Mathematics Test - 27

It is given that,

$a{\sin }^2\theta +b{\cos }^2\theta =c$

So,

$\sqrt{\frac{c-b}{a-c}}=\sqrt{\frac{a{\sin }^2\theta +b{\cos }^2\theta -b}{a-a{\sin }^2\theta -b{\cos }^2\theta }}$

$\begin{array}{l}=\sqrt{\frac{a{\sin }^2\theta -b(1-{\cos }^2\theta )}{a(1-{\sin }^2\theta )-b{\cos }^2\theta }}\\ =\sqrt{\frac{a{\sin }^2\theta -b{\sin }^2\theta }{a{\cos }^2\theta -b{\cos }^2\theta }}\\ =\sqrt{\frac{{\sin }^2\theta (a-b)}{{\cos }^2\theta (a-b)}}\\ =\frac{\sin \theta }{\cos \theta }=\tan \theta \end{array}$

Equation of the circle,

$2x^2+2y^2-3x+5y-7=0$

$\Rightarrow x^2+y^2-\frac{3}{2}x+\frac{5}{2}y-\frac{7}{2}=0$

Where,$2g=-\frac{3}{2}$and$2f=\frac{5}{2},c=-\frac{7}{2}$

$\therefore$$g=-\frac{3}{4},f=\frac{5}{4},c=-\frac{7}{2}$

Center of the circle$=(-g,-f)=(\frac{3}{4},-\frac{5}{4})$

Radius of the circle$=\sqrt{g^2+f^2-c}$

$=\sqrt{{\left(\frac{-3}{4}\right)}^2+{\left(\frac{5}{4}\right)}^2-(-\frac{7}{2})}$

$=\sqrt{\frac{9}{16}+\frac{25}{16}+\frac{7}{2}}$

$=\sqrt{\frac{9+25+56}{16}}$

$=\sqrt{\frac{90}{16}}=\frac{3\sqrt{10}}{4}$

If \(\vec{a}\) and \(\vec{b}\) be two vectors, then \(|\vec{a} \times \vec{b}|\) represents the area of a parallelogram with sides \(\vec{a}\) and \(\vec{b}\)

Therefore, statement 1 is false.

If \(\vec{a} \times \vec{b}=\overrightarrow{0}\) and it is given that \(\vec{a} \neq \overrightarrow{0}, \vec{b} \neq \overrightarrow{0}, \Rightarrow \vec{a}\) and \(\vec{b}\) are parallelto each other i.e \(\vec{a}=\lambda \vec{b}\)

Therefore. statement 2 is true.

Put,t = tan-1x

$\Rightarrow dt=\frac{1}{1+x^2}dx$

then

${\int }_0^1\frac{{\tan }^{-1}x}{1+x^2}dx$

$={\int }_0^{\frac{\mathrm{\pi }}{4}}tdt$

$={\left[\frac{t^2}{2}\right]}_0^{\frac{\mathrm{\pi }}{4}}$

$=\frac{{\mathrm{\pi }}^2}{32}$

Probability is the chance of an occurring event.

On tossing a coin, certainty of occurrence of a head \((\mathrm{H})\) or tail \((\mathrm{T})\) on the top.

So, probability of occurrence of head (or tail) \(=\frac{1}{2}\)

Here, a coin tossed repeatedly, but the result of fourth toss is independent of first three tosses.

\(\therefore\) Probability of the tail appearing on the last (fourth) toss \(=\frac{1}{2}\)

Given,

\(\cot ^{-1}(\sqrt{3})\)

Let \(\cot ^{-1}(\sqrt{3})=\theta\)

\(\Rightarrow \cot \theta=\sqrt{3}=\cot \frac{\pi}{6}\)

\(\therefore \theta=\frac{\pi}{6}\)

Given that

Point \(\mathrm{R}\) lie on the line segment \(\mathrm{PQ}\).

And \(x\) - Coordinate of \(R\) is 4.

Let Point \(\mathrm{R}\) be \((4, \mathrm{~b}, \mathrm{c})\).

Let \(R\) divide line segment PR in the ratio \(\mathrm{k}: 1\).

