Mathematics Test - 32

The Given planes are,

\(p_1\) : x + y + z = 1

\(p_2\) : 2x + 3y + 4z = 7

Therefore, equation of plane passing through intersection of planes \(p_1\) and \(p_2\) is

x + y + z -1 + k (2x + 3y + 4z - 7) = 0

\(\Rightarrow\) x + y + z - 1 + 2kx + 3ky + 4kz - 7k = 0

\(\Rightarrow\) x(1+2k) + y(1+3k) + z(1+4k) - 1- 7k = 0

This is perpendicular to x- 5y+ 3z=5

\(\Rightarrow\) x - 5y + 3z - 5 = 0

\(\Rightarrow\) 1(1+2 k) - 5(1+3k) + (1+4k) = 0

\(\Rightarrow\) 1 + 2k - 5- 15k + 3 + 12k = 0

\(\Rightarrow-k-1=0 \Rightarrow k=-1\)

\(\therefore\) Equation of plane is x + y + z - 1 - 1(2x + 3y + 4z - 7)=0

\(\Rightarrow\) x + y + z - 1 - 2x - 3y - 4z + 7 = 0

-x - 2y - 3z + 6 = 0

\(\Rightarrow\) x + 2y + 3z - 6 = 0

Curve 1: \(y=\sin x=f(x)\) (say)

Curve 2: Lines \({x}=-\frac{\pi}{3}\) and \({x}=\frac{\pi}{3}\)

It can be drawn as follows:

According to the figure the sum of area curve \(\mathrm{OAB}\) and curve OCD.

Here, \(\mathrm{OAB}\) and \(\mathrm{OCD}\) are equal and limit 0 to \(\frac{\pi}{3}\).

So,  Area \(=2 \times\) area of \(\mathrm{OAB}\).

The area between the curves \(y_{1}=f(x)\) and \(y_{2}=g(x)\) is given by: 

Area enclosed \(=\left|\int_{\mathrm{x}_{1}}^{\mathrm{x}_{2}}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right) \mathrm{dx}\right|\)

Where, \(\mathrm{x}_{1}\) and \(\mathrm{x}_{2}\) are the intersections of curves \(\mathrm{y}_{1}\) and \(\mathrm{y}_{2}\)

Now, the required area (A) is,

Area of \(O A B=\left|\int_{\mathrm{x}_{1}}^{\mathrm{x}_{2}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\right|\)

\(=\left|\int_{0}^{\frac{\pi}{3}} \sin x ~d x\right|\)

\(=\left|[-\cos x]_{0}^{\frac{\pi}{3}}\right|\)

\(=\left|-\cos \frac{\pi}{3}+\cos 0^{\circ}\right|\)

\(=\left|1-\frac{1}{2}\right|\) \(=\frac{1}{2}\)

\(\therefore\) Shaded Area \(=2 \times \frac{1}{2}=1\)

\(\mathrm{Z}=3 \mathrm{x}+2 \mathrm{y}\)

Subject to the constraints

\(x+2 y \leq 10 \)

\(3 x+y \leq 15\)

Convert these inequalities into equations

\(\mathrm{x}+2 \mathrm{y}=10 \).....(i)

\(3 \mathrm{x}+\mathrm{y}=15 \text {......... (ii) }\)

From (i), we get

\(\mathrm{x}=0 \) when \( \mathrm{y}=5\) and \(\mathrm{y}=0\) when \(\mathrm{x}=10\)

So, the points \((0,5)\) and \((10,0)\) lie on the line given in (i).

From (ii), we get the points \((0,15)\) and \((5,0)\)

Let's plot these point and we get the graph in which, shaded part shows the feasible region.

Lines (i) and (ii) intersect at \((4,3)\) and other corner points of the region are \((0,5),(5,0)\) and \((0,0)\).

To find the maximum value of \(Z\), we need to find the value of \(Z\) at the corner points.

\(\begin{array}{lc}\text { Corner points } & \mathrm{Z}=3 \mathrm{x}+2 \mathrm{y} \\ (0,0) & 0 \\ (5,0) & 15 \\ (0,5) & 10 \\ (4,3) & 18\end{array}\)

Thus, \(Z\) is maximum at \((4,3)\) and its maximum value is 18.

Given,

Two parallel line\(p(x+y)+q=0\) and \(p(x+y)-r=0\)

The distance between the parallel lines \(ax+by+c_{1}=0\) and \(a x+b y+c_{2}=0\) is given by:

\(d=\left|\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\right|\)

Here, we have to find the distance between the parallel lines \(p(x+y)+q=0\) and \(p(x\) \(+y)-r=0\)

The given equations of line can be re-written as: \(p x+p y+q=0\) and \(p x+p y-r=0\)

By comparing the equations of the given line with \(ax+by+\mathrm{c}_{1}=0\) and \(ax+by+\) \(c_{2}=0\) we get,

\(a=p, b=p, c_{1}=q\) and \(c_{2}=-r\)

As we know that, the distance between the parallel lines is given by:

\(d=\left|\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\right|\)

\(d=\left|\frac{q+r}{\sqrt{p^{2}+p^{2}}}\right|\)

\(=\frac{|q+r|}{\sqrt{2} p}\)

Distance between the parallel lines \(p(x+y)+q=0\) and \(p(x+y)-r=0\) is\(\frac{|q+r|}{\sqrt{2} p}\).

