Mathematics Test - 7

If any two adjacent rows or columns of a determinant are interchanged in position, the value of the determinant changes its sign.

For example,

$A=$$\left|\begin{array}{lll}1& 2& 3\\ 4& 5& 6\\ 2& 1& 2\end{array}\right|$

The value of determinant is:

$A=1[5\times 2–6\times 1]–2[4\times 2–6\times 2]+3[4\times 1–5\times 2]$

$A=1[10–6]–2[8–12]+3[4–10]$

$A=4+8–18=-6$

Now changing the row R­₁ with R₂,

$A=$$\left|\begin{array}{lll}4& 5& 6\\ 1& 2& 3\\ 2& 1& 2\end{array}\right|$

The value of determinant is:

$A=4[2\times 2–3\times 1]–5[1\times 2–3\times 2]+6[1\times 1–2\times 2]$

$A=4[4–3]–5[2–6]+6[1–4]$

$A=4+20–18=6$

Now changing the column C₁ with C₂,

$A=$$\left|\begin{array}{lll}2& 1& 3\\ 5& 4& 6\\ 1& 2& 2\end{array}\right|$

The value of determinant is:

$A=2[4\times 2–6\times 2]–1[5\times 2–6\times 1]+3[5\times 2–4\times 1]$

$A=2[8–12]–1[10–6]+3[10–4]$

$A=-8–4+18=6$

Given,

The planes \(2 x-y-3 z-7=0\) and \(4 x-2 y+5 k z+9=0\) are parallel

We know that,

If plane \(a_{1} x+b_{1} y+c_{1} z+d_{1}=0\) and \(a_{2} x+b_{2} y+c_{2} z+d_{2}=0\) are a parallel i.e.,

\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \neq \frac{d_{1}}{d_{2}}\)

Now,

If plane are parallel than ratio of coefficient of \(x, y\) and \(z\) are equal.

\(\frac{2}{4}=\frac{-1}{-2}=\frac{-3}{5 k}\)

\(\Rightarrow \frac{1}{2}=\frac{-3}{5 k}\)

\(\Rightarrow 5 {k}=-6\)

So, \(k=\frac{-6}{5}\)

Now,

\(5 k+7\)

\(=5 \times(\frac{-6}{5})+7\)

\(=(-6)+7\)

\(=1\)

Let, \(\vec{\mathrm{A}}=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{j}}\)

\(\therefore|\vec{\mathrm{A}}|=\sqrt{\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}}\)

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

\(=\sqrt{1}\)

\(=1\)

It is a unit vector.

Given,

Total number of cards \(=52\)

Number of king card \(=4\)

Now, 7 cards are drawn from 52 cards.

\(P(3 \text { cards are king })=\frac{{ }^{4} C_{3} \times{ }^{48} \mathrm{C}_{4}}{{ }^{52} \mathrm{C}_{7} }\)

\(=\frac{4 \times\frac{(48 \times 47 \times 46 \times 45)}{(4 \times 3 \times 2 \times 1)\ }}{\frac{(52 \times 51 \times 50 \times 49 \times 48 \times 47 \times 46)}{(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} }\)

\(=\frac{4 \times(48 \times 47 \times 46 \times 45) \times(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}{(4 \times 3 \times 2 \times 1) \times(52 \times 51 \times 50 \times 49 \times 48 \times 47 \times 46)}\)

\(=\frac{(7 \times 6 \times 5 \times 4 \times 45)}{(52 \times 51 \times 50 \times 49)}\)

\(=\frac{(6 \times 5 \times 4 \times 45)}{(52 \times 51 \times 50 \times 7)}\)

\(=(\frac{(6 \times 4 \times 45)}{(7 \times 52 \times 51 \times 10)}\)

\(=\frac{(6 \times 45)}{(7 \times 13 \times 51 \times 10)}\)

\(=\frac{(6 \times 3)}{(7 \times 13 \times 17 \times 2) }\)

\(=\frac{(3 \times 3) }{(7 \times 13 \times 17) }\)

\(=\frac{9}{1547}\)

Given

Objective function

Maximize, \(Z=X_{1}+2 X_{2}\)

Constraints:

\(X_{1} \leq 2 \ldots \ldots  (1)\)

\(X_{2} \leq 2 \ldots \ldots  \ldots(2) \)

\(X_{1}+X_{2} \leq 2 \ldots\ldots(3)\)

Non neagative constarints

\(X_{1}, X_{2} \geq 0\)

The above equations can be written as

\(\frac{X_{1}}{2} \leq 1(4) \)

\(\frac{X_{2}}{2} \leq 1(5) \)

\(\frac{X_{1}}{2}+\frac{X_{2}}{2} \leq 1(6)\)

Plot the above equations on \(X_{1}-X_{2}\) graph and find out the solution space,

Now, find out the value of the objective function at every extreme point of solution space.

