Mathematics Test - 68

Given,

The length of the perpendicular from the point \((4,1)\) on the line \(3 x-4 y+k=0\) is 2 units.

The perpendicular distance \(d\) from \(P\left(x_{1}, y_{1}\right)\) to the line \(a x+b y+c=0\) is given by

\(d=\left|\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\right|\).........(i)

Let \(P=(4,1)\)

Here \(x_{1}=4, y_{1}=1, a=3, b=-4\) and \(d=2\)

Now substitute \(x_{1}=4\) and \(y_{1}=1\) in the equation (i) we get,

\(\Rightarrow d=\left|\frac{3 × x_{1}-4 × y_{1}+k}{\sqrt{3^{2}+(-4)^{2}}}\right|=2\)

\(\Rightarrow d=\left|\frac{3 × 4-4 × 2+k}{\sqrt{25}}\right|=2\)

\(\Rightarrow|\frac{8+\mathrm{k}}{5}|=2\)

\(\Rightarrow|8+\mathrm{k}|\) = \(10\)

\(\Rightarrow \mathrm{k}=2 \text { or }-18\)

Given that:

All boys are to sit together.

So, all boys can be considered as a single group.

\(\therefore\) Total no. of students \(=4\) Girls \(+1\) Group \(=5\)

Since, this is a cyclic permutation.

We know that, for cyclic permutation:

\({}^nP_r=(n-1)!\)

Therefore,

The number of ways of arranging 5 students in a round table is \((5-1) !=4 !\)

Now, no. of the ways of arranging 5 boys is \(5 !\).

Therefore, the total number of ways \(=4 ! 5 !\)

Given,

\(|{x}|<-5\)

Now, LHS \(\geq 0\) and RHS \(<0\)

Since LHS is non-negative and RHS is negative So, \(|x|<-5\) does not posses any solution.

Given:

Curve \(y^{2}=16 x\)

Differentiating w.r.t \(x\) on both sides:

\(\frac{d}{d x} y^{2}=\frac{d}{d x} 16 x\)

\(2 y \frac{d y}{d x}=16\)

The point on which tangent touches the curve is (1,4).

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{8}{4}=2\)

Slope of tangent \(m=2\)

Now equation of the tangent:

\(\left(\mathrm{y}-\mathrm{y}_{1}\right)=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_{1}\right)\)

\(y-4=2(x-1)\)

\(y-2 x-2=0\)

So, distance of the origin (0,0) from normal is:

\(D=\left|\frac{a p+b q+c}{\sqrt{a^{2}+b^{2}}}\right|\)

\(D=\left|\frac{1 \times 0+(-2) \times 0+(-2)}{\sqrt{1^{2}+(-2)^{2}}}\right|\)

\(D=\left|\frac{-2}{\sqrt{5}}\right|=\frac{2}{\sqrt{5}}\)

\(\cos ^{-1}\left(4 x^3-3 x\right)\)

Put \(x=\cos \theta \)

\(\Rightarrow \theta=\cos ^{-1} x\)

\(\cos ^{-1}\left(4 x^3-3 x\right)\)

\(=\cos ^{-1}\left(4 \cos ^3 \theta-3 \cos \theta\right)\)

\(=\cos ^{-1}(\cos 3 \theta) \quad\left(\because \cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta\right)\)

\(=3 \theta \quad\left(\because \cos ^{-1} \cos \theta=\theta\right)\)

\(=3 \cos ^{-1} x\)

Given, curve \(y=f(x)\) passes through \((1,2)\) and satisfies

\(x \frac{d y}{d x}+y=b x^4 \\\)

\( \Rightarrow x \frac{d y}{d x}+y=b x^4 \\\)

\( \Rightarrow \frac{d y}{d x}+\frac{y}{x}=b x^3 \\\)

\( \quad I F=e^{j \frac{1}{x} \cdot d x}=x \\\)

\( y x=\int b x^4 d x=\frac{b x^5}{5}+C \\\)

\( \therefore y=\frac{b x^4}{5}+\frac{C}{x}=f(x) \ldots\)

\(\because\) This curve passes through \((1,2)\).

