Mathematics Test - 4

Given: Equation of the curve \(y=x^2 \ldots\)(1)

Let's find Slope of tangent at any point on curve \((x, y)\)

\(y=x^2\).......(1)

Differentiating with respect to \(x\), we get

\(\Rightarrow \frac{ dy }{ dx }=2 x\)

According to question, Slope of the tangent \(=y\)-coordinate of the point

\( \Rightarrow 2 x = y \)

\(\Rightarrow 2 x = x ^2 \)

\(\Rightarrow x ^2-2 x =0\)

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

\( \therefore x =0 \text { or } 2\)

Put the value of \(x\) inn equation \(1^{\text {st }}\), we get

\(y=0\)  or \(4\)

Therefore, the required points are \((0,0)\) and \((2,4)\)

\(\begin{aligned} & =\int \frac{\left(1-\frac{1}{\sqrt{3}}\right)(\cos x-\sin x)}{\left(1+\frac{2}{\sqrt{3}} \sin 2 x\right)} d x \\ & =\int \frac{\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right) \sqrt{2} \sin \left(\frac{\pi}{4}-x\right)}{\left(\frac{2}{\sqrt{3}}\right)\left(\sin \frac{\pi}{3}+\sin 2 x\right)} d x \\ & =\int \frac{\frac{(\sqrt{3}-1)}{\sqrt{2}} \sin \left(\frac{\pi}{4}-x\right)}{\left(\sin \frac{\pi}{3}+\sin 2 x\right)} d x \\ & =\int \frac{\frac{\sqrt{3}-1}{2 \sqrt{2}} \sin \left(\frac{\pi}{4}-x\right)}{\sin \left(\frac{\pi}{6}+x\right) \cos \left(\frac{\pi}{6}-x\right)} d x \\ & =\frac{1}{2} \int \frac{2 \sin \frac{\pi}{12} \sin \left(\frac{\pi}{4}-x\right)}{\sin \left(\frac{\pi}{6}+x\right) \cos \left(\frac{\pi}{6}-x\right)} d x \\ & =\frac{1}{2} \int \frac{\cos \left(\frac{\pi}{6}-x\right)-\cos \left(\frac{\pi}{3}-x\right)}{\sin \left(\frac{\pi}{6}+x\right) \cos \left(\frac{\pi}{6}-x\right)} d x \\ & =\frac{1}{2}\left[\int \operatorname{cosec}\left(\frac{\pi}{6}+x\right) d x-\int \sec \left(\frac{\pi}{6}-x\right) d x\right] \\ & =\frac{1}{2}\left[\ln \left|\tan \left(\frac{\pi}{12}+\frac{x}{2}\right)\right|-\int \operatorname{cosec}\left(\frac{\pi}{3}-x\right) d x\right] \\ & =\frac{1}{2}\left[\ln \left|\tan \left(\frac{\pi}{12}+\frac{x}{2}\right)\right|-\ln \left|\frac{\pi}{6}+\frac{x}{2}\right|\right]+C \\ & =\frac{1}{2} \ln \left|\frac{\tan \left(\frac{\pi}{12}+\frac{x}{2}\right)}{\tan \left(\frac{\pi}{6}+\frac{x}{2}\right)}\right|+C \\ & \end{aligned}\)

Given,

\(A =\left[\begin{array}{ccc}1 & -1 & 0 \\ 3 & 2 & -1\end{array}\right]\) and \(B =\left[\begin{array}{l}1 \\ 3 \\ 5\end{array}\right]\)

\(\Rightarrow AB =\left[\begin{array}{ccc}1 & -1 & 0 \\ 3 & 2 & -1\end{array}\right] \times\left[\begin{array}{l}1 \\ 3 \\ 5\end{array}\right]\)

\(\Rightarrow AB =\left[\begin{array}{c}1-3+0 \\ 3+6-5\end{array}\right]\)

\(\Rightarrow AB =\left[\begin{array}{c}-2 \\ 4\end{array}\right]\)

As we know,

The new matrix obtained by interchanging the rows and columns of the original matrix is called as the transpose of the matrix. It is denoted by \(A'\) or \(A^T\).

