In a quadratic function, the value of the product of the roots \((\alpha, \beta)\) is \(4\). Find the value of \[\frac{\alpha^{n} + \beta^{n}}{\alpha^{-n} + \beta^{-n}}.\]
Options:
\(n^4\)
\(4^n\)
\(2^{2n-1}\)
\(4^{n-1}\)
Solution
\[\frac{\alpha^{n} + \beta^{n}} {\alpha^{-n} + \beta^{-n}} = \frac{\alpha^{n} + \beta^{n}} {\frac{1}{\alpha^{n}} + \frac{1}{\beta^{n}}}\]
\[= \frac{\alpha^{n}\beta^{n}(\alpha^{n} + \beta^{n})} {\alpha^{n} + \beta^{n}}\]
\[= (\alpha\beta)^n = 4^n\]
Answer: B
Given the following polynomial equation \[s^{3} + 5.5s^{2} + 8.5s + 3 = 0,\] the number of roots of the polynomial which have real parts strictly less than \(-1\) is .
Solution
\[s^{3} + 5.5s^{2} + 8.5s + 3 = 0\]
Let \[s = z - 1.\]
Substitute: \[(z - 1)^3 + 5.5(z - 1)^2 + 8.5(z - 1) + 3 = 0\]
Expand: \[z^3 + 2.5z^2 + 0.5z - 1 = 0.\]
Routh table:
\[\begin{array}{c|cc} z^{3} & 1 & 0.5 \\ z^{2} & 2.5 & -1 \\ z^{1} & 0.9 & 0 \\ z^{0} & -1 & \\ \end{array}\]
There is one sign change in the first column.
\[\text{Therefore, two roots lie to the left of } s = -1.\]
Answer: 2
Let \(A\) be a \(4 \times 3\) real matrix with rank \(2\). Which one of the following statements is TRUE?
Rank of \(A^{T}A\) is less than 2.
Rank of \(A^{T}A\) is equal to 2.
Rank of \(A^{T}A\) is greater than 2.
Rank of \(A^{T}A\) can be any number between 1 and 3.
Solution
Option 2 is correct: \[\operatorname{rank}(A^{T}A) = \operatorname{rank}(A) = \operatorname{rank}(A^{T}) = \operatorname{rank}(AA^{T}).\]
Since \(\operatorname{rank}(A) = 2\), \[\operatorname{rank}(A^{T}A) = 2.\]
Answer: B
Let the eigenvalues of a \(2 \times 2\) matrix \(A\) be \(1\) and \(-2\) with eigenvectors \(x_{1}\) and \(x_{2}\) respectively. Then the eigenvalues and eigenvectors of the matrix \(A^{3}\) would, respectively, be:
Options:
\(1, -8 : x_{1}, x_{2}\)
\(-1, -2 : x_{1} + x_{2},\, x_{1} - x_{2}\)
\(1, -2 : x_{1}, x_{2}\)
\(2, 0 : x_{1} + x_{2},\, x_{1} - x_{2}\)
Solution
If \(\lambda\) is an eigenvalue of \(A\) with eigenvector \(x\), then for \(A^{3}\) the corresponding eigenvalue is: \[\lambda^{3}\]
Given eigenvalues: \[1,\; -2\]
So eigenvalues of \(A^{3}\) are: \[1^{3} = 1,\qquad (-2)^{3} = -8\]
Eigenvectors remain the same: \[x_{1},\; x_{2}\]
Answer: Option 1
Candidates were asked to come to an interview with 3 pens each. Black, blue, green, and red were the permitted pen colours. The probability that a candidate comes with all 3 pens having the same colour is .
