Mathematics Test - 41

Given,

\(\sin \left(-1125^{\circ}\right)\)

\(=-\sin \left(1125^{\circ}\right)\) \(\quad\quad (\because\) \(\sin (-x)=-\sin x)\)

\(=-\sin \left(360^{\circ} \times 3+45^{\circ}\right)\)\(\quad\quad(\because\) \(\sin (2 n \pi+x)=\sin x)\)

\(=-\sin 45^{\circ}\)

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

\(\therefore\) \(\sin \left(-1125^{\circ}\right)\) \(=\frac{-1}{\sqrt{2}}\).

Given:

\((2,4),(4,6)\) and \((5,6)\) are three elements in \(A \times B\).

Also, \(A \times B\) has 6 elements.

As we know that,

\(n(A \times B)=n(A) \times n(B)\)

\(\Rightarrow {n}({A} \times {B})=6\)

\(=3 \times 2\)

By the definition of Cartesian Product,

\(A \times B=\{(a, b): a \in A\) and \(b \in B)\}\)

\(\Rightarrow {n}({A})=3\) and \({n}({B})=2\)

Thus, the elements of set \({A}\) is \(\{2,4,5\}\) and set \({B}\) is \(\{4,6\}\).

Now,

\(B \times A=\{4,6\} \times\{2,4,5\}\)

\(=\{(4,2),(4,4),(4,5),(6,2),(6,4),(6,5)\}\)

Therefore,

\({B} \times {A}=\{(4,2),(4,4),(4,5),(6,2),(6,4),(6,5)\}\)

Given equation is \(y ^2+( x - b )^2= c\)

There are two constants \(b\) and \(c\) so differentiate two times

Differentiating w.r.t \(x\)

\(2 y \frac{d y}{d x}+2(x-b)=0 \)

\(y \frac{d y}{d x}=b-x\)

Differentiating again w.r.t \(x\)

\(\left(\frac{d y}{d x}\right)^2+y \frac{d^2 y}{d x^2}=-1 \)

\(y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2+1=0\)

As we know,

If three points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\) and \(\left(x_{3}, y_{3}\right)\) are collinear then the area of the triangle determined by the three points is zero.

\(\left|\begin{array}{lll} x _{1} & y _{1} & 1 \\ x _{2} & y _{2} & 1 \\ x _{3} & y _{3} & 1\end{array}\right|=0\)

Given,

The points \((-2,-5),(2,-2)\) and \((8, a)\) are collinear.

\(\left|\begin{array}{ccc}-2 & -5 & 1 \\ 2 & -2 & 1 \\ 8 & a & 1\end{array}\right|=0\)

\(\Rightarrow-2(-2-a)-(-5)(2-8)+1(2 a+16)=0\)

\(\Rightarrow 4+2 a -30+2 a +16=0\)

\(\Rightarrow 4 a -10=0\)

\(\Rightarrow 4 a =10\)

\(\therefore a =\frac{5}{2}\)

Given,

\(y=a e^{3 x} \cos (x+b)\)

There are two constants a and b so differentiate two times Differentiating wr.t \(x\),

We get,

\(y^{\prime}=3 a e^{3 x} \cos (x+b)-a e^{3 x} \sin (x+b)\)

\(\Rightarrow y^{\prime}=3 y-a e^{3 x} \sin (x+b)\)

\(\Rightarrow a e^{3 x} \sin (x+b)=3 y-y^{\prime}\)

Differentiating again w.r.t \({x}\),

We get,

\( 3 {ae}^{3 x} \sin (x+b)+a e^{3 x} \cos (x+b)=3 y^{\prime}-y^{\prime \prime}\)

\(\Rightarrow 3\left(3 y-y^{\prime}\right)+y-3 y^{\prime}-y^{n}=0 \)

\(\Rightarrow 9 y-3 y^{\prime}+y-3 y^{\prime}-y^{n}=0 \)

\(\Rightarrow y^{\prime \prime}+6 y^{\prime}-10 y=0\)

Given that:

\(A=\{x, y, z\}\)

\(n(A)=3\)

And,

\(B=\{1,2,3,4,5\}\)

\(n(B)=5\)

We know that:

If \(n(A)=x\) and \(n(B)=y\), then:

\(n(A \times B)=n(A) \times n(B)\)

