Mathematics Test-3

Total prime numbers from \(1\) to \(100\) are:

\(2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97\)

That means \(25\) out of \(100\)

Probability \(= \frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ outcomes}\)

So probability is:

\(P(\text {prime})=\frac{25}{100}\)

\(\Rightarrow {P}(\text {prime})=\frac{1}{4}\)

\(\sin 5 \theta=\cos 4 \theta\)

\(\Rightarrow \cos \left(90^{\circ}-5 \theta\right)=\cos 4 \theta\)

Comparing we get,

\(90^{\circ}-5 \theta=4 \theta\)

\(\Rightarrow 90^{\circ}=4 \theta+5 \theta=9 \theta\)

\(\therefore \quad \theta=\frac{90^{\circ}}{9}=10^{\circ}\)

\(\text { Now } 2 \sin 3 \theta-\sqrt{3} \tan 3 \theta\)

\(=2 \sin \left(3 \times 10^{\circ}\right)-\sqrt{3} \tan \left(3 \times 10^{\circ}\right)\)

\(=2 \sin 30^{\circ}-\sqrt{3} \tan 30^{\circ}\)

\(=2 \times \frac{1}{2}-\sqrt{3} \times \frac{1}{\sqrt{3}} \quad\left\{\begin{array}{l}\because \sin 30^{\circ}=\frac{1}{2} \\ \tan 30^{\circ}=\frac{1}{\sqrt{3}}\end{array}\right\}\)

\(=1-1=0\)

Given:

\(I=\int_{1}^{2} \frac{\log x}{x^{2}}\)

\(I=\int_{1}^{2}\left(\log x \times \frac{1}{x^{2}}\right) d x\)

Integration by parts:

\(\int f(x) g(x) d x=f(x) \int g(x) d x-\int\left[f^{\prime}(x) \int g(x) d x\right] d x\)

Where \(f\) is first function and \(g\) is second function.

\(I=\log x \int_{1}^{2}\frac{1}{x^{2}} d x-\int_{1}^{2}\left\{\frac{d}{d x}(\log x) \cdot \int_{1}^{2}\frac{1}{x^{2}} d x\right\} d x\)

\(=\log x\left(-\frac{1}{x}\right)-\int\left(\frac{1}{x}\right) \cdot\left(-\frac{1}{x}\right) d x\)

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

\(=-\frac{\log x}{x}-\frac{1}{x}\)

\(=-\left(\frac{\log x+1}{x}\right)\)

Putting limits:

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

\(I=\left[-\left(\frac{\log 2+1}{2}\right)\right]-\left[-\left(\frac{\log 1+1}{1}\right)\right]\)

\(I=\left(\frac{1-\log 2}{2}\right)\)

\(=\left(\frac{\log e-\log 2}{2}\right)\)

\(I=\frac{1}{2} \log \frac{e}{2}\)

Concept:

The equation of a line passing through the point \(\left(x_{1}, y_{1}\right)\) and having the slope ' \(m\) ' is given as:

\(y-y_{1}=m \cdot\left(x-x_{1}\right)\)

Given: The point of intersection of diagonals of a square ABCD is at the origin and one of its vertices is at A(4, 2).

So, the diagonal AC passes through the origin,

As we know that, the slope of the line joining the points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is: \(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

The slope of line \(A C\) is given by

\(\frac{2-0}{4-0}=\frac{1}{2}\)

In square \(ABCD\), the diagonals \(AC\) and \(BD\) are perpendicular to each other.

\(\Rightarrow \text { Slope of } A C \times \text { slope of } B D=-1\)

So, the slope of BD is - 2 .

As we know that, the equation of a line passing through the point \(\left(x_{1}, y_{1}\right)\) and having the slope ' \(m\) ' is given as:

\(y-y_{1}=m \cdot\left(x-x_{1}\right)\)

The equation of \(B D\) whose slope is - 2 and passes through origin is given by:

\(y-0=(-2) \cdot(x-0)\)

\(\Rightarrow 2 x+y=0\)

Let \({r}\) be the common ratio of the given GP

Now,

\(\left|\begin{array}{lll}\ln a_{1} & \ln a_{2} & \ln a_{3} \\ \ln a_{4} & \ln a_{5} & \ln a_{6} \\ \ln a_{7} & \ln a_{8} & \ln a_{9}\end{array}\right|\)

Apply, \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}\) and \(\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}\)