Options:
\(0.1\)
\(0.2\)
\(0.25\)
\(0.3\)
Solution
Total number of available colours: \[n = 4 \quad ( \text{Black, Blue, Green, Red} )\]
Since a candidate brings 3 pens, and repetition is allowed, the number of ways to choose colours with repetition is: \[\binom{n + r - 1}{r} = \binom{4 + 3 - 1}{3} = \binom{6}{3} = 20\]
Favourable cases: all three pens having the same colour. There are 4 such possibilities: \[\{\text{BBB, BBBlu, GGG, RRR}\}\]
Required probability: \[\frac{4}{20} = 0.2\]
Answer: B
The value of \[\int_{-\infty}^{+\infty} e^{-t}\,\delta(2t - 2)\,dt,\] where \(\delta(t)\) is the Dirac delta function, is:
Options:
\(\dfrac{1}{2e}\)
\(\dfrac{2}{e}\)
\(\dfrac{1}{e^{2}}\)
\(\dfrac{1}{2e^{2}}\)
Solution
Given: \[I=\int_{-\infty}^{\infty} e^{-t}\,\delta(2t-2)\,dt\]
Rewrite the delta function: \[2t - 2 = 2(t - 1)\] Using the scaling property: \[\delta(2(t-1)) = \frac{1}{|2|}\,\delta(t-1)\]
Thus: \[I=\int_{-\infty}^{\infty} e^{-t}\,\frac{1}{2}\,\delta(t-1)\,dt\]
Apply the shifting property of the delta function: \[I = \frac{1}{2} \, e^{-t}\Big|_{t=1}\]
\[I = \frac{1}{2} \cdot e^{-1} = \frac{1}{2e}\]
Answer: A
A function \(y(t)\), such that \(y(0)=1\) and \(y(1)=3e^{-1}\), is a solution of the differential equation \[\frac{d^{2}y}{dt^{2}} + 2\frac{dy}{dt} + y = 0.\] Then \(y(2)\) is
\(5e^{-1}\)
\(5e^{-2}\)
\(7e^{-1}\)
\(7e^{-2}\)
Solution
The differential equation \[y'' + 2y' + y = 0\] has the general solution: \[y(t) = e^{-t}(C_{1} + C_{2}t).\]
Using \(y(0)=1\): \[C_{1}=1.\]
Using \(y(1)=3e^{-1}\): \[e^{-1}(1 + C_{2}) = 3e^{-1} \quad\Rightarrow\quad 1 + C_{2} = 3 \quad\Rightarrow\quad C_{2}=2.\]
Thus, \[y(2) = e^{-2}(1 + 2\cdot 2) = 5e^{-2}.\]
Answer: Option B
The value of the integral \[\oint_{C} \frac{2z + 5}{\left(z - \tfrac{1}{2}\right)\,(z^{2} - 4z + 5)}\,dz\] over the contour \(\lvert z \rvert = 1\), taken in the anti-clockwise direction, is:
\(\dfrac{24\pi i}{13}\)
\(\dfrac{48\pi i}{13}\)
\(\dfrac{24}{13}\)
\(\dfrac{12}{13}\)
Solution
Poles of the integrand:
\[z = \frac{1}{2}, \qquad z^{2} - 4z + 5 = (z-2)^2 + 1 \Rightarrow z = 2 \pm i.\]
Only the pole at \[z = \frac{1}{2}\] lies inside the contour \(\lvert z \rvert = 1\).
Residue at \(z=\tfrac12\):
\[\operatorname{Res}_{z=\frac12} = \frac{2z+5}{z^{2}-4z+5}\Bigg|_{z=\frac12}.\]
Compute numerator:
\[2\left(\tfrac12\right)+5 = 6.\]
Compute denominator:
\[\left(\tfrac12\right)^2 - 4\left(\tfrac12\right) + 5 = \frac14 - 2 + 5 = \frac{13}{4}.\]
Thus residue:
\[\text{Residue} = \frac{6}{\tfrac{13}{4}} = \frac{24}{13}.\]
Integral value:
\[\oint_{C} f(z)\,dz = 2\pi i \left(\frac{24}{13}\right) = \frac{48\pi i}{13}.\]
Answer: Option B
Consider a \(3 \times 3\) matrix with every element being equal to \(1\). Its only non-zero eigenvalue is .
Solution
Let \[A=\begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & 1 \end{bmatrix}.\]
All rows are identical, so \(\operatorname{rank}(A)=1\). Hence, the matrix has exactly one non-zero eigenvalue.
Since every row sums to \(3\), the vector \[\begin{bmatrix}1\\1\\1\end{bmatrix}\] is an eigenvector with eigenvalue \(3\). The remaining two eigenvalues are \(0\).
Thus the only non-zero eigenvalue is: \[\boxed{3}\]
Answer: 3
The maximum value attained by the function \[f(x) = x(x-1)(x-2)\] in the interval \([1,2]\) is .
