{r}} \cdot{ }^{10} \mathrm{C}_{\mathrm{r}}\left(\sqrt{\frac{\mathrm{x}}{3}}\right)^{10-\mathrm{r}} \cdot\left(\frac{\sqrt{3}}{\mathrm{x}^{2}}\right)^{\mathrm{r}}\)

\(=(-1)^{\mathrm{r}}{ }^{10} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{5-\frac{5 \mathrm{r}}{2}} 3^{\mathrm{r}-5}\).....(1)

For term to be independent of \(\mathrm{x}\),

\(5-\frac{5 \mathrm{r}}{2}=0\)

\(\Rightarrow \mathrm{r}=2\)

Substituting the values in (1) we get:

\(\mathrm{T}_{3}=(-1)^{2}{ }^{10} \mathrm{C}_{2} 3^{-3}\)

\(=\frac{10(9)}{2(27)}\)

\(=\frac{5}{3}\)