Engineering Mathematics

\begin{aligned} I&=\int_{0}^{4}\int_{-x}^{x}(3x^2+3y^2)dydx\\ &=\int_{0}^{4}\left [ 3x^y+\frac{3y^3}{3} \right ]_{-x}^xdx\\ &=\int_{0}^{4}[3x^2(2x)+2x^3]dx\\ &=\int_{0}^{4}8x^3 dx\\ &=2 \times 4^4\\ &=512 \end{aligned}

\begin{aligned} \Sigma u_n&=\Sigma \frac{n}{2^n}\\ &\text{Ratio test; }\\ \lim_{n \to \infty }\frac{u_{n+1}}{u_n}&=\lim_{n \to \infty }\frac{n+1}{2^{n+1}} \times \frac{2^n}{n}=\frac{1}{2}\\ \frac{1}{2}& \lt 2 \end{aligned}
\therefore By ratio test, \Sigma u_n is convergent.
(A) c=1, d=-1
\Sigma u_n=\Sigma \frac{1}{n} is divergent by P-test
(B) c=0.5, d=-10
\begin{aligned} \Sigma u_n&=\Sigma \frac{n^{-10}}{(0.5)^n}\\ &\text{Ratio test; }\\ \lim_{n \to \infty }\frac{u_{n+1}}{u_n}&=\lim_{n \to \infty }\frac{(n+1)^{-10}}{(0.5)^{n+1}} \times \frac{(0.5)^n}{n^{10}}\\ &=\frac{1}{0.5}=2\\ 2& \gt 1 \end{aligned}
\therefore By ratio test, \Sigma u_n is divergent.
(D) c=1,d=-2 \Sigma u_n=\Sigma \frac{n^{-2}}{(1)^n}=\Sigma \frac{1}{n^2}
\Sigma u_n is convergent by P-test.

After multiply by v_1 on both sides,
\begin{aligned} Av_1&=\alpha v_1v_1^Tv_1+ \beta v_2v_2^Tv_1 Av_1&=\alpha v_1 \end{aligned}
\alpha is an eigen value of A.

After multiply by v_2 on both sides,
\begin{aligned} Av_2&=\alpha v_1v_1^Tv_2+ \beta v_2v_2^Tv_2 Av_2&=\beta v_2 \end{aligned}
\beta is an eigen value of A.
v_1v_1^T and v_2v_2^T both are singular matrieces.
Therefore, A is also a singular matrix.
|A|=0\Rightarrow \lambda _A=0
Hence \lambda _A= 0,\alpha ,\beta

f(x)=8 \log _e x-x^2+3 when x \in [1,e]
Differentiating both side,
\begin{aligned} f(x)&=\frac{8}{x}-2x=0 \; where\; x \gt 0\\ f'(x)&=0\\ \frac{8}{x}-2x&=0\\ 8-2x^2&=0\\ x^2&=4\\ x&=\pm 2 \end{aligned}
\begin{aligned} f''(x)&=\frac{-8}{x^2}-2 \\ f''(2)&=-6 \lt 0\\ \end{aligned}
f(x) is maximum for x = 2
Minimum of f(x) will be in [1, e] = min [f(1), f(e)]
f(e)=8 \ln e -e^2+3=3.61
Hence, minimum value of f(x) occurs at x=1

\begin{aligned} Roots\;\; (z-1)(z+1)&=0\\ z&=\pm 1\\ \because \; z&=-1 \text{ is in contour} \end{aligned}
\begin{aligned} \Rightarrow \oint _c \frac{2^z}{z^2-1}&=\left [ \lim_{z \to -1} \frac{(z+1)2^z}{(z+1)(z-1)} \right ] \times 2 \pi i\\ &=\frac{2^{-1}}{(-1-1)} \times 2 \pi i\\ &=-\frac{1}{2} \pi i\\ \Rightarrow \;\; A&=1/2=0.5 \end{aligned}