We know that

Coordinate of Point that divides line in the ratio \(\mathrm{m}: \mathrm{n}\) is \(\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}, \frac{m z_{2}+n z_{1}}{m+n}\right)\)

Here,

\(m=k, n=1\)

\(x_{1}=2, y_{1}=-3, z_{1}=4\)

\(x_{2}=8 , y_{2}=0, z_{2}=10\)

Putting values

\(\mathrm{R}=\left(\frac{k(8)+1(2)}{k+1}, \frac{k(0)+1(-3)}{k+1}, \frac{k(10)+1(4)}{k+1}\right) \)

\((4, \mathrm{~b}, \mathrm{c})=\left(\frac{8 k+2}{k+1}, \frac{0-3}{k+1}, \frac{10 k+4}{k+1}\right) \)

\((4, \mathrm{~b}, \mathrm{c})=\left(\frac{8 k+2}{k+1}, \frac{-3}{k+1}, \frac{10 k+4}{k+1}\right)\)

\({x}\) - coordinate

\(4=\frac{8 k+2}{k+1}\)

\(4(k+1)=8 k+2\)

\(4k+4=8 k+2\)

\(4 k-8 k=2-4\)

\(-4 k=-2\)

\(k=\frac{-2}{-4}\)

\(k=\frac{1}{2}\)

\(y\) - Coordinate

\(b=\frac{-3}{k+1}\)

\(b(k+1)=-3\)

Putting \(k=\frac{1}{2}\)

\(b\left(\frac{1}{2}+1\right)=-3\)

\(\mathrm{b}\left(\frac{1+2}{2}\right)=-3\)

\(\mathrm{~b}\left(\frac{3}{2}\right)=-3\)

\(\mathrm{~b}=\frac{-3 \times 2}{3}\)

\(\mathrm{~b}=-2\)

\(z\) - Coordinate

\(\mathrm{C}=\frac{10 \mathrm{k}+4}{k+1}\)

\(c(k+1)=10 k+4\)

Putting \(k=\frac{1}{2}\)

\(c\left(\frac{1}{2}+1\right)=10\left(\frac{1}{2}\right)+4\)

\(c\left(\frac{1+2}{2}\right)=5+4\)

\(c\left(\frac{3}{2}\right)=9\)

\(\mathrm{c}=\frac{9 \times 2}{3}\)

\(c=6\)

Thus, \(a=4, b=-2, c=6\)

So, coordinates of point \(\mathrm{R}=(\mathrm{a}, \mathrm{b}, \mathrm{c})=(4,-\mathbf{2}, 6)\).

Given that:

\(\int\sqrt{1+\sin 2 x d x}\)

Let \(I=\int\sqrt{1+\sin 2 x d x}\)

Now, we know that, \(\sin ^{2} x+\cos ^{2} x=1\).

So, we get:

\(I=\int \sqrt{\sin ^{2} x+\cos ^{2} x+\sin 2 x} d x\)

Now, we know that \(2 \sin x \cos x=\sin 2 x\).

So, we can get:

\(I=\int \sqrt{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x} d x\)

Now, we know that, \((a+b)^{2}=a^{2}+b^{2}+2 a b\).

So, for \(a=\sin x\) and \(b=\cos x\), we get:

\(\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=(\sin x+\cos x)^{2}\)

Therefore, we can write \(I\) as,

\(I=\int \sqrt{(\sin x+\cos x)^{2}} d x\)

\(I=\int(\sin x+\cos x) d x\)

\(I=\int \sin x d x+\int \cos x d x\)

We also know that, \(\int \sin x d x=-\cos x+c_{1}\) and \(\int \cos x d x=\sin x+c_{2}\).

Therefore, we can write \(I\) as,

\(I=-\cos x+\sin x+c_{1}+c_{2}\)

\(I=\sin x-\cos x+c\)

Given:

\(\int \frac{d x}{2^{x}-1}\)

Let \(2^{x}=t\)

On solving, we get-

\(2^{x} \log _{e} 2 d x=d t\)

\(d x=\frac{d t}{t \log _{e} 2}\)

\(\int \frac{d t}{t \log _{e} 2(t-1)}\)

\(\frac{1}{\log _{e} 2} \int \frac{d t}{t(t-1)}=\frac{1}{\log _{e} 2}\left[\int \frac{-1 d t}{t}+\int \frac{1 d t}{t-1}\right]\)

\(=\frac{1}{\log _{e} 2}[-\log t+\log (t-1)]+C\)

\(=\frac{1}{\log _{e} 2}\left[\log t^{-1}+\log (t-1)\right]+C\)

\(=\frac{1}{\log _{e} 2}\left[\log t^{-1}(t-1)\right]+C\)

\(=\frac{1}{\log _{e} 2}\left[\log \left(1-t^{-1}\right)\right]+C\)

\(=\frac{1}{\log _{e} 2}\left[\log \left(1-2^{-x}\right)\right]+C\)

\(=\frac{\log \left(1-2^{-x}\right)}{\ln 2}+C\)