Mathematics Test - 46

We have,

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

Applying \(R_{1} \rightarrow c R_{1}\), we have:

\(\Delta=\frac{1}{c}\left|\begin{array}{ccc} 0 & a c & -b c \\ -a & 0 & -c \\ b & c & 0\end{array}\right|\)

Applying \(R_{1} \rightarrow R_{1}-b R_{2}\), we have,

\(\Delta=\frac{1}{c}\left|\begin{array}{ccc}a b & a c & 0 \\ -a & 0 & -c \\ b & c & 0\end{array}\right|\)

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

Here, the two rows \(R_{1}\) and \(R_{3}\) are identical.

\(\therefore \Delta=0\)

\(\frac{d y}{d x}=\sec \left(\frac{y}{x}\right)+\frac{y}{x}\)

Let \(\frac{ y }{ x }= t\)

\(\Rightarrow y = xt\)

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

\(\Rightarrow \frac{ dy }{ dx }= x \frac{ dt }{ dx }+ t\)

Now,

\( x \frac{ dt }{ dx }+ t =\sec t + t\)

\( \Rightarrow x \frac{ dt }{ dx }=\sec t \)

\( \Rightarrow \frac{ dt }{\sec t }=\frac{ dx }{ x }\)

Integrating both sides, we get

\( \Rightarrow \int \frac{d t}{\sec t}=\int \frac{d x}{x} \)

\(\Rightarrow \int \cos t d t=\int \frac{d x}{x} \)

\(\Rightarrow \sin t=\log x+\log c \)

\( \Rightarrow \sin t=\log (c x) \quad(\because \log m+\log n=\log (m n))\)

\(\therefore \sin \left(\frac{y}{x}\right)=\log (c x)\)

Given:

Arrival rate \((\lambda)=\frac{5}{8} \frac{\text { Jobs }}{\text { Hour }}\)

Service time for one job is \(40\) min

Therefore, Service rate \((\mu)=\frac{3}{2} \frac{\text { Jobs }}{\text { Hour }}\)

Utilization factor \(\rho=\frac{\lambda}{\mu}\)

Idle time \(=1-\rho\)

\(\lambda=\frac{5}{8} \frac{\text { Jobs }}{\text { Hour }}\)

\(\mu=\frac{3}{2} \frac{\text { Jobs }}{\text { Hour }}\)

\(\rho=\frac{\lambda}{\mu}=\frac{\frac{5}{8}}{\frac{3}{2}}=\frac{5}{12}\)

Idle time \(=1-\rho\)

\(=1-\frac{5}{12}\)

\(=\frac{7}{12}\) hour

Therefore, Idle time for \(8\) hour shift\(=\frac{7}{12} \times 8\)

\(=\frac{14}{3}\)hours.

Mean Deviation from Mean \(=\frac{\sum|x-\bar{x}|}{n}\)

Mean Deviation from Median \(=\frac{\sum|x-M|}{n}\)

Here,

\(\sum\) represents the summation

\(x=\) Observations given

\(\bar{x}=\) Mean

\(n =\) The number of observations

\(M =\) Median

Mean \(=\frac{(1+0+2+3+1+1+15+1+3) }{9}=\frac{27}{9}=3\)

Mean Deviation (M.D) about mean will be = Subtract Mean from each given data set values

Example, \(0-3=3\) (don't consider the sign here, only focus on the values)

\(1-3=2,2-3=1,3-3=0,15-3=12\)

M.D (Mean) \(=\frac{(3+2+2+2+2+1+0+12)}{9}=\frac{24}{9}=\frac{8}{3}\)

As we have calculated above the value of Mean Deviation(M.D) \(=3,2,2,2,2,1,12\)

The value of Variance \(=\frac{\left(3^{2}+2^{2}+2^{2}+2^{2}+2^{2}+1^{2}+12^{2}\right)}{9}=\frac{\sqrt{170}}{\sqrt{9}}={\frac{\sqrt{170}}{3}}\)

\(\therefore\) Standard Deviation (S.D) \(=\sqrt{\left(\frac{170}{9}\right)^{2}}=\frac{170}{9}\)

The slope of the tangent to the curve:

The slope of curve \(y = f ( x )\) at some point \(P \left( x _{1}, y _{1}\right)\) means the slope of the tangent to the curve at \(P \left( x _{1}, y _{1}\right)\).

The slope is given as \(m\), where \(m =\frac{ dy }{ dx }\) at \(\left( x _{1}, y _{1}\right)\).