\(=x y\)

\(\therefore n(A \times B)=n(A) \times n(B)\)

\(=3 \times 5\)

\(=15\)

Given that:

Direction cosines of a line are \(\left(\frac{1}{k}, \frac{2}{k},-\frac{2}{k}\right)\)

So, \(l=\frac{1}{k}, m=\frac{2}{k}\) and \(n=-\frac{2}{k}\)

We know that:

Sum of squares of the direction cosines of a line is equal to unity, i.e.,

\(l ^{2}+ m ^{2}+ n ^{2}=1\)

\(\Rightarrow \frac{1}{k^{2}}+\frac{4}{k^{2}}+\frac{4}{k^{2}}=1\)

\(\Rightarrow \frac{9}{k^{2}}=1\)

\(\Rightarrow k ^{2}=9\)

\(\therefore k =\pm 3\)

\(n(A)=40 \%\) of \(10000=4000\)

\(n(B)=20 \%\) of \(10000=2000\)

\(n(C)=10 \%\) of \(10000=1000\)

\(n(A \cap B)=5 \%\) of \(10000=500\)

\(n(B \cap C)=3 \%\) of \(10000=300\)

\(n(A \cap C)=4 \%\) of \(10000=400\)

\(n(A \cap B \cap C)=2 \%\) of \(10000=200\)

\(=n\left[A \cap(B \cup C)^{C}\right]\)

\(=n(A)-n[A \cap(B \cup C)]\)

\(=n(A)-[n(A \cap B) \cup n(A \cap C)]\)

\(=n(A)-[n(A \cap B)+n(A \cap C)-n(A \cap B \cap C)]\)

\(=4000-[500+400-200]=3300\)

Given:

\(|\vec{a}|=3,|\vec{b}|=4\) and \(|\vec{a}-\vec{b}|=5\)

We know that,

\(|\vec{a}-\vec{b}|^{2}+|\vec{a}+\vec{b}|^{2}=2 \times\left(|\vec{a}|^{2}+|\vec{b}|^{2}\right)\)

\(\Rightarrow 5^{2}+|\vec{a}+\vec{b}|^{2}=2 \times\left(3^{2}+4^{2}\right)\)

\(\Rightarrow|\vec{a}+\vec{b}|^{2}=50-25=25\)

\(\therefore|\vec{a}+\vec{b}|=5\)

Let, \(E_1=\) Event in which spade is missing

\( P\left(E_1\right)=\frac{1}{4} \ldots( i ) \)

\(P\left(\overline{E_1}\right)=1-P\left(E_1\right) \)

\(P\left(\overline{E_1}\right)=\frac{3}{4} \ldots( ii )\)

\(E =\) Event in which drawn two cards are spade.

\( P(E)=\frac{\left(\frac{1}{4}\right)\left(\frac{{ }^{12} C_2}{{ }^{51} C_2}\right)+\left(\frac{3}{4}\right)\left(\frac{{ }^{13} C_2}{{ }^{51} C_2}\right)-\left(\frac{3}{4}\right)\left(\frac{{ }^{13} C_2}{{ }^{51} C_2}\right)}{\left(\frac{1}{4}\right)\left(\frac{{ }^{12} C_2}{{ }^{51} C_2}\right)+\left(\frac{3}{4}\right)\left(\frac{{ }^{13} C_2}{{ }^{51} C_2}\right)} \)

\(P(E)=\frac{\left(\frac{1}{4}\right)\left(\frac{12 \times 11}{51 \times 50}\right)+\left(\frac{3}{4}\right)\left(\frac{13 \times 12}{51 \times 50}\right)-\left(\frac{3}{4}\right)\left(\frac{13 \times 12}{51 \times 50}\right)}{\left(\frac{1}{4}\right)\left(\frac{12 \times 11}{51 \times 50}\right)+\left(\frac{3}{4}\right)\left(\frac{13 \times 12}{51 \times 50}\right)} \)

\( P(E)=\frac{(12 \times 11)+(3)(13 \times 12)-(3)(13 \times 12)}{(12 \times 11)+(3)(13 \times 12)} \)

\( P(E)=\frac{11}{50}\)

Required probability=1-P(E)

\( =1-\frac{11}{50} =\frac{39}{50}\)