\(Z_{0}=0+2 \times 0=0 \)

\(Z_{A}=0+2 \times 2=4\)

\(Z_{B}=2+2 \times 0=2\)

Since the value of the objective function is maximum at A. There \(A(0,2)\) is the optimal solution.

The area of a triangle with vertices \((-3,0),(3,0)\) and \((0, k)\) is 9 sq units

Area of a triangle with vertices \(\left(x_1, y_1\right),\left(x_2, y_2\right)\) and \(\left(x_3, y_3\right)\) is given by,

\(\Delta=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\x_2 & y_2 & 1 \\x_3 & y_3 & 1\end{array}\right|\)

\(\therefore \Delta=\frac{1}{2}\left|\begin{array}{ccc}-3 & 0 & 1 \\3 & 0 & 1 \\0 & k & 1\end{array}\right|\)

\(9=\frac{1}{2}[-3(-k)-0+1(3 k)]\)

\(\Rightarrow 18=3 k+3 k=6 k\)

\(\therefore k=\frac{18}{6}=3\)

Given: 

\(\sqrt{y}=\sin x+\cos x\)  

Squaring both sides, we get:

\(\Rightarrow y=(\sin x+\cos x)^{2}\)

\(\Rightarrow y=\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x\)

\(\Rightarrow y=1+\sin 2 x \quad(\because 2 \sin x \cos x=\sin 2 x)\)

Differentiating above with respect to x, we get:

\(\frac{{dy}}{{dx}}=0+2 \times \cos 2 {x}\)

\(=2 \cos 2 {x}\)

Let \(\mathrm{x}=\frac{\sqrt{22}+\sqrt{10}}{\sqrt{22}-\sqrt{10}} ; \mathrm{y}=\frac{\sqrt{22}-\sqrt{10}}{\sqrt{22}+\sqrt{10}}\)

We know that, \(x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)\)

\(\Rightarrow(\mathrm{x}+\mathrm{y})=\frac{\left[(\sqrt{22}+\sqrt{10})^{2}+(\sqrt{22}-\sqrt{10})^{2}\right]}{22-10}\)

\(\Rightarrow(x+y)=\frac{22+10+22+10}{ 12}\)

\(\Rightarrow(x+y)=\frac{64 }{ 12}\)

\(\Rightarrow(x+y)=\frac{16 }{ 3}\) and \(x y=1\)

\(\Rightarrow x^{2}+y^{2}+2 x y=(\frac{16 }{ 3})^{2}\)

\(\Rightarrow x^{2}+y^{2}+2(1)=\frac{256 }{ 9}\)

\(\Rightarrow x^{2}+y^{2}=\frac{256 }{ 9}-2\)

\(\Rightarrow x^{2}+y^{2}=\frac{238 }{ 9}\)

\(x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)\)

\(\Rightarrow x^{3}+y^{3}=\left(\frac{16}{3}\right) \times\left(\frac{238}{9}-1\right)\)

\(\Rightarrow x^{3}+y^{3}=\left(\frac{16}{3}\right) \times\left(\frac{229}{9}\right)\)

\(\Rightarrow\left(\frac{\sqrt{22}+\sqrt{10}}{\sqrt{22}-\sqrt{10}}\right)^{3}+\left(\frac{\sqrt{22}-\sqrt{10}}{\sqrt{22}+\sqrt{10}}\right)^{3}=\left(\frac{229}{27}\right) \times 16\)

\(\Rightarrow\left(\frac{229}{27}\right) \times 16=\frac{229 a}{27}\)

\(\therefore a=16\)

Let A = Students who play cricket, B = Students who play football

⇒ n (A) = 50% and n (B) = 40%

⇒ n (A ∩ B) = 10%

⇒ n (A ∪ B) = n (A) + n (B) - n (A ∩ B) = 50 + 40 - 10 = 80%

As we know that (A ∪ B)’ = A’ ∩ B’

⇒ n ((A ∪ B)’) =n (U) - n (A ∪ B) = 100 - 80 = 20% = n (A’ ∩ B’)

Given,

\(P(n)=\left(\frac{n+1}{2}\right)^{n} \geq n !\)

Put, \(n=1\)

\(P(1)=\left(\frac{1+1}{2}\right)^{1} \geq 1 !\)

\(=1 \geq 1\)

Put, \(n=2\)

\(P(2)=\left(\frac{2+1}{2}\right)^{2} \geq 2 !=\left(\frac{3}{2}\right)^{2} \geq 2 \times 1\)

\(=2.25 \geq 2\)

Put, \(n=3\)

\(P(3)=\left(\frac{3+1}{2}\right)^{3} \geq 3 !\)

\(=8 \geq 3 \times 2 \times 1\)

\(=8 \geq 6\)

Thus, the given expression \(P(n)\) is true for \(n\geq 1\).