Applied Mathematics
1. The smallest positive integer (mod 11) to which 282 is congruent, is:
Solution \(282 \div 11 = 25\) with remainder \(7\). Thus \(282 \equiv 7 \pmod{11}\).
2. A man can row \(6~km/h\) in still water. It takes him twice as long to row up as to row down the river. Then, the speed of the stream is:
Solution Let stream speed be \(y\). Speed down = \(6+y\), Speed up = \(6-y\).
Time up = 2 \(\times\) Time down \(\Rightarrow \frac{D}{6-y} = 2 \frac{D}{6+y}\).
\(6+y = 12-2y \Rightarrow 3y=6 \Rightarrow y=2\).
Time up = 2 \(\times\) Time down \(\Rightarrow \frac{D}{6-y} = 2 \frac{D}{6+y}\).
\(6+y = 12-2y \Rightarrow 3y=6 \Rightarrow y=2\).
3. If \(\frac{|x+1|}{x+1}>0\), \(x \in R\), then:
Solution For the fraction to be positive, the numerator and denominator must have the same sign. \(|x+1|\) is always positive (for \(x \neq -1\)). Thus, \(x+1 > 0 \Rightarrow x > -1\).
4. If \(P=\begin{bmatrix}1&0\\ 2&1\end{bmatrix}\), \(Q=\begin{bmatrix}x&0\\ 1&1\end{bmatrix}\) and \(P=Q^{2}\) then x equals:
Solution \(Q^2 = \begin{bmatrix}x&0\\ 1&1\end{bmatrix}\begin{bmatrix}x&0\\ 1&1\end{bmatrix} = \begin{bmatrix}x^2&0\\ x+1&1\end{bmatrix}\).
Equating to P: \(x^2=1\) and \(x+1=2\). From second eq, \(x=1\).
Equating to P: \(x^2=1\) and \(x+1=2\). From second eq, \(x=1\).
5. If A is an invertible matrix, then which of the following is NOT true?
Solution All options listed are mathematically True .
6. The system of linear equations \(2x+ky=7\) and \(3x+2y=7\) will be consistent, if:
Solution For a unique solution (consistent), \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\).
\(\frac{2}{3} \neq \frac{k}{2} \Rightarrow k \neq \frac{4}{3}\).
\(\frac{2}{3} \neq \frac{k}{2} \Rightarrow k \neq \frac{4}{3}\).
7. If \(y=x^{y}\), then \(\frac{dy}{dx}\) is:
Solution Taking log: \(\ln y = y \ln x\). Diff w.r.t x: \(\frac{1}{y}y' = y'\ln x + \frac{y}{x}\).
\(y'(\frac{1-y\ln x}{y}) = \frac{y}{x} \Rightarrow y' = \frac{y^2}{x(1-y\ln x)}\).
\(y'(\frac{1-y\ln x}{y}) = \frac{y}{x} \Rightarrow y' = \frac{y^2}{x(1-y\ln x)}\).
8. The function \(f(x)=a^{x}\) is increasing on R, if:
Solution Exponential functions \(a^x\) increase when base \(a > 1\).
9. A function \(f:R\rightarrow R\) is defined as \(f(x)=x^{3}+1\). The function f has:
Solution Cubic functions on R range from \(-\infty\) to \(+\infty\), so they have neither a global maximum nor minimum.
10. The relation between "Marginal Cost (MC)" and "Average Cost (AC)" of producing 'x' units is:
Solution Since \(AC = \frac{C(x)}{x}\), differentiating gives \(\frac{x C'(x) - C(x)}{x^2} = \frac{1}{x}(MC - AC)\).
11. For a random variable X, \(E(X)=3\) and \(E(X^{2})=11\). The variance of X is:
Solution \(Var(X) = E(X^2) - (E(X))^2 = 11 - (3)^2 = 11 - 9 = 2\).
12. If mean and SD of a binomial distribution are 12 and 2 respectively, then parameter p is:
Solution Mean \(np = 12\), SD \(\sqrt{npq} = 2 \Rightarrow npq = 4\).
\(12q = 4 \Rightarrow q = 1/3\). Thus \(p = 1 - q = 2/3\).
\(12q = 4 \Rightarrow q = 1/3\). Thus \(p = 1 - q = 2/3\).
13. If the variance of a Poisson distribution is 2, then \(P(X=2)\) is:
Solution For Poisson, Variance \(\lambda = 2\).
\(P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!} \Rightarrow P(X=2) = \frac{e^{-2}2^2}{2!} = 2e^{-2}\).
\(P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!} \Rightarrow P(X=2) = \frac{e^{-2}2^2}{2!} = 2e^{-2}\).
14. Normal distribution is symmetric about:
Solution A normal distribution curve is always symmetric about its Mean (\(\mu\)).
15. Using flat rate method, the EMI to repay a loan of ₹20,000 in \(2\frac{1}{2}\) years at 8% p.a. is:
Solution Interest \(I = 20000 \times 0.08 \times 2.5 = 4000\). Total Amt = 24000.
Months = \(2.5 \times 12 = 30\). EMI = \(24000 / 30 = 800\).
Months = \(2.5 \times 12 = 30\). EMI = \(24000 / 30 = 800\).
16. The graph of the inequality \(3x+2y>6\) is the:
Solution At origin (0,0), \(0 > 6\) is False. So origin is not included. It is a strict inequality, so the line itself is not included.
17. A straight line trend is represented by which equation?
Solution The standard equation for a linear (straight line) trend in time series is \(y = a + bx\).
18. For a t-test of significance, if a random sample of size (n) 34 is taken from a normal population, the degrees of freedom (v) is:
Solution Degrees of freedom for a single sample t-test is \(n - 1\). Here, \(34 - 1 = 33\).
19. Assertion (A): The solution set of inequality \(|3x-2|\le\frac{1}{2}, x\in R\) is \([\frac{1}{2}, \frac{5}{6}]\).
Reason (R): \(|x-a|\le r \Leftrightarrow x\le a-r\) or \(x\ge a+r\).
Reason (R): \(|x-a|\le r \Leftrightarrow x\le a-r\) or \(x\ge a+r\).
Solution Assertion is True: \(-\frac{1}{2} \le 3x-2 \le \frac{1}{2} \Rightarrow 1.5 \le 3x \le 2.5 \Rightarrow 0.5 \le x \le 5/6\).
Reason is False: The correct property is \(a-r \le x \le a+r\). The given reason describes the exterior region (\(|x-a| \ge r\)).
Reason is False: The correct property is \(a-r \le x \le a+r\). The given reason describes the exterior region (\(|x-a| \ge r\)).
20. Assertion (A): Matrix \(A=\begin{bmatrix}0&-6&7\\ 6&5&-1\\ -7&1&0\end{bmatrix}\) is a skew-symmetric matrix.
Reason (R): A matrix A is skew-symmetric if \(A^{\prime}=-A\).
Reason (R): A matrix A is skew-symmetric if \(A^{\prime}=-A\).
Solution For a skew-symmetric matrix, all diagonal elements must be 0. Here, \(A_{22}=5\), so Assertion is False. The Reason is the correct definition.