\(\therefore( AB )^{T}=\left[\begin{array}{ll}-2 & 4\end{array}\right]\)

Given, \(\int_{0}^{\frac{\pi}{4}} \sin ^{3} 2 t \cos 2 t d t\)

Let,

\(\mathrm{F}(x)=\int \sin ^{3} 2 t \cos 2 t d t\)

Let \(\sin 2 t=u\)

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

\(\frac{d(\sin 2 t)}{d t}=\frac{d u}{d t}\)

\(2 \cos 2 t=\frac{d u}{d t}\)

\(d t=\frac{d u}{2 \cos 2 t}\)

Putting value of \(u\) and \(du\) in our integral

\(\int \sin ^{3} 2 t \cos 2 t d t =\int u^{3} \cos 2 t \times \frac{d u}{2 \cos 2 t} \)

\(=\frac{1}{2} \int u^{3} d u \)

\(=\frac{1}{2} \frac{u^{3+1}}{3+1}=\frac{1}{2} \frac{u^{4}}{4}=\frac{u^{4}}{8}\)

Putting back \(u=\sin 2 t\)

\(=\frac{1}{8} \sin ^{4} 2 t\)

Hence, \(F(t)=\frac{1}{8} \sin ^{4} 2 t\)

Now,

\(\int_{0}^{\frac{\pi}{4}} \sin ^{3} 2 t \cos 2 t =F\left(\frac{\pi}{4}\right)-F(0) \)

\(=\frac{1}{8} \sin ^{4} 2\left(\frac{\pi}{4}\right)-\frac{1}{8} \sin ^{4} 2(0) \)

\(=\frac{1}{8} \sin ^{4} \frac{\pi}{2}-\frac{1}{8} \sin ^{4}(0) \)

\(=\frac{1}{8} \times 1^{4}-\frac{1}{8} \times 0^{4} \)

\(=\frac{1}{8} \times 1-0 \)

\(=\frac{1}{8}\)

Two digit numbers which are divisible by 4 are 12, 16, 20, ..., 96 forms an AP with first term a = 12, common difference d = 4 and n^th term a_n = 96.

⇒ a_n = a + (n - 1) × d 

⇒ 12 + (n - 1) × 4 = 96

⇒ n = 22

Given:

\(\alpha\) is any vector

Let \(\alpha=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\)

As we know that, if \(\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\) and \(\vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\) then

\(\vec{a} . \vec{b}=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\)

\((\alpha. \hat{i}) \hat{i}+(\alpha.\hat{j}) \hat{j}+(\alpha . \hat{k}) \hat{k}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\)

\((\alpha.\hat{i}) \hat{i}+(\alpha. \hat{j}) \hat{j}+(\alpha. \hat{k}) \hat{k}=\alpha\)

Given,

\(x=\tan ^{-1}(\frac{1}{5})\)

\(\tan x=\frac{1}{5}\)

As we know that, \(\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}\)

\(\therefore \sin 2 x=\frac{2 \tan x}{1+\tan ^{2} x}\)

\(=\frac{2 \times \frac{1}{5}}{1+\left(\frac{1}{5}\right)^{2}} \)

\(=\frac{\left(\frac{2}{3}\right)}{\frac{s+1}{25}} \)

\(=\frac{2}{5} \times \frac{25}{26}=\frac{10}{26}\)

\(=\frac{5}{13}\)

Given: \(x=\) M.D., \(y=\) S.D

We know that,

M.D \(=\frac{4}{5}\) S.D

Where, M.D is mean deviation and S.D is standard deviation

\( x =\frac{4}{5} y\)

\(\therefore x < y\)

$P(x=r)={}^{n}c_r p^r{q}^{n-r}$

$P(X=x)=k(x+1){\left(\frac{1}{5}\right)}^x$

$x=0,1,2,3,...$

$\sum _{x=0}^{\infty }p(X=x)=1$

$k\sum _{x=0}^{\infty }(x+1){\left(\frac{1}{5}\right)}^x=1$

$k\left[1+2\times \left(\frac{1}{5}\right)+3\times {\left(\frac{1}{5}\right)}^2+....\right]=1$

$a+(a+d)r+(a+d)r^2+\dots =\frac{a}{1-r}+\frac{dr}{(1-r{)}^2}$

$\Rightarrow k\left[\frac{1}{1-\frac{1}{5}}+\frac{1\times \frac{1}{5}}{{\left(1-\frac{1}{5}\right)}^2}\right]=1$

$\Rightarrow k\left(\frac{1-\frac{1}{5}+\frac{1}{5}}{1+\frac{1}{25}-\frac{2}{5}}\right)=1$

$k\left[\frac{1}{\frac{16}{25}}\right]=1$

$\Rightarrow \frac{25}{16}k=1$

$k=\frac{16}{25}$

Given that: 

\({ }^{n} C_{15}={ }^{n} C_{8}\)

As we know that, 

If \({ }^{n} C_{x}={ }^{n} C_{y}\), then, 

\(x+y=n\)

Therefore, 

\(n=15+8=23\)