Sum of frequency of No. of wickets = 5 +7 + 8 + 25 + 20 +17 + 8 + 4 + 3 + 2 + 1 = 100
Mean = Average of the 50th & 51th matches = 4

Given
\begin{aligned} f(x,y)&=e^{(\xi x+\eta y)}\\ \frac{\partial f(x,y)}{\partial x}&=\xi \cdot e^{(\xi x+\eta y)}\\ \frac{\partial^2 f(x,y)}{\partial x^2}&=\xi ^2\cdot e^{(\xi x+\eta y)}\\ \end{aligned}
Similarly,
\frac{\partial^2 f(x,y)}{\partial y^2}=\eta ^2\cdot e^{(\xi x+\eta y)}
Now, as given
a \frac{\partial^2 f(x,y)}{\partial x^2}+b\frac{\partial^2 f(x,y)}{\partial y^2}=f(x,y)
or
a \times \xi ^2\cdot e^{(\xi x+\eta y)}+b \times \eta ^2\cdot e^{(\xi x+\eta y)}= e^{(\xi x+\eta y)}
a\cdot \xi ^2+b\cdot \eta ^2=1
Thus, \xi =\frac{1}{\sqrt{2a}}, \eta = \frac{1}{\sqrt{2b}}
(\xi =\frac{1}{\sqrt{a}}, \eta = 0) and (\xi =0, \eta = \frac{1}{\sqrt{b}}) satisfy the above result.

Here equation will be
x-\sqrt{2}y+3z=1
-x+\sqrt{2}y-3z=3
therefore inconsistant solution i.e. there will not be any solution.

\oint \vec{F} (x,y)\cdot [dx\vec{i}+dy\vec{j}]
Given \vec{F} (x,y)=x\vec{i}+y\vec{j}
\therefore \int_{c}xdx+ydy=0
Because here vector is conservative.
If the integral function is the total derivative over the closed contoure then it will be zero

\begin{aligned} \text{Given,}\qquad P(G)&=\frac{1}{2} \text { and } P(B)=\frac{1}{2} \\ P(C)&=\frac{1}{4} \text { and } P(S)=\frac{3}{4} \end{aligned}
Let probability of selected science student is a boy is P(B/S) = x
Given that Commerce students are two time more likely to be a boy than are Science students.
Then, probability of selected Commerce student is a boy
P\left(\frac{B}{C}\right)=2 x
We have to find probability of randomly selected girl studies Commerce i.e.
\begin{aligned} P\left(\frac{C}{G}\right)&=\frac{P(C \cap G)}{P(G)}=\frac{P(C) P(G / C)}{P(G)} \\ P\left(\frac{C}{G}\right)&=\frac{\frac{1}{4} \times P\left(\frac{G}{C}\right)}{\frac{1}{2}} &\ldots(i) \end{aligned}
To find P(G / C), first we have to find P(B/C).
The probability of selected student is a boy
\begin{aligned} P(B) &=P(S) P(B / S)+P(C) \times P(B / C) \\ \frac{1}{2} &=\frac{3}{4} \times x+\frac{1}{4} \times 2 x \\ \Rightarrow\qquad\qquad x &=\frac{2}{5}\\ \text{Then},\qquad \qquad P(B / C)&=2 x=\frac{4}{5}\\ \text{We know that},\qquad \\ P\left(\frac{B}{C}\right)+P\left(\frac{G}{C}\right)&=1 \\ P\left(\frac{G}{C}\right) =1- \frac{4}{5}&=\frac{1}{5} \\ \text{From equation (i),} \\ P\left(\frac{G}{C}\right) = \frac{1}{4} \times \frac{1}{\frac{5}{1/2}} &= \frac{1}{10} \end{aligned}
The amount of information gained in knowing that a randomly selected girl student studies Commerce
I\left ( \frac{C}{G} \right )=\log _2 \frac{1}{P\left ( \frac{C}{G} \right )}= \log _2 10=3.32