Mathematics Test - 70

Let the matrix \(B\) has an order \(p \times q\), i.e, \(p\) rows and \(q\) columns.

Transpose of \(B\) is \(B'\) will have order \(q \times p\).

For matrix, \(A B^{T}=\left[A_{(2 \times 3)} \times B_{(q \times p)}^{T }\right]\) is defined.

\(\therefore q=3\)

Given,

\(AB ^{T}\) is a square matrix.

\(AB ^{T}=\left[ A _{(2 \times 3)} \times B _{(3 \times p )}^{T}\right]\)

\(\therefore p=2\)

So, the order of \(B=p \times q=2 \times 3\)

Given, 

\(\frac{d y}{d x}=e^{x-y}\)

\(\Rightarrow \frac{d y}{d x}=\frac{e^{x}}{e^{y}}\)

\(\Rightarrow e^{y} d y=e^{x} d x\)

Now,

Integrating both sides, we get

\(\int {e}^{{y}} {dy}=\int {e}^{{x}} {dx}\)

\(\Rightarrow {e}^{{y}}={e}^{{x}}+{c}\)

We know that:

Mean \(=\sum \frac{x_{i}}{N}\)

\(G M=N \sqrt{x}_{1} \times x_{2} \times \ldots \ldots x_{N}\)

\(H M=\frac{N}{\Sigma}(\frac{1}{x_{i}})\)

Where, N is the number of observations

Let, us take an observation \(4,6,2\)

Arithmetic mean \(=\frac{(4+6+2)}{3}\)

\(\therefore\) Mean \(=\frac{12}{3}=4\)

Harmonic mean \(=\) It is reciprocal of AM

\(\therefore HM =\frac{1}{4}\)

\(GM =\sqrt{ (AM \times HM )}\)

\({ GM =\sqrt{\left(4 \times \frac{1}{4}\right)}}\)

\({\therefore GM =1}\)

From this result we can say that \(AM > GM > HM\)

When we take some other observation which follows \(AM = GM = HM\)

\(\therefore\) The relation between \(AM , GM\), and \(HM\) is \(AM \geq GM \geq HM\)

Given:

Probability of getting \(3=\frac{1}{6}\)

Probability of not getting \(3=1-\frac{1}{6}=\frac{5}{6}\)

The random variable \(X\) does represent number of throws required for getting \(3\).

So,

\(X=1, P(1)=\frac{1}{6}\)

\(X=2, P(2)=\frac{5}{6} \times \frac{1}{6}\)

\(X=3, P(3)=\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\)

\(X=4, P(4)=\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\)

\(E[X]=\sum X P(X)\)

\(=(1)\left(\frac{1}{6}\right)+(2)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)+3\left(\frac{5}{6}\right)^{2}\left(\frac{1}{6}\right)+4\left(\frac{5}{6}\right)^{3}\left(\frac{1}{6}\right) \ldots\)

\(=\frac{1}{6}\left[1+(2)\left(\frac{5}{6}\right)+3\left(\frac{5}{6}\right)^{2}+4\left(\frac{5}{6}\right)^{3} \ldots\right]\)

Let,

\(S-1+2\left(\frac{5}{6}\right)+3\left(\frac{5}{6}\right)^{2}+4\left(\frac{5}{6}\right)^{3}\quad \quad ...(i)\)

\(\left(\frac{5}{6}\right) S=\left(\frac{5}{6}\right)+2\left(\frac{5}{6}\right)^{2}+3\left(\frac{5}{6}\right)^{3}+4\left(\frac{5}{6}\right)^{4}\quad \quad ...(ii)\)

Equation \((i)\) \(-\) \((ii)\),

\(S-\frac{5}{6} S=1+\left(\frac{5}{6}\right)(2-1)+\left(\frac{5}{6}\right)^{2}+(3-2)+\left(\frac{5}{6}\right)^{3}(4-3) \ldots\)

\(\frac{1}{6} S=1+\left(\frac{5}{6}\right)+\left(\frac{5}{6}\right)^{2}+\left(\frac{5}{6}\right)^{3} \ldots\)

\(\frac{1}{6} S=\frac{1}{1-\frac{5}{6}}=6\)

\(S=36\)

\(\therefore\) The expected value of \(X\):

\(E[X]=\frac{36}{6}=6\)

Given: \(f(x)=\log (1+x)-\frac{x}{(1+x)}\)

Here we have to find the interval in which \(f(x)\) is decreasing.

