Mathematics Test-23

Given,

There are \(7\) consonants and \(4\) vowels

Here, we have to find how many words can be formed such that it has \(3\) consonants and \(2\) vowels.

No. of ways to select \(2\) vowels out of \(4\) vowels \(={ }^{4} C_{2}\)

No. of ways to select \(3\) consonants out of \(7\) consonants \(={ }^{7} C_{3}\)

\(\therefore\) No. of words that can be formed which contains \(3\) consonants and \(2\) vowels \(={ }^{4} C_{2} \times{ }^{7} C_{3}\)

As we know that,

\({ }^{n} C_{r}=\frac{n !}{r ! \times(n-r) !}\)

\(\Rightarrow{ }^{4} C_{2} \times{ }^{7} C_{3}=6 \times 35=210\)

The no. of ways to arrange words containing \(3\) consonants and \(2\) vowels \(=210 \times 5 !=25200\).

Given two lines are \(x-6 y=0\) and \(x-y=0\).

We know \(\frac{0+x_1+x_2}{3}=1\)

\(\Rightarrow x_1+x_2=3 \).....(i)

and \( y_1+y_2=0\).........(ii)

Also, \(x_1-6 y_1=0 \).............(iii)

\(x_2-y_2=0\)......(iv)

[Since the points \(\left(x_1, y_1\right)\) and \(\left(x_2, y_2\right)\) lie on the lines \(A B\) and \(A C\) respectively]

Now on solving, the co-ordinates of \(B\) and \(C\) are \(\left(\frac{18}{5}, \frac{3}{5}\right)\) and \(\left(\frac{-3}{5}, \frac{-3}{5}\right)\) respectively.

Hence, the equation of third side i.e., \(B C\) is \(2 x-7 y-3=0\).

Let \(x^{2}+x+1=t\)

Differentiating both sides with respect to \(x\).

\(2 x+1+0 =\frac{d t}{d x}\)

\(2 x+1 =\frac{d t}{d x}\)

\(d x =\frac{d t}{2 x+1}\)

Integrating the function,

\(\int(4 x+2) \sqrt{x^{2}+x+1} \cdot d x\)

\(=\int 2(2 x+1) \sqrt{x^{2}+x+1} \cdot d x\)

Putting \(x^{2}+x+1=t \) and \(d x=\frac{d t}{2 x+1}\)

\(=\int 2(2 x+1) \sqrt{t} \cdot \frac{d t}{2 x+1}\)

\(=\int 2 \sqrt{t} \cdot d t\)

\(=2 \int \sqrt{t} \cdot d t\)

\(=2 \int t^{\frac{1}{2}} \cdot d t\)

\(=2 \cdot \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C\)

\([\)\(\because\) \(\int x^{n} \cdot d x=\frac{x^{n+1}}{n+1}+C\)\(]\)

\(=2 \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+C\)

\(=\frac{2 \cdot 2}{3} \cdot t^{\frac{3}{2}}+C\)

\(=\frac{4}{3} \cdot t^{\frac{3}{2}}+C\)

\(=\frac{4}{3} \cdot\left(x^{2}+x+1\right)^{\frac{3}{2}}+C\)

\([\)\(\because\) \(x^{2}+x+1=t\)\(]\)

We have \((x+y)^{n}={ }^{n} C_{0} x^{n}+{ }^{n} C_{1} x^{n-1} . y+{ }^{n} C_{2} x^{n-2}. y^{2}+\ldots+{ }^{n} C_{n} y^{n}\)

General term: General term in the expansion of \((x+y)^{n}\) is given by:

\(\mathrm{T}_{(\mathrm{r}+1)}={ }^{n} C_{\mathrm{r}} \times x^{\mathrm{n}-\mathrm{r}} \times y^{\mathrm{r}}\)

We have to find term independent of \(x\) in \(\left(x^{2}-\frac{1}{x^{3}}\right)^{10}\).

