Mathematics Test-1

If the angle of the base radius is the same then the corresponding length of sides will be the same.

AB = 4 cm

Therefore, BC = 4 cm

Choose \(P\) as the origin of the reference.

\(\overrightarrow{ PR }=\vec{a}+\vec{b}\)

\(\overrightarrow{ PS }=-(\vec{a}-\vec{b})=\vec{b}-\vec{a}\)

\(\overrightarrow{ PM }=\overrightarrow{ a }+\frac{\overrightarrow{ b }}{2}\)

Given:\(\overrightarrow{ SX } =\frac{4}{5}\overrightarrow{ SM }\)

\(\overrightarrow{ PX }-(\overrightarrow{ b }-\overrightarrow{ a })=\frac{4}{5}\left\{\overrightarrow{ a }+\frac{\overrightarrow{ b }}{2}-(\overrightarrow{ b }-\overrightarrow{ a })\right\}\)

\(\overrightarrow{ PX }=\frac{3}{5}\overrightarrow{ a }-(-\frac{6}{10}\overrightarrow{ b })\)

\(\Rightarrow \overrightarrow{ PX } =\frac{3}{5}(\overrightarrow{ a }+\overrightarrow{ b })\)

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

Any square matrix can be expressed as sum of symmetric and skew symmetric as.

\(A=\frac{1}{2}\left(A+A^T\right)+\frac{1}{2}\left(A-A^T\right)\)

Where the first term is symmetric and second term is skew-symmetric here \(S = P + Q\)

To find \(P\) i.e. the symmetric term is given by,

\(P=\frac{1}{2}\left(S+S^{ T }\right)\)

\(P=\frac{1}{2}\left(\left[\begin{array}{ll}1 & 6 \\ 7 & 2\end{array}\right]+\left[\begin{array}{ll}1 & 7 \\ 6 & 2\end{array}\right]\right)\)

\(P=\frac{1}{2}\left[\begin{array}{cc}2 & 13 \\ 13 & 4\end{array}\right]\)

\(P=\left[\begin{array}{cc}1 & \frac{13}{2} \\ \frac{13}{2} & 2\end{array}\right]\)

Given, \(\left \{\frac{(1+i)}{(1-i)}\right\}^{n}\)

\(=\left \{\frac{(1+i) \times(1+i)}{(1-i) \times(1+i)}\right\}^{n}\)

\(=\left[\frac{\left\{(1+i)^{2}\right\}}{\left\{\left(1-i^{2}\right)\right\}}\right]^{n}\)

\(=\left[\frac{\left(1+i^{2}+2 i\right) }{\{1-(-1)\}}\right]^{n}\)

\(=\left[\frac{\left\{1-1+2 i\right\}}{\left\{1+1\right\}}\right]^{n}\)

\(=\left[\frac{2 i}{ 2}\right]^{ n }\)

\(= i ^{ n }\)

Now, in is real when \(n =2\) [since \(i^ 2=-1\)]

So, the least value of \(n\) is \(2\)

The given series:

5 + 9 + 13 + … + 49

First term, a = 5

Common difference, d = 4

The given series is an arithmetic progression.

Let's say that the last term 49 is the nth term.

\(\therefore a+(n-1) d=49\)

\(\Rightarrow 5+4(n-1)=49\)

\(\Rightarrow 4 n=48\)

\(\Rightarrow n=12\)

And, the sum of this AP is:

\(\mathrm{S}_{12}=\left(\frac{\text { First Term }+\text { Last Term }}{2}\right) \times 12\)

\(=\left(\frac{5+49}{2}\right) \times 12=54 \times 6=324\)

Given:

\(I=\int \frac{d x}{\sec ^{2}\left(\tan ^{-1} x\right)}\)

As we know, \(\sec ^{2} x-\tan ^{2} x=1\)

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

\(\Rightarrow I=\int \frac{d x}{1+\left[\tan \left(\tan ^{-1} x\right)\right]^{2}}\)

\(\Rightarrow I=\int \frac{d x}{1+x^{2}}\)

We know that, \(\int \frac{d x}{1+x^{2}}=\tan ^{-1} x+C\) where \(C\) is constant.