Given:

\(\mu_{1}=60.125\)

\(\mu_{2}=61.345\)

\(\mu_{3}=62.688\)

Skewness coefficient is denoted by \(\beta_{1}\)

\(\beta_{1}=\frac{(\mu_{3})^{2}}{(\mu_{2})^{3}}\)

\(=\frac{(62.688)^{2}}{(61.345)^{3}}\)

\(\Rightarrow \beta_{1}=0.0170\)

Absolute skewness measurement \(=\gamma_{1}\)

\(\Rightarrow \gamma_{1}=\sqrt{( \beta_{1})}\)

\(=\sqrt{ 0.0170}\)

\(=0.130\)

\(\therefore\) Absolute skewness measurement is\(0.130\).

Shortest distance of a point \(\left(x_1, y_1\right)\) from line \(a x+b y=c\) is \(d=\left|\frac{a x_1+b y_1-c}{\sqrt{a^2+b^2}}\right|\)

Now shortest distance of \(P(1,2)\) from \(3 x+4 y=9\) is \(P C=d=\left|\frac{3(1)+4(2)-9}{\sqrt{3^2+4^2}}\right|=\frac{2}{5}\) Given that \(\triangle \mathrm{APB}\) is an equilateral triangle Let ' \(\mathrm{a}\) ' be its side then \(\mathrm{PB}=\mathrm{a}, \mathrm{CB}=\frac{\mathrm{a}}{2}\) Now, In \(\triangle \mathrm{PCB},(\mathrm{PB})^2=(\mathrm{PC})^2+(\mathrm{CB})^2\)

(By Pythagoras theoresm)

\(\begin{aligned}& a^2=\left(\frac{2}{5}\right)^2+\frac{a^2}{4} \\& a^2-\frac{a^4}{4}=\frac{4}{25} \Rightarrow \frac{3 a^2}{4}=\frac{4}{25} \\& a^2=\frac{16}{75} \Rightarrow a=\sqrt{\frac{16}{75}}=\frac{4}{5 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{4 \sqrt{3}}{15}\end{aligned}\)

\(\therefore\) Length of Equilateral triangle (a) \(=\frac{4 \sqrt{3}}{15}\)

Given integral is,

\(I=\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{d x}{\left(1+x^{2}\right)\left(1+x^{3}\right)}\)

Let \(\tan ^{-1} x=\theta\)

\(\Rightarrow d x=\sec ^{2} \theta d \theta\)

\(\therefore I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d \theta}{1+\tan ^{3} \theta}\)

\(=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos ^{3} \theta}{\sin ^{3} \theta+\cos ^{3} \theta} d \theta\)

Applying \((a+b-x)\) property and adding, we get ,

\(2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos ^{3} \theta+\sin ^{3} \theta}{\sin ^{3} \theta+\cos ^{3} \theta} d \theta\)

\(2 I=[\theta]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\)

\(\Rightarrow 2 I=\frac{\pi}{6}\)

\(\Rightarrow I=\frac{\pi}{12}\)

Let \(y=\log _{10} x\) and \(z=\log _{x} 10\)

Now, \(y z=\left(\log _{10} x\right) \times\left(\log _{x} 10\right)\)

\(\Rightarrow y z=1\)

Differentiating with respect to \(z\), we get

\(y\left(\frac{d z}{d z}\right)+z\left(\frac{d y}{d z}\right)=0\)

\(y+z\left(\frac{d y}{d z}\right)=0\)

\(\frac{d y}{d z}=-\frac{y}{z}\)

\(=-\frac{\log _{10} x}{\log _{x} 10}\)

\(=-\frac{\log _{10} \mathrm{x}}{\frac{1}{\log _{10} x}}\)

\(=-\left(\log _{10} x\right)^{2}\)

\(\mathrm{S}(\mathrm{K})=1+3+5+\ldots+(2 \mathrm{~K}-1)=3+\mathrm{K}^2\)

\(\mathrm{S}(1): 1=3+1\), which is not true

\(\because \mathrm{S}(1)\) is not true.

\(\therefore\) P.M.I cannot be applied

Let \(\mathrm{S}(\mathrm{K})\) is true, i.e. \(\mathrm{S}(\mathrm{K}): 1+3+5 \ldots .+(2 \mathrm{~K}-1)=3+\mathrm{K}^2\)

Adding \(2 \mathrm{~K}+1\) on both sides

\(\begin{aligned}& \Rightarrow 1+3+5 \ldots+(2 \mathrm{~K}-1)+2 \mathrm{~K}+1 \\& =3+\mathrm{K}^2+2 \mathrm{~K}+1=3+(\mathrm{K}+1)^2=\mathrm{S}(\mathrm{K}+1) \\& \therefore \mathrm{S}(\mathrm{K}) \Rightarrow \mathrm{S}(\mathrm{K}+1)\end{aligned}\)