GATE CE 2016 SET-2

Median speed is the speed at the middle value in series of spot speeds that are arranged in ascending order. 50% of speed values will be greater than the median 50% will be less than the median.
Ascending order order of spot speed studies are 32, 39, 45, 51, 53, 56, 60, 62, 66, 79
Median speed=\frac{53+56}{2}=54.5 km/hr

\begin{aligned} {f}'&=0 \\ \Rightarrow \;\; 2x-4&=0 \\ \Rightarrow \;\; x&=2 \text{ (stationary point)}\\ {f}''\left ( x \right )&=2 \gt 0 \\ \Rightarrow\;\; f(x)& \text{ is minimum at } x=2\end{aligned}
i.e., \left ( 2 \right )^{2}-4\left ( 2 \right )+2=-2
\therefore The optimum value of f(x) is -2 (minimum).

The function is f(x)=0
-p\lt x\leq 0
=p-x,\, 0 \lt x \lt \pi
And Fourier series is,
f\left ( x \right )=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{\cos x}{1^{2}}+\frac{\cos 3x}{3^{2}}+\frac{\cos 5x}{5^{2}}+... \right ]+\left [ \frac{\sin x}{1}+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+... \right ] ...\left ( i \right )
At x=0, (a point of discontinuity), the fourier series converges to \frac{1}{2}\left [ f\left ( 0^{-1} \right )+f\left ( 0^{+} \right ) \right ]
where f\left ( 0^{-} \right )=\lim_{x\rightarrow 0}\left ( \pi -x \right )=\pi
Hence, eq. (i), we get,
\frac{\pi }{2}=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{1}{1^{2}}+\frac{1}{3^{2}}+... \right ]
\Rightarrow \;\; \frac{1}{1}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+...\frac{\pi ^{2}}{8}

\; \; \; \; P\left ( X\: \cup \: Y^{c} \right )=0.7
\Rightarrow \; \; P\left ( X \right )+P\left ( Y^{c} \right )-P\left ( X \right )P\left ( Y^{c} \right )=0.7
(Since X, Y are independent events)
\Rightarrow \; \; P\left ( X \right )+1-P\left ( Y \right )-P\left ( X \right )\left \{ 1-P\left ( Y \right ) \right \}=0
\Rightarrow \; \; P\left ( X \right )-P\left ( X\: \cap \: Y \right )=0.3\; \; \; \; \; \; ...\left ( i \right )
\; \; \; \; P\left ( X\: \cup \: Y \right )=P\left ( X \right )+P\left ( Y \right )-P\left ( X\: \cap \: Y \right )
\; \; \; \; =0.4+0.3=0.7

(i) \lim_{x\rightarrow \infty }\frac{xy}{x^{2}+y^{2}}\lim_{y\rightarrow \infty }\left ( \frac{0}{0^{2}+y^{2}} \right )=0
(i.e., put x=0 and then y=0)
(ii) \lim_{x\rightarrow 0 y\rightarrow 0}\frac{xy}{x^{2}+y^{2}}\lim_{x\rightarrow 0}\left ( \frac{0}{x^{2}+0} \right )=0
( i.e., put y=0 and then x=0)
(iii)\lim_{x\rightarrow 0 y\rightarrow 0}\frac{xy}{x^{2}+y^{2}}\lim_{x\rightarrow 0}\frac{x\left ( mx \right )}{x^{2}+m^{2}x^{2}}
(i.e., put y=mx)
\lim_{x\rightarrow \infty }\left ( \frac{m}{1+m^{2}} \right )=\frac{m}{1+m^{2}}
which depends on m.

Kinematic indeterminacy,
\begin{aligned} D_{k}&=2 j-r_{e} \\ &=2 \times 7-3=11 \end{aligned}

\begin{aligned} \sum M_{A}&=0\\ \Rightarrow \; R_{D}\times 2 L-P\times L&=0\\ \Rightarrow R_{D}&=\frac{P}{2} \end{aligned}
At joint D:


\begin{aligned} \sum F_{y}&=0\\ \Rightarrow \; T \cos 45^{\circ}&=\frac{P}{2}\\ \therefore \; T&=\frac{P}{\sqrt{2}} \end{aligned}

As per IS 1077 : 1992 clause 4.1, minimum strength of burnt clay bricks is 3.5 Mpa

Total float can be determined once the activity times i.e. EST, EFT, LST and LFT are known.
total float,\begin{aligned} F_{T}&=LST-EST\\ &=LFT-EFT \end{aligned}

For activity 5-7,
\begin{aligned} EST&=38\\ EFT&=63\\ LFT&=63\\ LST&=38\\ F_{T}&=0 \end{aligned}
for activity 11-12,
\begin{aligned} EST&=80\\ EFT&=81\\ LFT&=81 \\ LST&=80 \\ F_T&=0 \end{aligned}
\textbf{Note}:It can be seen directly that since the slack of all events are zero, there is not margin left for the occurence of events and therefore.
Maximum available line=Time required for completion of activity
\therefore \; F_{t} for all activities is zero.