\(\therefore 2 \times 1=\frac{\mathrm{b} \times(1)^5}{5}+\mathrm{C} \\\)

\( \Rightarrow 2=\frac{\mathrm{b}}{5}+\mathrm{C} \ldots \text { (ii) }\)

Also, \(\int_1^2 f(x) d x=\frac{62}{5}\)

\(\Rightarrow \int_1^2\left(\frac{b x^4}{5}+\frac{C}{x}\right) d x=\frac{62}{5} \\\)

\( \Rightarrow\left[b \times \frac{x^5}{25}+C \log x\right]_1^2=\frac{62}{5} \\\)

\( \Rightarrow\left[\left(\frac{b \times 32}{25}+C \log 2\right)-\left(\frac{b}{25}+c \log 1\right)\right]=\frac{62}{5} \\\)

\( \Rightarrow \frac{b \times 32}{25}+C \log 2-\frac{b}{25}=\frac{62}{5}+0 \log 2\)

[from Eq. (i)] On comparing, we get

\( \frac{b}{25} \times 31=\frac{62}{5} \text { and } c=0 \\\)

\( b=\frac{62 \times 25}{31 \times 5} \\\)

\( b=10\)

Hence, the required value of \(b=10\).

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)

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

\( \because \vec{a}=\hat{i}+\hat{j}+\hat{k} \Rightarrow|\vec{a}|=\sqrt{3} \\\)

\( \& \vec{c}=\hat{j}-\hat{k} \Rightarrow|\vec{c}| \sqrt{2} \\\)

\( \text { Now, } \vec{a} \times \vec{b}=\vec{c} \text { (Given) } \\\)

\( \Rightarrow|\vec{a}||\vec{b}| \sin \theta=|\vec{c}| \\\)

\( \Rightarrow|\vec{a}||\vec{b}| \sin \theta=\sqrt{2} \ldots \ldots . . \text { (i) }\)

Also \(\vec{a} \cdot \vec{b}=3 \Rightarrow|\vec{a}||\vec{b}| \cos \theta=3\)

Dividing [i] by [ii], we get

\(\tan \theta=\frac{\sqrt{2}}{3} \therefore \sin \theta=\frac{\sqrt{2}}{\sqrt{11}}\)

Substituting value of \(\sin \theta\) in [i] we get

\(\sqrt{3}|\vec{b}| \frac{\sqrt{2}}{\sqrt{11}}=\sqrt{2} \\\)

\(|\overrightarrow{\mathrm{b}}|=\frac{\sqrt{11}}{\sqrt{3}}\)

Given that,

\(\left|\begin{array}{ccc}x+a & b & c \\ a & x+b & c \\ a & b & x+c\end{array}\right|=0\)

Applying\(C_1 \rightarrow C_1+C_2+C_3\)we get

\(\begin{aligned} & \Rightarrow\left|\begin{array}{ccc}x+a+b+c & b & c \\ x+a+b+c & x+b & c \\ x+a+b+c & b & x+c\end{array}\right|=0 \\ & \Rightarrow(x+a+b+c)\left|\begin{array}{ccc}1 & b & c \\ 1 & x+b & c \\ 1 & b & x+c\end{array}\right|=0\end{aligned}\)

Applying\(R_2 \rightarrow R_2-R_1\)and\(R_3 \rightarrow R_3-R_1\)weget

\(\Rightarrow(x+a+b+c)\left|\begin{array}{lll}1 & b & c \\ 0 & x & 0 \\ 0 & 0 & x\end{array}\right|=0\)

Now,expanding along\(C_1\)

\(\begin{aligned} & (x+a+b+c) \cdot 1 \cdot\left(x^2\right)=0 \\ & \Rightarrow x^2 \cdot(x+a+b+c)=0 \\ & \Rightarrow x=0 \text { or } x=-(a+b+c)\end{aligned}\)