Mathematics Test - 1

We know that:

Area bounded by function f(x) and g(x) is given as,

Area \(=\int_{\mathrm{a}}^{\mathrm{b}}[\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})] \mathrm{dx}=\int_{\mathrm{a}}^{\mathrm{b}}[\) Top \(-\) bottom \(] \mathrm{dx}\)

Given: 

\(y=\log x\)

Then,

Area \(=\int_{1}^{2} \log \mathrm{x} \mathrm{~dx}\)

Applying by parts rule, we get:

\(=[\log x \mathrm{x}]_{1}^{2}-\int_{1}^{2} \frac{1}{x} \times \mathrm{x} \mathrm{dx}\)

\(=[\mathrm{x} \log \mathrm{x}]_{1}^{2}-[\mathrm{x}]_{1}^{2}\)

\(=[2 \log 2-\log 1]-[2-1]\)

\(=2 \log 2-1\)

\(=\log 2^{2}-\log \mathrm{e}\)

\(=\log 4-\log \mathrm{e}\)

\(=\log \left(\frac{4}{e}\right)\) sq. unit

Given, \(|x+1|+\sqrt{(x-1)}=0\), where each term is non-negative.

So, \(|x+1|=0\) and \(\sqrt{(x-1)}=0\) should be zero simultaneously.

i.e. \(x=-1\) and \(x=1\), which is not possible.

So, there is no value of \(x\) for which each term is zero simultaneously.

Given:

\(\vec{\alpha}=\mathrm{k}\) and \(\vec{\gamma}=2 \mathrm{i}+3 \mathrm{j}+4 \mathrm{k}\)

\(\vec{\beta}\) is perpendicular to both \(\vec{\alpha}\) and \(\vec{\gamma}\)

\(\therefore \vec{\alpha} \times \vec{\gamma}=\vec{\beta}\)

\(\vec{\alpha} \times \vec{\gamma}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 1 \\ 2 & 3 & 4\end{array}\right|\)

\(=\hat{i}(0-3)-\hat{j}(0-2)+\hat{k}(0)\)

\(=-3 \hat{i}+2 \hat{j}\)

Given:

\(A =\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]\)

\(A + A ^{T}=1\)

The transpose of matrix \(A\) is given by,

\(A^{T}=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]\)

As, \(A + A ^{T}=I\)

\(\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]+\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]=I\)

\(\Rightarrow\left[\begin{array}{cc}2 \cos 2 \theta & 0 \\ 0 & 2 \cos 2 \theta\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\)

As we know that,

If two matrices \(A\) and \(B\) are equal then their corresponding elements are also equal.

\(\therefore 2 \cos 2 \theta=1\)

\(\Rightarrow \cos 2 \theta=\frac{1}{2}\)

\(\Rightarrow \cos 2 \theta=\cos 60^\circ\)

\(\Rightarrow 2 \theta=\frac{\pi}{3}\quad[\because \cos 60^\circ=\frac{\pi}{3}]\)

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

So, the value of \(\theta\) is \(\frac{\pi}{6}\).

Maximum of \(Z\) occurs at \((25,20)\) and at \((0,30)\).

So, equating the vales of \(Z\) at these points, we get

\(25 p+20 q=30 q\)

\( 25 p=10 q\)

\(\therefore 5 p=2 q\)

This is the required relation.

Also as \(p, q>0\), the value of \(Z\) is always positive and hence, is greater at \((25,20)\) and at \((0,30)\) than at \((0,10)\) and \((5,5)\).

Let \(f(x)\) be any function.

\(f(x)\) is onto if range of \(f(x)=\) Co-domain

The function \(f\) is said to be many-one functions if there exist two or more than two different elements in \(X\) having the same image in \(Y\).

Given function is:

\(f: R \rightarrow\{0,1\}\), such that:

\(f(x)=\left\{\begin{array}{c}1 \text { if } x \text { is rational } \\0 \text { if } x \text { is irrational }\end{array}\right.\) 

Co-domain \(=\{0,1\}\)

Since, on taking a straight line parallel to the \(x\)-axis, the group of given function intersect it at many points.

\(\Rightarrow f(x)\) is many-one.

