Mathematics Test - 29

Given:

Number of house \((\mathrm{m})=3\)

Number of people applying for house(n) \(=3\)

Total number of way three people can apply for house \(=3^{3}=27\)

Number of way in which all people apply for same house \(=3\)

\(\mathrm{P}(\) all three applying for same house \()=\frac{3}{27}=\frac{1}{9}\)

Let's first draw the diagram of the given information:

Given: 

AB = 1 km, ∠ MAN = 30°, ∠ MBN = 75°

In Δ MAN, tan 30° = $\frac{MN}{AM}$

⇒ MN = AM tan 30°             ...(1)

And in Δ MBN, tan 75° = $\frac{MN}{BM}$

⇒ MN = BM tan 75°            ...(2)

Using (1) and (2), AM tan 30° = BM tan 75°

⇒ $\left(AB+BM\right)\frac{1}{\sqrt{3}}=BM\left(2+\sqrt{3}\right)$

⇒ $\left(1+BM\right)=2\sqrt{3}BM+3BM$

⇒ $BM=\frac{1}{2+2\sqrt{3}}$

And, MN = BM tan 75°

⇒ $MN=\frac{1}{2+2\sqrt{3}}\left(2+\sqrt{3}\right)=\frac{2+\sqrt{3}}{2+2\sqrt{3}}\times \frac{2-2\sqrt{3}}{2-2\sqrt{3}}$

⇒ $MN=\frac{4-4\sqrt{3}+2\sqrt{3}-6}{4-12}=\frac{2+2\sqrt{3}}{8}$

⇒ $MN=\frac{1}{4}\left(\sqrt{3}+1\right)$ km = 250 $\left(\sqrt{3}+1\right)$ m.

Event A is choosing a yellow pencil first, and Event \(\mathrm{B}\) is choosing a yellow pencil second.

Initially, there are 12 pencils, 7 of which are yellow.

Probability the first pencil is yellow \(=\mathrm{P}(\mathrm{A})=\frac{7}{12}\)

If a yellow pencil is chosen, there will be 11 pencils left, 6 of which are yellow.

Probability the second pencil is yellow \(=\mathrm{P}(\mathrm{B})=\frac{6}{11}\)

Two pencils are chosen at random from the box without replacement.

So events are independent of each other.

Probability they are both yellow:

\(=P(A \cap B)=P(A) \times P(B)\)

\(=\frac{7}{12} \times \frac{6}{11}=\frac{7}{22}\)

Here, \(f(x)=|\cos x-\sin x|\)

\(\left\{\begin{array}{l}\cos x-\sin x, \text { if } 0 \leq x \leq \frac{\pi}{4} \\ \sin x-\cos x, \text { if } \frac{\pi}{4}

\(\Rightarrow f^{\prime}(x)=\left\{\begin{array}{c}-\sin x-\cos x, \text { if } 0 \leq x \leq \frac{\pi}{4} \\ \cos x+\sin x, \text { if } \frac{\pi}{4}

Here, \(x=\frac{\pi}{4}>0\)

\(\Rightarrow f^{\prime}\left(\frac{\pi}{4}\right)=-\sin \left(\frac{\pi}{4}\right)-\cos \left(\frac{\pi}{4}\right)=-\sqrt{2}\)

The given equation can be simplified as,

\(\frac{d x}{d y}+\frac{x}{2 y}=\frac{5}{2} y\)

On comparing eqn (i) with standard eqn,\(\frac{d x}{d y}+P x=Q\),we get,

\(P=\frac{1}{2 y}\) and \(Q=\frac{5}{2} y\)

\(\therefore I F=e^{\int P d y}=e^{\int \frac{1}{2 y} d y}\)

\(\Rightarrow I F=e^{\frac{1}{2} \log y}=e^{\log y^{\frac{1}{2}}}\)

\(\Rightarrow I F=\sqrt{y} \cdot\left(\because e^{a \log x}=x^{a}\right)\)

\(\frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^{2}-2 \mathrm{ax}+48>0\)

\((\because \mathrm{y}\) is an increasing function)

\(\therefore\) Discriminant, \(\mathrm{D} \leq 0\)

\(\Rightarrow 4 \mathrm{a}^{2}-4 \times 3 \times 48<0\)

\(\Rightarrow \mathrm{a}^{2}-144<0\)

\(\Rightarrow \mathrm{a} \epsilon(-12,12)\)

Given:

\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2\end{array}\right|\)

\(=\hat{i}(-2-15)-(-4-9) \hat{j}+(10-3) \hat{k}=-17 \hat{i}+13 \hat{j}+7 \hat{k}\)

\(|\vec{a} \times \vec{b}|=\sqrt{(-17)^{2}+(13)^{2}+(7)^{2}}=\sqrt{507}\)

Given,

\(A=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]\)

\(A^2=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]=2\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]=2 A\)

\(A^3=2^2\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right], A^4=2^3\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]\)

\(\therefore A^n=2^{n-1}\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]\)

\(\Rightarrow A^{100}=2^{100-1} A\)

\(\therefore A^{100}=2^{99} A\)

\(\lim _{x \rightarrow-1^{+}} \frac{\sqrt{\pi}-\sqrt{\cos ^{-1} x}}{\sqrt{x+1}}\)

Let \(x=\cos \theta\)

as \(x \rightarrow-1^{+}, \theta \rightarrow \pi^{-}\)

\(\lim _{x \rightarrow \pi}-\frac{\sqrt{\pi}-\sqrt{\theta}}{\sqrt{2} \cos \theta / 2}\left(\frac{0}{0}\right)\)

using L Hospital rule

\(\lim _{x \rightarrow \pi^{-}} \frac{-\frac{1}{2 \sqrt{\theta}}}{-\frac{\sqrt{2}}{2} \sin \frac{\theta}{2}}=\frac{1}{\sqrt{2 \pi}}\)