Solution
Evaluate at the endpoints:
\[f(1) = 1(0)(-1) = 0, \qquad f(2) = 2(1)(0) = 0.\]
For \(x \in (1,2)\), \[x>0,\quad (x-1)>0,\quad (x-2)<0,\] so \[f(x) < 0.\]
Thus on the entire interval \([1,2]\), the function is never positive and attains its maximum value at the endpoints.
\[\boxed{ \max_{x\in[1,2]} f(x) = 0 }\]
Answer: 0
Question : Consider a linear time invariant system \(\dot{x} = Ax\) with initial condition \(x(0)=\alpha\) at \(t=0\). Suppose \(\alpha\) and \(\beta\) are eigenvectors of \((2 \times 2)\) matrix \(A\) corresponding to distinct eigenvalues \(\lambda_1\) and \(\lambda_2\) respectively. Then the response \(x(t)\) of the system due to initial condition \(x(0)=\alpha\) is
\(e^{\lambda_1 t}\alpha\)
\(e^{\lambda_2 t}\beta\)
\(e^{\lambda_2 t}\alpha\)
\(e^{\lambda_1 t}\alpha + e^{\lambda_2 t}\beta\)
Solution:
Given \(\dot{x} = Ax\), taking Laplace transform:
\[sX(s) - x(0) = AX(s) \Rightarrow X(s) = (sI - A)^{-1}\alpha\]
Thus, \[x(t) = e^{At}\alpha\]
Since \(\alpha\) is the eigenvector corresponding to eigenvalue \(\lambda_1\),
\[e^{At}\alpha = e^{\lambda_1 t} \alpha\]
\[\boxed{x(t) = e^{\lambda_1 t}\alpha}\]
Correct Option: A
Question : Let the probability density function of a random variable \(X\) be given as: \[f_X(x) = \frac{3}{2}e^{-3x}u(x) + ae^{4x}u(-x)\] where \(u(x)\) is the unit step function. Then the value of \(a\) and \(\Pr(X \le 0)\) respectively are:
\(2, \frac{1}{2}\)
\(4, \frac{1}{2}\)
\(2, \frac{1}{4}\)
\(4, \frac{1}{4}\)
Solution:
For a valid PDF, \[\int_{-\infty}^{\infty} f_X(x)\,dx = 1\]
\[\int_{-\infty}^{0} ae^{4x}dx + \int_{0}^{\infty} \frac{3}{2}e^{-3x}dx = 1\]
\[a\left[\frac{e^{4x}}{4}\right]_{-\infty}^{0} + \frac{3}{2}\left[\frac{e^{-3x}}{-3}\right]_{0}^{\infty} = 1\]
\[a \cdot \frac{1}{4}(1 - 0) + \frac{3}{2} \cdot \left(0 - (-\frac{1}{3})\right) = 1\]
\[\frac{a}{4} + \frac{1}{2} = 1 \quad \Rightarrow \quad \frac{a}{4} = \frac{1}{2} \Rightarrow a = 2\]
Now, \[P(X \le 0) = \int_{-\infty}^{0} 2e^{4x}dx = 2\left[\frac{e^{4x}}{4}\right]_{-\infty}^{0} = \frac{1}{2}(1 - 0) = \frac{1}{2}\]
\[\boxed{a = 2,\; \Pr(X \le 0) = \frac{1}{2}}\]
Correct Option: A
Question : Let \(P=\begin{bmatrix}3 & 1\\[4pt]1 & 3\end{bmatrix}\). Consider the set \(S\) of all vectors \(\begin{pmatrix}x\\[2pt]y\end{pmatrix}\) such that \(a^2+b^2=1\) where \(\begin{pmatrix}a\\[2pt]b\end{pmatrix}=P\begin{pmatrix}x\\[2pt]y\end{pmatrix}\). Then \(S\) is
A circle of radius \(\sqrt{10}\).
A circle of radius \(\dfrac{1}{\sqrt{10}}\).
An ellipse with major axis along \(\begin{pmatrix}1\\[2pt]1\end{pmatrix}\).
An ellipse with minor axis along \(\begin{pmatrix}1\\[2pt]1\end{pmatrix}\).
