Mathematics Test - 59

Using the triangle law for addition of vectors:

\(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}}\)

\(\Rightarrow \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}}-\overrightarrow{\mathrm{AB}}=(2 \overrightarrow{\mathrm{i}}+\overrightarrow{\mathrm{j}}+4 \overrightarrow{\mathrm{k}})-(\overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}})=\overrightarrow{\mathrm{i}}+2 \overrightarrow{\mathrm{j}}+4 \overrightarrow{\mathrm{k}}\)

Let's say that \(A D\) is the median from \(A\).

\(\therefore \overrightarrow{\mathrm{BD}}=\frac{1}{2} \overrightarrow{\mathrm{BC}}=\frac{1}{2}(\overrightarrow{\mathrm{i}}+2 \overrightarrow{\mathrm{j}}+4 \overrightarrow{\mathrm{k}})=\frac{1}{2} \overrightarrow{\mathrm{i}}+\overrightarrow{\mathrm{j}}+2 \overrightarrow{\mathrm{k}}\)

And \(\overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BD}}\)

\(\Rightarrow \overrightarrow{\mathrm{AD}}=(\overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}})+\left(\frac{1}{2} \overrightarrow{\mathrm{i}}+\overrightarrow{\mathrm{j}}+2 \overrightarrow{\mathrm{k}}\right)=\frac{3}{2} \overrightarrow{\mathrm{i}}+0 \overrightarrow{\mathrm{j}}+2 \overrightarrow{\mathrm{k}}\)

\(\therefore|\mathrm{AD}|=\sqrt{\left(\frac{3}{2}\right)^{2}+0^{2}+2^{2}}=\sqrt{\frac{9}{4}+4}=\sqrt{\frac{25}{4}}=\frac{5}{2}=2.5\)

Given,

\(Z=1+i\)

We have to find the modulus of \(Z+\frac{2}{Z}\)

\(\Rightarrow(1+i)+\frac{2}{1+i}\)

On rationalizing the second term, we get 

\(\Rightarrow(1+i)+\frac{2}{1+i} \times \frac{1-i}{1-i}\)

\(\Rightarrow(1+ i )+\frac{2 \times(1- i )}{1- i ^{2}}\)

\(\Rightarrow(1+ i )+\frac{2 \times(1- i )}{1-(-1)}\)

\(\Rightarrow(1+ i )+\frac{2 \times(1- i )}{2}\)

\(\Rightarrow 1+ i +1- i\)

\(\Rightarrow 2\)

\(\therefore\) The modulus of \(Z +\frac{2}{ Z }=2\).

Given here,

\(R=\{(a, b): a, b \in Z\) and \((a+b)\) is even \(\}\)

We know that:

R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.

Therefore,

1. Since, \(a+a=2 a\), which is even, so \(R\) is reflexive.

2. If \(a+b\) is even, then \(b+a\) will also be even. So, \(R\) is symmetric.

3. Let, \(a=3, b=5\), and \(c=7\), then:

\(a+b=3+5=8\) (which is even),

b \(+c=5+7=12\) (which is again even)

and, \(a+c=3+7=10\) (which is also even)

So, R is transitive.

Therefore, \(R\) is an equivalence relation on \(Z\).

Let's say that \(\sin ^{-1} 5 x=\theta\)

\(\Rightarrow \sin \left(\sin ^{-1} 5 x\right)=\sin \theta \)

\(\Rightarrow \sin \theta=5 x\)

Since, \(-1 \leq \sin \theta \leq 1\)

\(\Rightarrow-1 \leq 5 x \leq 1\)

\(\Rightarrow-\frac{1}{5} \leq x \leq \frac{1}{5} \)

\(\Rightarrow x \in\left[-\frac{1}{5}, \frac{1}{5}\right]\)

The domain of the function is the closed interval \(\left[-\frac{1}{5}, \frac{1}{5}\right]\).

The probability that exactly \(4\) vessels arrive safely is, \(={ }^{5} C_{4} \times\left(\frac{9}{10}\right)^{4}\left(\frac{1}{10}\right)\)

The probability that all \(5\) arrive safely is \(\left(\frac{9}{10}\right)^{5}\)

The probability that at-least \(4\) vessels arrive safely,

\(={ }^{5} C_{4} \times\left(\frac{9}{10}\right)^{4}\left(\frac{1}{10}\right)+\left(\frac{9}{10}\right)^{5}\)

\(=\frac{14 \times 9^{4}}{10^{5}}\)

Given,

\(y=a \sin (\lambda x+a) \ldots { (i) }\)

Now,

Differentiating both sides by equation \((i)\)

We get,

\(\frac{d y}{d x}=\frac{d}{d x}(a \sin (\lambda x+a\)

\(\Rightarrow \frac{d y}{d x}=a \cos (\lambda x+\alpha) \times \frac{d}{d x}(\lambda x+\alpha)\)

\(\Rightarrow \frac{d y}{d x}=a \lambda \cos (\lambda x+\alpha)\)

Again differentiating both sides we get,

\(\frac{{d}^{2} {y}}{{dx}^{2}}=-{a} \lambda^{2} \sin (\lambda x+{a})\)

From equation \((i)\) we get,

\(\frac{d^{2} y}{d x^{4}}=-\lambda^{2} y \)

\(\therefore \frac{d^{2} y}{d x^{2}}+\lambda^{2} y=0\)

Given function:

\(y=f(x)=\sqrt{1+x^{2}}\) and \(x=0 =\mathrm{a}, ~ x=4 =\mathrm{b}\)

We know that:

\(V=\pi \int_{a}^{b}[f(y)]^{2} d y\)

Therefore,

Volume, \((V)=\pi \int_{0}^{4}\left(\sqrt{1+x^{2}}\right)^{2} d x\)

\(=\pi \int_{0}^{4}\left(1+x^{2}\right) d x\)

\(=\pi\left[x+\frac{x^{3}}{3}\right]_{0}^{4}\)

\(=\pi\left[(4-0)+\left(\frac{64}{3}-\frac{0}{3}\right)\right]\)

\(=\pi\left[\frac{12+64}{3}\right]\)

\(=\frac{76 \pi}{3}\)