Question 1 |
The spot speeds (expressed in km/hr) observed at a road section are 66, 62, 45, 79, 32, 51, 56, 60, 53, and 49. The median speed (expressed in km/hr) is ________.
(Note: answer with one decimal accuracy)
(Note: answer with one decimal accuracy)
54.5 | |
51.5 | |
53.5 | |
56 |
Question 1 Explanation:
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
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
Question 2 |
The optimum value of the function f(x)=x^{2}-4x+2 is
2 (maximum) | |
2 (minimum) | |
-2 (maximum) | |
-2 (minimum) |
Question 2 Explanation:
\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).
i.e., \left ( 2 \right )^{2}-4\left ( 2 \right )+2=-2
\therefore The optimum value of f(x) is -2 (minimum).
Question 3 |
The Fourier series of the function,
\begin{matrix} f(x) & =0 & -\pi \lt x \leq 0 \\ f(x) &=\pi-x & 0 \lt x \lt \pi \end{matrix}
in the interval [-\pi ,\pi ] is
f(x)=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{\cos x}{1^{2}}+\frac{\cos 3x}{3^{2}}+\cdots\: \cdots \ \cdots \right ] + \left [ \frac{\sin x}{1}+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+\cdots \: \cdots\: \cdot \right ]
The convergence of the above Fourier series at x = 0 gives
\begin{matrix} f(x) & =0 & -\pi \lt x \leq 0 \\ f(x) &=\pi-x & 0 \lt x \lt \pi \end{matrix}
in the interval [-\pi ,\pi ] is
f(x)=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{\cos x}{1^{2}}+\frac{\cos 3x}{3^{2}}+\cdots\: \cdots \ \cdots \right ] + \left [ \frac{\sin x}{1}+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+\cdots \: \cdots\: \cdot \right ]
The convergence of the above Fourier series at x = 0 gives
\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6} | |
\sum_{n=1}^{\infty }\frac{(-1)^{n+1}}{n^{2}}=\frac{\pi ^{2}}{12} | |
\sum_{n=1}^{\infty }\frac{1}{(2n-1)^{2}}=\frac{\pi ^{2}}{8} | |
\sum_{n=1}^{\infty }\frac{(-1)^{n+1}}{(2n-1)}=\frac{\pi}{4} |
Question 3 Explanation:
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\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}
Question 4 |
X and Y are two random independent events. It is known that P(X)=0.40 and
P(X\cup Y^{C})=0.7. Which one of the following is the value of P(X\cup Y) ?
0.7 | |
0.5 | |
0.4 | |
0.3 |
Question 4 Explanation:
\; \; \; \; 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
\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
Question 5 |
What is the value of \lim_{\begin{matrix} x\rightarrow 0\\ y\rightarrow 0 \end{matrix}} \frac{xy}{x^{2}+y^{2}} ?
1 | |
-1 | |
0 | |
Limit does not exist |
Question 5 Explanation:
(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.
(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.
There are 5 questions to complete.