Solution:
We have \[\begin{pmatrix}a\\ b\end{pmatrix} = \begin{bmatrix}3 & 1\\[4pt]1 & 3\end{bmatrix}\begin{pmatrix}x\\[2pt]y\end{pmatrix} \quad\Rightarrow\quad a=3x+y,\; b=x+3y.\] Condition \(a^2+b^2=1\) gives \[(3x+y)^2+(x+3y)^2=1\] which expands to \[10x^2+12xy+10y^2=1.\]
Introduce rotated coordinates \[u=x+y,\qquad v=x-y.\] In these variables \[x=\frac{u+v}{2},\qquad y=\frac{u-v}{2},\] and the quadratic becomes \[8u^2+2v^2=1.\] Divide both sides to obtain the standard ellipse form: \[\frac{u^2}{1/8}+\frac{v^2}{1/2}=1.\]
Since \(1/2>1/8\), the semi-axis along the \(v\)-direction (which is the line \(x-y=0\), i.e. direction vector \(\begin{pmatrix}1\\[2pt]-1\end{pmatrix}\)) is larger; the semi-axis along the \(u\)-direction (line \(x+y=0\), direction \(\begin{pmatrix}1\\[2pt]1\end{pmatrix}\)) is smaller. Hence the curve is an ellipse whose **minor axis** lies along \(\begin{pmatrix}1\\[2pt]1\end{pmatrix}\).
\[\boxed{\text{Correct Option: D. An ellipse with minor axis along } \begin{pmatrix}1\\[2pt]1\end{pmatrix}.}\]
Question : Let \(y(x)\) be the solution of the differential equation \[\frac{d^2 y}{dx^2} - 4\frac{dy}{dx} + 4y = 0\] with initial conditions \(y(0)=0\) and \(\left.\frac{dy}{dx}\right|_{x=0}=1\). Then the value of \(y(1)\) is __________.
\(e^2\)
\(2e^2\)
\(e\)
\(1\)
Solution:
Given: \[\frac{d^2 y}{dx^2} - 4\frac{dy}{dx} + 4y = 0, \quad y(0)=0,\quad y'(0)=1\]
Taking Laplace transform: \[s^2Y(s) - sy(0) - y'(0) - 4[sY(s)-y(0)] + 4Y(s) = 0\]
Substitute initial values \(y(0)=0\), \(y'(0)=1\): \[s^2Y(s) - 1 - 4sY(s) + 4Y(s) = 0\]
Factor: \[(s^2 - 4s + 4)Y(s) = 1\] \[Y(s) = \frac{1}{(s-2)^2}\]
Taking inverse Laplace: \[\mathcal{L}^{-1}\left\{\frac{1}{(s-2)^2}\right\} = te^{2t}\]
Thus, \[y(t) = te^{2t}\]
So, \[y(1) = 1\cdot e^{2} = e^2\]
\[\boxed{e^2}\]
Correct Option: A
Question : The value of the integral \[2\int_{-\infty}^{\infty} \left(\frac{\sin 2\pi t}{\pi t}\right) dt\] is equal to
0
0.5
1
2
Solution:
We use the Fourier transform property:
\[\int_{-\infty}^{\infty} x(t)\,dt = X(0)\]
Also recall the standard Fourier transform pair: \[\frac{\sin(at)}{\pi t} \quad \Longleftrightarrow \quad \begin{cases} 1, & |\omega| < a \\ 0, & |\omega| > a \end{cases}\]
Here \(a = 2\pi\), so \[\frac{\sin 2\pi t}{\pi t} \quad \Longleftrightarrow \quad X(\omega) = 1 \text{ for } |\omega| < 2\pi\]
Thus, \[\int_{-\infty}^{\infty} \frac{\sin 2\pi t}{\pi t}\,dt = X(0) = 1\]
Therefore, \[2\int_{-\infty}^{\infty} \left(\frac{\sin 2\pi t}{\pi t}\right) dt = 2(1) = 2\]
\[\boxed{2}\]
Correct Option: D
Question : The value of the line integral \[\int_{C}\bigl(2xy^{2}\,dx + 2x^{2}y\,dy + dz\bigr)\] along a path joining the origin \((0,0,0)\) and the point \((1,1,1)\) is
\(0\)
\(2\)
\(4\)
\(6\)
Solution:
Parameterize the straight line from \((0,0,0)\) to \((1,1,1)\) by \[x(t)=t,\qquad y(t)=t,\qquad z(t)=t,\qquad 0\le t\le 1,\] so \(dx=dy=dz=dt\).