Given: \(2 x^{3}+3 y=2 y^{3}+3 x\)

Differentiating with respect to \(x\)

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

\(6 x^{2}+3 \frac{d y}{d x}=6 y^{2} \frac{d y}{d x}+3\)

Taking \(\frac{ dy }{ dx }\) on one side, we get:

\(6 x^{2}-3=\left(6 y^{2}-3\right) \frac{d y}{d x}\)

\(\frac{d y}{d x}=\frac{6 x^{2}-3}{6 y^{2}-3}\)

We have to find the sum of the series 3 + 9 + 27 + 81 + ..... + 6561

Here,

a = 3

r = 3

Let, a_n = 6561

As we know that, the the general term of a GP is given by:

\(a_{n}=a r^{n-1}\)

\(\Rightarrow 6561=(3) \cdot(3)^{n-1}\)

\(\Rightarrow 3^{n}=3^{8}\)

\(\therefore \mathrm{n}=8\)

As we know that,

Sum of \(n\) terms of GP,

\(\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\);

where \(r>1\)

\(\therefore \mathrm{S}_{\mathrm{n}}=\frac{3\left(3^{8}-1\right)}{3-1}\)

\(=\frac{3\left(3^{8}-1\right)}{2}\)

Given lines are \(\frac{ x -2}{2 k }=\frac{ y -3}{3}=\frac{ z +2}{-1}\) and \(\frac{ x -2}{8}=\frac{ y -3}{6}=\frac{ z +2}{-2}\)

The direction ratio of the first line is \((2 k, 3,-1)\) and the direction ratio of second line is \((8,6,-2)\)

Lines are parallel;

So, \(\frac{2 k }{8}=\frac{3}{6}=\frac{-1}{-2}\)

\(\frac{ k }{4}=\frac{1}{2}=\frac{1}{2}\)

\(\therefore k =2\)

z1+z2=5
z13+z23=20+15i
z13+z23=(z1+z2)3−3z1z2(z1+z2)
z13+z23=125−3z1⋅z2(5)
⇒20+15i=125−15z1z2
⇒3z1z2=25−4−3i
⇒3z1z2=21−3i
⇒z1⋅z2=7−i
⇒(z1+z2)2=25
⇒z12+z22=25−2(7−i)
⇒11+2i
(z12+z22)2=121−4+44i
⇒z14+z24+2(7−i)2=117+44i
⇒z14+z24=117+44i −2(49−1−14i)
⇒|z14+z24|=75

Given:

\(\mathrm{P}(\mathrm{n}): \frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7 n}{15}\) is a natural number, for all \(n \in N\).

\(P(1): \) \(\frac{1^{5}}{5}+\frac{1^{3}}{3}+\frac{7(1)}{15}=\frac{3+5+7}{15}=\frac{15}{15}=1\), which is a natural number.

So, \(\mathrm{P}(1)\) is true.

Let \(P(n)\) be true for some \(n=k\).

Then \(\frac{k^{5}}{5}+\frac{k^{3}}{3}+\frac{7 k}{15}\) is natural number,

Now, \(\frac{(k+1)^{5}}{5}+\frac{(k+1)^{3}}{3}+\frac{7(k+1)}{15}\)

Let \(P(n)\) be true for some \(n=k+1\).

\(\frac{k^{5}+5 k^{4}+10 k^{3}+10 k^{2}+5 k+1}{5}+\frac{k^{3}+1+3 k^{2}+3 k}{3}+\frac{7 k+7}{15} \)

\(=\frac{k^{5}}{5}+\frac{k^{3}}{3}+\frac{7 k}{15}+k^{4}+2 k^{3}+3 k^{2}+2 k+1 \)

\(=P(k)+k^{4}+2 k^{3}+2 k^{2}+2 k+1 \)

= Natural number

Thus, \(P(k+1)\) is true whenever \(P(n)\) is true.

So, by the principle of mathematical induction, \(\mathrm{P}(\mathrm{n})\) is true for any natural number \(\mathrm{n}\).

Given,

\(\underset{{{x \rightarrow 0}}}{\lim} \frac{x \tan x}{1-\cos x}\)

We know that:

\(\underset{{{x \rightarrow a}}}{\lim}\left[\frac{f(x)}{g(x)}\right]=\frac{\underset{{{x \rightarrow a}}}{\lim} f(x)}{\underset{{{x \rightarrow a}}}{\lim} g(x)}\), provided \(\underset{{{x \rightarrow a}}}{\lim} g(x) \neq 0\)

\(\cos 2 x=1-2 \sin ^{2} x\)

\(\underset{{{x \rightarrow 0} }}{\lim}\frac{\tan x}{x}=1\)

\(\underset{{{x \rightarrow 0} }}{\lim} \frac{\sin x}{x}=1\)

\(\Rightarrow \underset{{{x \rightarrow 0}}}{\lim} \frac{x \tan x}{2 \sin ^{2} \frac{x}{2}}\)

\(=\frac{1}{2} \underset{{{x \rightarrow 0}}}{\lim} \left[ \frac{\tan x}{x} \times \frac{x \cdot x}{\frac{\sin ^{2} \frac{x}{2}}{\left(\frac{x}{2}\right)^{2}} \times\left(\frac{x}{2}\right)^{2}}\right]\)

\(=\frac{1}{2} \underset{{{x \rightarrow 0}}}{\lim}\left[\frac{\tan x}{x} \times \frac{4}{\frac{\sin ^{2} \frac{x}{2}}{\left(\frac{x}{2}\right)^{2}}}\right]\)

\(=\frac{1}{2} \left(\underset{{{x \rightarrow 0}}}{\lim} \left[\frac{\tan x}{x}\right]\right) \times \frac{4}{ \left(\underset{{{x \rightarrow 0}}}{\lim}\left[\frac{\sin ^{2} \frac{x}{2}}{\left(\frac{x}{2} \right)^{2}}\right]\right)} \)

\(=\frac{1}{2} \times 1 \times 4\)

\(=2\)