Range of function is \(\{0,1\}\)

As range of \(f(x)=\) Co-domain 

\(\Rightarrow f(x)\) is onto.

Therefore, \(f(x)\) is many-one onto.

Given scores 10, 12, 13, 15, 15, 13, 12, 10, x and mode = 15

The mode of the n observation is the number that has the highest frequency.

Frequency of score '12' = 2

Frequency of score '15' = 2

But for mode to be 15, x should be '15'.

Given equation is

\(y ^2= a \left( b ^2- x ^2\right)\)

Differentiating w.r.t \(x\)

\(\Rightarrow 2 y^{\prime}=a(-2 x) \)

\(\Rightarrow y^{\prime}=-a x\)

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

\(\Rightarrow yy ^{\prime \prime}+\left( y ^{\prime}\right)^2=- a\)

From (i) and (ii) we get

\(\Rightarrow y y^{\prime}=x\left(y y^{\prime \prime}+\left(y^{\prime}\right)^2\right) \)

\(\Rightarrow y y^{\prime}-x y y^{\prime \prime}-x\left(y^{\prime}\right)^2=0\)

\(\mathrm{y}=\mathrm{x}\) and \(\mathrm{y}=2 \mathrm{x}+6\)

Finding a point of intersection:

\(\Rightarrow \mathrm{x}=2 \mathrm{x}+6\)

\(\Rightarrow \mathrm{x}=-6\) Thus, y = - 6.

Let us draw the graph of the curve \(\mathrm{y}=\mathrm{x}\) and \(\mathrm{y}=2 \mathrm{x}+6\).

Let the enclosed area be A.

Using the formula of the area under the curve, 

\(\mathrm{A}=\left|\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x}) \mathrm{dx}\right|\)

\(\Rightarrow \mathrm{A}=\left|\int_{-6}^{0}(2 \mathrm{x}+6-\mathrm{x}) \mathrm{dx}\right|\)

\(=\left|\int_{-6}^{0}(\mathrm{x}+6) \mathrm{dx}\right|\)

\(=\left|\left[\frac{\mathrm{x}^{2}}{2}+6 \mathrm{x}\right]_{-6}^{0}\right|\)

Substitute the limit to evaluate the area:

\(\Rightarrow \mathrm{A}=\left|0+0-\frac{36}{2}+36\right|\) \(=18\)

Given:

\(\underset{{{x \rightarrow 0} }}{\lim}\frac{\log (1+2 x)}{\tan 2 x}\)

Dividing and multiplying the numerator and denominator by \(2x\), we get:

\(=\underset{{{x \rightarrow 0} }}{\lim} \frac{\frac{\log (1+2 x)}{2 x} \times 2 x}{\frac{t a n 2 x}{2 x} \times 2 x}\)

We know that:

\(\underset{{{x \rightarrow a}}}{\lim}\left[\frac{f(x)}{g(x)}\right]=\frac{\underset{{{x \rightarrow a}}}{\lim} f(x)}{\underset{{{x \rightarrow a}}}{\lim} g(x)}\), provided \(\underset{{{x \rightarrow a}}}{\lim} g(x) \neq 0\) 

\(=\frac{\underset{{{x \rightarrow 0} }}{\lim} \frac{\log (1+2 x)}{2 x}}{\underset{{{x \rightarrow 0} }}{\lim} \frac{\tan 2 x}{2 x}}\)

As we know that:

\(\underset{{{{x} \rightarrow 0}}}{\lim}\frac{\tan {x}}{{x}}=1\) 

and, \(\underset{{{{x} \rightarrow 0}}}{\lim} \frac{\log (1+{x})}{{x}}=1\)

Therefore, 

\(\underset{{{{x} \rightarrow 0}}}{\lim} \frac{\tan 2 {x}}{2 {x}}=1\) 

and, \(\underset{{{{x} \rightarrow 0}}}{\lim}\frac{\log (1+2 {x})}{2 {x}}=1\)

Therefore, 

\(\underset{{{{x} \rightarrow 0}}}{\lim} \frac{\log (1+2 {x})}{\tan 2 {x}}=\frac{1}{1}=1\)