Substitute into the integrand: \[2xy^{2}\,dx + 2x^{2}y\,dy + dz = 2t\cdot t^{2}\,dt + 2t^{2}\cdot t\,dt + 1\cdot dt = 2t^{3} + 2t^{3} + 1 \,=\, 4t^{3}+1.\]
Hence the integral is \[\int_{0}^{1} (4t^{3}+1)\,dt = \left[t^{4}+t\right]_{0}^{1} = 1+1 = 2.\]
\[\boxed{2}\]
Question : Let \(f(x)\) be a real, periodic function satisfying \(f(-x)=-f(x)\). The general form of its Fourier series representation would be
\(f(x)=a_0+\sum_{k=1}^{\infty} a_k\cos(kx)\)
\(f(x)=\sum_{k=1}^{\infty} b_k\sin(kx)\)
\(f(x)=a_0+\sum_{k=1}^{\infty} a_{2k}\cos(kx)\)
\(f(x)=\sum_{k=0}^{\infty} a_{2k+1}\sin((2k+1)x)\)
Solution:
Since \(f(-x)=-f(x)\), \(f\) is an odd function. The Fourier series of a real odd periodic function contains no cosine (even) terms and no constant (DC) term — only sine terms remain. Therefore the series has the form \[f(x)=\sum_{k=1}^{\infty} b_k\sin(kx).\]
Correct Option: B
Question : The solution of the differential equation, for \(t>0\), \(y''(t) + 2y'(t) + y(t) = 0\) with initial conditions \(y(0)=0\) and \(y'(0)=1\), is \((u(t)\) denotes the unit step function\()\)
\(te^{-t}u(t)\)
\((e^{-t}-te^{-t})u(t)\)
\((-e^{-t}+te^{-t})u(t)\)
\(e^{-t}u(t)\)
Solution:
Taking Laplace transform,
\[s^2Y(s) - sy(0) - y'(0) + 2[sY(s) - y(0)] + Y(s)=0\]
\[(s^2 + 2s +1)Y(s) -1 = 0 \Rightarrow Y(s)=\frac{1}{(s+1)^2}\]
Inverse Laplace:
\[y(t)=t e^{-t}u(t)\]
Correct Option: A
Question : A \(3\times 3\) matrix \(P\) is such that \(P^3 = P\). Then the eigenvalues of \(P\) are
\(1,\,1,\,-1\)
\(1,\,0.5+j0.866,\,0.5-j0.866\)
\(1,\,-0.5+j0.866,\,-0.5-j0.866\)
\(0,\,1,\,-1\)
Solution:
If \(\lambda\) is an eigenvalue of \(P\) with eigenvector \(v\neq 0\), then \[P^3v = Pv \;\Rightarrow\; \lambda^3 v = \lambda v\] so \[\lambda^3 = \lambda \quad\Rightarrow\quad \lambda(\lambda^2-1)=0.\] Thus \(\lambda\in\{0,1,-1\}\). For a \(3\times 3\) matrix the spectrum can consist of these values (with multiplicities summing to 3). Therefore the eigenvalues are \(0,1,-1\).
Correct Option: D
Question : Consider a causal LTI system characterized by the differential equation \[\frac{dy(t)}{dt} + \frac{1}{6}y(t) = 3x(t).\] The response of the system to the input \(x(t)=3e^{-t/3}u(t)\), where \(u(t)\) denotes the unit step function, is
\(9e^{-t/3}u(t)\)
\(9e^{-t/6}u(t)\)
\(9e^{-t/3}u(t) - 6e^{-t/6}u(t)\)
\(54e^{-t/6}u(t) - 54e^{-t/3}u(t)\)
Solution:
Take Laplace transforms (zero initial conditions): \[(s+\tfrac{1}{6})Y(s)=3\cdot X(s)=3\cdot\frac{3}{s+\tfrac{1}{3}}=\frac{9}{s+\tfrac{1}{3}}.\] Hence \[Y(s)=\frac{9}{(s+\tfrac{1}{6})(s+\tfrac{1}{3})}.\] Partial fractions: \[\frac{9}{(s+a)(s+b)}=\frac{9}{b-a}\Big(\frac{1}{s+a}-\frac{1}{s+b}\Big),\] with \(a=\tfrac{1}{6},\,b=\tfrac{1}{3}\), so \(b-a=\tfrac{1}{6}\) and \(\dfrac{9}{b-a}=54\). Thus \[Y(s)=54\Big(\frac{1}{s+\tfrac{1}{6}}-\frac{1}{s+\tfrac{1}{3}}\Big).\] Inverse Laplace gives \[y(t)=54\big(e^{-t/6}-e^{-t/3}\big)u(t).\]
Correct Option: D