We know that,

\(\mathrm{T}_{(\mathrm{r}+1)}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \times {x}^{\mathrm{n}-\mathrm{r}} \times \mathrm{y}^{\mathrm{r}}\)

\(\Rightarrow \mathrm{T}_{(\mathrm{r}+1)}={ }^{10} \mathrm{C}_{\mathrm{r}} \times\left({x}^{2}\right)^{10-\mathrm{r}} \times\left(\frac{-1}{{x}^{3}}\right)^{\mathrm{r}}\)

\(=(-1)^{\mathrm{r}} \times{ }^{10} \mathrm{C}_{\mathrm{r}} \times({x})^{20-2 \mathrm{r}} \times\left({x}^{3}\right)^{-\mathrm{r}}\)

\(=(-1)^{\mathrm{r}} \times{ }^{10} \mathrm{C}_{\mathrm{r}} \times({x})^{20-2 \mathrm{r}} \times({x})^{-3 \mathrm{r}}\)

\(=(-1)^{\mathrm{r}} \times{ }^{10} \mathrm{C}_{\mathrm{r}} \times({x})^{20-5 \mathrm{r}}\)

For the term independent of \(x\), power of \(x\) should be zero

Therefore, \(20-5 r=0\)

\(\Rightarrow \mathrm{r}=4\)

\(\mathrm{T}_{(4+1)}=(-1)^{4} \times{ }^{10} \mathrm{C}_{4}={ }^{10} \mathrm{C}_{4}\)

As we given

\((-2 \hat{i}-\hat{j}),(4 \hat{i}),(3 \hat{i}+3 \hat{j}),(-3 \hat{i}+2 \hat{j})\)

On converting the above vectors into Cartesian coordinates Then, consider,

\(P (-2,-1), Q (4,0), R (3,3), S (-3,2)\)

By distance formula

\(=\sqrt{\left( x _1- x _2\right)^2+\left( y _1+ y _2\right)^2}\)

\(d ( P Q)=\sqrt{(4+2)^2+(0+1)^2}=\sqrt{37}\)

Similarly,

\(d ( QR )=\sqrt{10}\)

\(d ( RS )=\sqrt{37}\)

\(d ( SP )=\sqrt{10}\)

Therefore,

\(d ( PQ )= d ( RS )\) and \(d ( QR )= d ( SP )\)

Hence,

\(PQRS\) is parallelogram

Now,

Slope \(=\frac{y_2-y_1}{x_2-x_1}\)

Therefore, let slope is \(S\) then,

\(S ( PQ )=\frac{0-(-1)}{4-(-2)}=\frac{0+1}{4+2}=\frac{1}{6}\)

Similarly,

\(S ( QR )=\frac{3}{-1}=-3\)

\(S ( RS )=\frac{-1}{-6}=\frac{1}{6}\)

\(S ( PS )=-3\)

\(S ( P Q )= S ( RS )\) this implies \(PQ \| RS\)

And

\(S ( QR )= S ( SP )\) this implies \(QR \| SP\)

\(S(P Q) = S(Q R)\) i.e. they aren't parallel

\(S ( PQ ) \times S ( QR ) =-1\) i.e. they aren't perpendicular.

Therefore

\(P Q R S\) is a parallelogram but it is nor a rhombus and not a rectangle.

Let U be the set of all students who took part in the survey.

Let T be the set of students taking tea.

Let C be the set of students taking coffee.

Given that:

Number of students \(=\mathrm{n}(\mathrm{U})=600\)

Number of students taking tea \(=\mathrm{n}(\mathrm{T})=150\)

Number of students taking coffee \(=\mathrm{n}(\mathrm{C})=225\)

Number of students taking both tea and coffee \(=\mathrm{n}(\mathrm{T} \cap \mathrm{C})=100\)

To find: Number of student taking neither tea nor coffee i.e., we have to find \(\mathrm{n}\left(\mathrm{T}^{\prime} \cap \mathrm{C}^{\prime}\right) .\)

\(\mathrm{n}\left(\mathrm{T}^{\prime} \cap \mathrm{C}^{\prime}\right)=\mathrm{n}(\mathrm{T} \cup \mathrm{C})^{\prime}\)

\(\mathrm{n}\left(\mathrm{T}^{\prime} \cap \mathrm{C}^{\prime}\right)=\mathrm{n}(\mathrm{U})-\mathrm{n}(\mathrm{T} \cup \mathrm{C})\)

