Mathematics Test - 10

\(A=\{1,-1, i,-i\}\)

\(B=\{1,-1\}\)

\(C=\{i,-i\}\)

Now, \(B \cup C=\{1,-1, i,-i\}=A\)

Given:

\(f(x)=|x-1|\)

\(\Rightarrow \mathrm{f}(\mathrm{x})=|\mathrm{x}-1|=\left\{\begin{array}{c}-(\mathrm{x}-1), \quad \mathrm{x}<1 \\ \mathrm{x}-1, \quad \mathrm{x} \geq 1\end{array}\right.\)

We have to find the derivative of \(|x-1|\) at \(x=2\)

It means we have to take \(\mathrm{f}(\mathrm{x})\) for \(\mathrm{x} \geq 1\)

Therefore, \(f(x)=x-1\)

Differentiating with respect to \(\mathrm{x}\), we get

\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=1\)

We know that, \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{b}}\) be the two vectors and the vector \(\overrightarrow{\mathrm{c}}\) perpendicular to both \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{b}}\)

So, \(\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\)

Given:

Vector \(\overrightarrow{\mathrm{c}}\) is perpendicular to both the vectors \(\hat{\imath}-\hat{\jmath}\) and\(\hat{\imath}\).

Therefore, \(\overrightarrow{\mathrm{c}}=(\hat{\imath}-\hat{\jmath}) \times \hat{i}\)

\(=(\hat{\imath} \times \hat{i})-(\hat{\jmath} \times \hat{i})\)

\(=0-(-\hat{k})\)

\(=\hat{\mathrm{k}}\)

Given:

\({ }^{9} \mathrm{P}_{5}+5 \cdot{ }^{9} \mathrm{P}_{4}={ }^{10} \mathrm{P}_{\mathrm{r}}\)

As we know,

\(={ }^{n} P_{r}+r \times{ }^{n} P_{r-1}\)

\(=\frac{n !}{(n-r) !}+r \times \frac{n !}{(n-r+1) ! }\)

\(=\frac{n !}{(n-r) !}+r \times \frac{n !}{(n-r+1) \times(n-r) !}\)

\(=\frac{n !}{(n-r) !}\left(1+\frac{r}{(n-r+1)}\right)\)

\(=\frac{n !}{(n-r) !}\left(\frac{n-r+1+r}{(n-r+1)}\right)\)

\(=\frac{(n+1) n !}{(n-r+1)(n-r) !}=\frac{(n+1) !}{(n-r+1) !}=n+ {}^1 P_{r}\)

\(\therefore^{n} P_{r}+r \times{ }^{n} P_{r-1}={ }^{n+1} P_{r}\)

Now, \({ }^{9} \mathrm{P}_{5}+5 .{ }^{9} \mathrm{P}_{4}={ }^{10} \mathrm{P}_{\mathrm{r}}\)

Compare with above results, we get

\(n=9\) and \(r=5\)

Circle is given as x² + y² = ax + by

or, x² + y² - ax - by = 0 ...(i)

Then line is given as cx - by + b² = 0

or, \(\mathrm{y}=\left(\frac{\mathrm{c}}{\mathrm{b}} \mathrm{x}+\mathrm{b}\right)\)...(ii)

Substituting the value of y from (ii) in (i), we get

\(x^{2}+\left(\frac{c}{b} x+b\right)^{2}-a x-b\left(\frac{c}{b} x+b\right)=0\)

or, \(\mathrm{x}^{2}+\frac{\mathrm{c}^{2}}{\mathrm{~b}^{2}} \mathrm{x}^{2}+2 \mathrm{cx}+\mathrm{b}^{2}-\mathrm{ax}-\mathrm{cx}-\mathrm{b}^{2}=0\)

or \(\mathrm{x}^{2}\left(1+\frac{\mathrm{c}^{2}}{\mathrm{~b}^{2}}\right)+\mathrm{x}(\mathrm{c}-\mathrm{a})=0\)

If it is a perfect square, then c - a = 0 or c = a

Given:

Position vectors of the points \(A\) and \(B\) are respectively \(3 \hat{\imath}-5 \hat{\jmath}+2 \hat{k}\) and \(\hat{\imath}+\hat{\jmath}-\hat{k}\)

\(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}\)

\(\Rightarrow \overrightarrow{\mathrm{AB}}=(\hat{\imath}+\hat{\jmath}-\hat{\mathrm{k}})-(3 \hat{\imath}-5 \hat{\jmath}+2 \hat{\mathrm{k}})\)

\(\overrightarrow{\mathrm{AB}}=-2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}\)

Now, length of \(\mathrm{AB}=|\overrightarrow{AB}|=\sqrt{x^{2}+y^{2}+z^{2}}\)

\(\Rightarrow|\overrightarrow{AB}|=\sqrt{(-2)^{2}+6^{2}+(-3)^{2}}=\sqrt{49}=7\)

We know that,

The order of a differential equation is the order of the highest derivative appearing in it.

The degree of a differential equation is the degree of the highest derivative occurring in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Given the differential equation is,

\(x\left(\frac{d^{2} y}{d x^{2}}\right)^{\frac{2}{3}}=y^{2}\left(\frac{d y}{d x}\right)^{\frac{3}{2}}\)

Taking cube on both sides, we get,

\({x}^{3}\left(\frac{{d}^{2} {y}}{{dx}^{2}}\right)^{2}={y}^{6}\left(\frac{{dy}}{{dx}}\right)^{\frac{9}{2}}\)

Taking square on both sides, we get,

\(x^{6}\left(\frac{d^{2} y}{d x^{2}}\right)^{4}=y^{12}\left(\frac{d y}{d x}\right)^{9}\)

Highest derivate is \(\frac{{d}^{2} {y}}{{dx}^{2}}\)

So, the order of the given differential equation \(=2\)

The power of the highest derivate \(=4\)

So, the degree of the given differential equation \(=4\)

First, arrange 5 men at the round table in \((5-1) !\) ways \(=4 !\)

Now, women sit on the 5 spaces generated between the men

This can be done in \({ }^{5} \mathrm{P}_{3}\) ways

Therefore, the total number of ways \(=4 ! \times{ }^{5} \mathrm{P}_{3}\)

At \(x=\frac{\pi}{6}, \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}\) and \(\sin \frac{\pi}{6}=\frac{1}{2}\)

\(\Rightarrow \cos x>\sin x\)

\(\Rightarrow \cos x-\sin x>0\)

\(\therefore f(x)=|\cos x-\sin x|=\cos x-\sin x\)

\(\Rightarrow f^{\prime}(x)=-\sin x-\cos x\)

And, \(\mathrm{f}^{\prime}\left(\frac{\pi}{6}\right)=-\sin \frac{\pi}{6}-\cos \frac{\pi}{6}\)

\(=-\frac{1}{2}-\frac{\sqrt{3}}{2}\)

\(=-\frac{1}{2}(1+\sqrt{3})\)