\(\begin{aligned}&=\left|\begin{array}{lll}\ln a_{1} & \ln a_{2}-\ln a_{1} & \ln a_{3}-\ln a_{2} \\ \ln a_{4} & \ln a_{5}-\ln a_{4} & \ln a_{6}-\ln a_{5} \\ \ln a_{7} & \ln a_{8}-\ln a_{7} & \ln a_{9}-\ln a_{8}\end{array}\right| \\&=\left|\begin{array}{lll}\ln a_{1} & \ln r & \ln r \\ \ln a_{4} & \ln r & \ln r \\ \ln a_{7} & \ln r & \ln r\end{array}\right| \quad\left[\text { Since, } \frac{a_{p+1}}{a_{p}}=r \text { and } \log a_{p+1}-\log a_{p}=\log \frac{a_{p+1}}{a_{p}}=\log \right]\end{aligned}\)

\(=0\qquad \qquad{\left[\text { Since, } \text{C}_{3}=\text{C}_{2}\right]}\)

Given:\(\frac{x^{3 / 2}-1}{x+1+x^{12}}-\frac{x^{32}+1}{x+1-x^{12}}\)

\(=\frac{\left(x^{12}\right)^{3}-1}{\left(x^{12}\right)^{2}+x^{12}+1}-\frac{\left(x^{1 / 2}\right)^{3}+1}{\left(x^{12}\right)^{2}-x^{12}+1} \)

\(=\frac{\left(x^{12}-1\right)\left\{\left(x^{12}\right)^{2}+x^{1 / 2}+1\right\}}{\left(x^{12}\right)^{2}+x^{12}+1}-\frac{\left(x^{1 / 2}+1\right)\left\{\left(x^{12}\right)^{2}-x^{12}+1\right\}}{\left(x^{12}\right)^{2}-x^{12}+1} \)

\(=\left(x^{12}-1\right)-\left(x^{12}+1\right) \)\(=-2\)

So, \(\left(\frac{x^{3 / 2}-1}{x+1+x^{1 / 2}}-\frac{x^{3 / 2}+1}{x+1-x^{1 / 2}}\right)^{5}=(-2)^{5}\)\(=-32\)

So, the expansion has only one term which is \(-32\) and independent of \(x\).

Given,

\( \vec{r}=3 i+2 j-4 k+\lambda(i+2 j+2 k) \)

\( \vec{r}=(5 j-2 k)+\mu(3 i+2 j+6 k) \)

We know that, angle between \(r=a_1+\lambda b_1\) and \(r=a_2+\lambda b_2\) is given by,

\(\cos \theta=\frac{\overrightarrow{b_1} \cdot \overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_1}\right|}\)

\(\overrightarrow{b_1}=1+2 j+2 k\) and \(\overrightarrow{b_2}=3 i+2 j+6 k\) respectively.

\(=\frac{(i+2 j+2 k)(3 i+2 j+6 k)}{|i+2 j+2 k||3 i+2 j+6 k|}\)

\(=\frac{3+4+12}{\sqrt{1+4+4} \sqrt{9+4+36}}\)

\(=\frac{19}{21}\)

\(\Rightarrow \theta=\cos ^{-1}\left(\frac{19}{21}\right)\)

Given:

Quadratic equation \(\sqrt{3} \mathrm{x}^{2}-\sqrt{2} \mathrm{kx}+2 \sqrt{3}=0\), has distinct real roots.

As we know that, if a quadratic equation \(a x^{2}+b x+c=0\) has distinct real roots then \(\Delta\) \(>0\) where \(\Delta=b^{2}-4 a c\).

Here, \(a=\sqrt{3}, b=-\sqrt{2} k\) and \(c=2 \sqrt{3}\)

\(\Rightarrow(-\sqrt{2} \mathrm{k})^{2}-4 \times \sqrt{3} \times 2 \sqrt{3}>0\)

\(\Rightarrow 2 \mathrm{k}^{2}-24>0 \)

\(\Rightarrow \mathrm{k}^{2}>12\)

\(\because \mathrm{k}\) is smallest positive integer. so \(\mathrm{k}=4\)

Given:

\(f(x)=\frac{\sin x}{x}\)

Function is continuous at \(x=0\)

Therefore, \(\mathrm{f}(0)=\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})\)

(Form \(=\lim _{\mathrm{x} \rightarrow 0} \frac{\sin \mathrm{x}}{\mathrm{x}}\)

Apply L-Hospital Rule,

\(=\lim _{x \rightarrow 0} \frac{\cos x}{1}\)

\(=\frac{\cos 0}{1}=1\)