Let's first calculate \(f^{\prime}(x)\)

As we know that \(\frac{d}{d x}\{f(x) \pm g(x)\}=\frac{d\{f(x)\}}{d x} \pm \frac{d\{g(x)\}}{d x}, \frac{d(\log x)}{d x}=\frac{1}{x}\), for \(x>0\) and \(\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x) \cdot \frac{d(f(x))}{d x}-f(x) \cdot \frac{d(g(x))}{d x}}{[g(x)]^2}\)

\(\Rightarrow f^{\prime}(x)=\left\{\frac{1}{1+x}-\frac{(1+x) \cdot 1-x \cdot 1}{(1+x)^2}\right\}=\frac{x}{(1+x)^2}\)

As we know that for a decreasing function say \(f(x)\) we have \(f^{\prime}(x) \leq 0\)

\( \Rightarrow \frac{x}{(1+x)^2} \leq 0 \)

\( \Rightarrow x \leq 0--\left(\because(1+ x )^2 \geq 0\right)\)

Hence, the given function is decreasing for the interval \((-\infty, 0]\)

Given:

\(\mathrm{P}(\mathrm{n})=\mathrm{n}^{2}-\mathrm{n}+41\)

For n = 2.

\(\mathrm{P}(2)=2^{2}-2+41=43\) (prime true)

For n = 3.

\(\mathrm{P}(3)=3^{2}-3+41=47\) (prime true)

For n = 41.

\(\mathrm{P}(41)=41^{2}-41+41=(41)^{2}\)

\(\therefore \mathrm{P}(41)\) is not true.

If \(f^{\prime}(x) \geq 0\) at each point in an interval, then the function is said to be increasing.

Given, \(f(x)=2 x^{3}-15 x^{2}+36 x+12\)

Differentiating, we get:

\(f^{\prime}(x)=6 x^{2}-30 x+36\)

\(f' ( x )\) is increasing function:

\(f^{\prime}(x) \geq 0\)

\(=6 x^{2}-30 x+36 \geq 0\)

\(=x^{2}-5 x+6 \geq 0\)

\(=(x-2)(x-3) \geq 0\)

Thus, \(x \in(-\infty, 2] \cup[3, \infty)\)

The interval in which the function \(f (x)=2 x^{3}-15 x^{2}+36 x+10\) is increasing in \((-\infty, 2] \cup\) \([3, \infty)\)

Since \(a, b, c\) are in G.P.

\(\therefore b^2=\mathrm{acc}\)

Given equation is, \(\mathrm{ax}^2+2 \mathrm{bx}+\mathrm{c}=0\)

\(\Rightarrow x^2+2 \sqrt{a c} x+c=0 \Rightarrow(\sqrt{a x}+\sqrt{c})^2=0 \\\)

\(\Rightarrow x=-\sqrt{\frac{c}{a}}\)

Also, given that \(\mathrm{ax}^2+2 \mathrm{bx}+\mathrm{c}=0\) and \(\mathrm{dx}^2+2 \mathrm{ex}+\mathrm{f}=0\) have a common root.

\(\Rightarrow x=-\sqrt{\frac{c}{a}} \text { must satisfy } d x^2+2 e x+f=0 \\\)

\( \Rightarrow d \cdot \frac{c}{a}+2 e\left(-\sqrt{\frac{c}{a}}\right)+f=0 \\\)

\(\frac{d}{a}-\frac{2 e}{\sqrt{a c}}+\frac{f}{c}=0 \Rightarrow \frac{d}{a}-\frac{2 e}{b}+\frac{f}{c}=0 \\\)

\(\Rightarrow \frac{2 e}{b}=\frac{d}{a}+\frac{f}{c} \Rightarrow \frac{d}{a},\frac{e}{b},\frac{f}{c} \text { are in A.P. }\)

Given,

the objective function is,

\(z=3 x+9 y\)

At \((5,5)\),

\(z=3 \times 5+9 \times 5=60\)

At \((0,20)\),

\(z=3 \times 0+9 \times 20=180\)

At \((15,15)\),

\(z=3 \times 15+9 \times 15=180\)

At \((0,10)\),

\(z=3 \times 0+9 \times 10=90\)

The minimum value of \({z}\) occurs at\((5,5)\).