\(I=\int \frac{d x}{\sec ^{2}\left(\tan ^{-1} x\right)}\)

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

Given,

\(\angle POQ=45^{\circ}\), AM=PQ

In \(\triangle AOM\) and \(\triangle POQ\) OA, OM and OP, OQ are radius of circle.

\(\therefore~ \)\(\triangle AOM\) and \(\triangle POQ\) are Congruent triangles

Since, two Congruent triangles have all three sides and angles equal in measurement, we conclude that \(\angle AOM=45^{\circ}\)

Given-

$5x^2–10x+p=0$

On comparing it with$a{x}^2+bx+c=0$, we get-

$\Rightarrow a=5$

$\Rightarrow b=-10$

$\Rightarrow c=p$

Sum of roots (s)$=\frac{-b}{a}$

$=\frac{10}{5}$

$=2$

Product of roots (p)$=\frac{c}{a}$

$=\frac{p}{5}$

According to the question,

The product of roots of this equation is 4 times of sum of the roots.

$\Rightarrow \frac{p}{5}=4\times 2$

$\Rightarrow p=40$

\(\sec \theta-\tan \theta=k\)

\(\frac{1}{\cos \theta}-\frac{\sin \theta }{ \cos \theta}=k\)

\(\frac{(1-\sin \theta)}{ \cos \theta}=\mathrm{k}\)

Squaring both sides, we get

\((\frac{(1-\sin \theta)}{ \cos \theta})^2=(\mathrm{k})^2\)

\(\Rightarrow \frac{(1-\sin \theta)^2 }{ \cos ^2 \theta}=\mathrm{k}^2\)

\(=\frac{(1-\sin \theta)^2 }{\left(1-\sin ^2 \theta\right)}=\mathrm{k}^2\)

\(\Rightarrow \frac{(1-\sin \theta)^2 }{(1+\sin \theta)(1-\sin \theta)}=k^2\)

\(=\frac{(1-\sin \theta)}{(1+\sin \theta)}=\mathrm{k}^2\)

\(\Rightarrow\frac{(1+\sin \theta) }{(1-\sin \theta)}=\frac{1 }{k^2}\)

\(\Rightarrow \sqrt{\frac{(1+\sin \theta)}{1-\sin \theta}}=\frac{1 }{ \mathrm{k}}\)

Multiplying and dividing by \((1+\sin \theta)\)

\(=\sqrt{\frac{(1+\sin \theta)^2}{1-\sin ^2 \theta}}=\frac{1 }{ \mathrm{k} }\)

\(\Rightarrow \sqrt{\frac{(1+\sin \theta)^2}{\cos ^2 \theta}}=\frac{1 }{ \mathrm{k}}\)

\(\frac{(1+\sin \theta) }{\cos \theta}=\frac{1 }{\mathrm{k} }\)

\(=\frac{1}{\cos \theta}+\frac{\sin \theta }{\cos \theta}=\frac{1 }{ \mathrm{k}} \)

\(\Rightarrow \sec \theta+\tan \theta=\frac{1 }{ \mathrm{k}}\)

Given: There are 6 teachers and 8 students out of which we need to form a committee of 11 members such that the committee has exactly 4 teachers in it.

If out 11 members of committee 4 are teachers then 7 members are students.

So, the committee compromises of 4 teachers and 7 students.

No. of ways in which 4 teachers can be selected from 6 teachers \(={ }^{6} C_{4}\)

No. of ways in which 7 students can be selected from 8 students \(={ }^{8} C_{7}\)

So, the number of ways in which the committee can be formed \(={ }^{6} C_{4} \times{ }^{8} C_{7}\)

As we know that, \({ }^{n} C_{r}=\frac{n !}{r ! \times(n-r) !}\)

\(\Rightarrow{ }^{6} C_{4} \times{ }^{8} C_{7}\)

\(=\left(\frac{6 !}{(6-4) ! 4 !} \times \frac{8 !}{(8-7) ! 7 !}\right)\)

\(=\left(\frac{6 !}{(2) ! 4 !} \times \frac{8 !}{(1) ! 7 !}\right)\)

\(=120\)

So, the committee can be formed in 120 ways.