\(=\mathrm{n}(\mathrm{U})-[\mathrm{n}(\mathrm{T})+\mathrm{n}(\mathrm{C})-\mathrm{n}(\mathrm{T} \cap \mathrm{C})]\)

\(=600-[150+225-100]\)

\(=600-275\)

\(=325\)

Given,

\(U=\left[\begin{array}{lll}2 & -3 & 4\end{array}\right], X=\left[\begin{array}{lll}0 & 2 & 3\end{array}\right], V=\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right], Y=\left[\begin{array}{l}2 \\ 2 \\ 4\end{array}\right]\)

Then,

\(U V+X Y=\left[\begin{array}{lll}2 & -3 & 4\end{array}\right]\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right]+\left[\begin{array}{lll}0 & 2 & 3\end{array}\right]\left[\begin{array}{l}2 \\ 2 \\ 4\end{array}\right]\)

\(U V+X Y=\left[\begin{array}{ll}6-6+4\end{array}\right]+[0+4+12]\)

\(=[4]+[16]\)

\(=[20]\)

Given: \(\left(2^{\frac{1}{3}}+\frac{1}{2(3)^{\frac{1}{3}}}\right)^{10}\)

Here, \({n}=10, {a}=2^{\frac{1}{3}}\) and \({b}=\frac{1}{2(3)^{\frac{1}{3}}}\)

As we know that, in the binomial expansion of \(({a}+{b})^{{n}}\), the \({r}^{\text {th }}\) term the end is \(({n}-{r}+2)^{\text {th }}\) term from the beginning.

\(\Rightarrow 5^{\text {th }}\) from the end in the binomial expansion of \(\left(2^{\frac{1}{3}}+\frac{1}{2(3)^{\frac{1}{3}}}\right)^{10}\) is \((10-5+2=7)^{\text {th }}\) term from the beginning.

\(\Rightarrow\) We need to find \(T_{5}: T_{7} .\)

As we know that, the general term in the binomial expansion of \((a+b)^{n}\) is given by: \(T_{r+1}={ }^{n} C_{r} \times a^{n-r} \times b^{r}\)

\(\Rightarrow \frac{T_{5}}{T_{7}}=\frac{{ }^{10} C_{4} \times\left(2^{\frac{1}{3}}\right)^{6} \times\left(\frac{1}{2(3)^{\frac{1}{3}}}\right)^{4}}{{ }^{10} C_{6} \times\left(2^{\frac{1}{3}}\right)^{4} \times\left(\frac{1}{2(3)^{\frac{1}{3}}}\right)^{6}}\)

As we know that, \({ }^{n} C_{r}={ }^{n} C_{n-r}\)

\(\Rightarrow{ }^{10} C_{6}={ }^{10} C_{4}\)

\(\Rightarrow \frac{T_{5}}{T_{7}}=\frac{2^{\frac{6}{3}} \times\left(2(3)^{\frac{1}{3}}\right)^{6}}{2^{\frac{4}{3}} \times\left(2(3)^{\frac{1}{3}}\right)^{4}}=4 \times(36)^{\frac{1}{3}}\)

Perpendicular distance of a point from a plane:

Let us consider a plane given by a cartesian equation,

\(A x+B y+C z=d\), and a point whose coordinates are \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\).

So,

Distance \(=\left|\frac{A x_1+B y_1+C z_1-d}{\sqrt{A^2+B^2+C^2}}\right|\)

We have to find the distance of point \((2,3,4)\) from the plane \(3 x-6 y+2 z+11=0\)

\(\Rightarrow\) Distance \(=\left|\frac{3 \times 2+(-6) \times 3+2 \times 4+11}{\sqrt{3^2+(-6)^2+2^2}}\right|\)

\(\Rightarrow\) Distance \(=\left|\frac{6-18+8+11}{\sqrt{9+36+4}}\right|\)

\(\Rightarrow\) Distance \(=\left|\frac{7}{\sqrt{49}}\right|\)

\(\Rightarrow\) Distance \(=1\)

So, The distance of the point \((2,3,4)\) from the plane \(3 x-6 y+2 z+11=0\) is 1 unit.