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GATE-2012 ECE Q12 (math)

Question 12 on math from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q12. With initial condition ${x(1)=0.5}$ the solution of the differential equation,

${t\frac{dx}{dt}+x= t}$ is

(A) ${x=t-\frac{1}{2}}$

(B) ${x=t^2-\frac{1}{2}}$

(C) ${x=\frac{t^2}{2}}$

(D) ${x=\frac{t}{2}}$

Solution

From the product rule used to find the derivative of product of two or more functions,

${\frac{d}{dx}(f.g)=f\frac{dg}{dx}+g\frac{df}{dx}}$

Applying this to the above equation, we can be seen that,

$\frac{d}{dt}(t.x)=t\frac{dx}{dt}+x\frac{dt}{dt}=t\frac{dx}{dt}+x$

Plugging this in and integrating both sides,

$\begin{array}t\frac{dx}{dt}+x&=&t\\\frac{d}{dt}(t.x)&=&t\\\int\frac{d}{dt}(t.x)&=&\int t\\tx&=&\frac{t^2}{2}+C\end{array}$ .

Using the initial condition ${x(1)=0.5}$ , we can solve for the unknown $C$ , i.e.
$\begin{array}t.x&=&\frac{t^2}{2}+C\\1*0.5&=&\frac{1^2}{2}+C\\C&=&0\end{array}$ .

So the solution to the differential equation is,

${x=\frac{t}{2}}$

Based on the above, the right choice is (D) ${x=\frac{t}{2}}$ .

References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

[2] Wiki entry on Product rule

GATE-2012 ECE Q47 (math)

Question 47 on math from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q47. Given that

${A=\[\begin{array}{rr}-5&-3\\2&0\end{array}\]}$ and ${I=\[\begin{array}{rr}1&0\\0&1\end{array}\]}$ , the value of ${A^3}$ is

(A) ${15A+12I}$

(B) ${19A+30I}$

(C) ${17A+15I}$

(D) ${17A+21I}$

Solution

To answer this question, we need to refer to Cayley Hamilton Theorem. This is discussed briefly in Pages 310-311 of Introduction to Linear Algebra, Glibert Strang (buy from Amazon.com, buy from Flipkart.com)

From the wiki entry on Cayley Hamilton theorem,

If $A$ is a given $\mbox{n x n}$ matrix, and $I_n$ is the $\mbox{n x n}$ identity matrix, the characteristic polynomial of $A$ is defined as,

$p(\lambda) = \det $ A-\lambda I_n$$ .

The Cayley Hamilton theorem states that substituting matrix $A$ for $\lambda$ in this polynomial results in a zero matrix, i.e.

$p(A) = 0$

This theorem allows for $A^n$ to be expressed as linear combination of the lower matrix powers of $A$ .

For a general 2×2 matrix the theorem is relatively easy to prove.

Let ${A=\[\begin{array}{rr}a&b\\c&d\end{array}\]}$

The characteristic polynomial is

$\begin{array}{lll}p(\lambda)&=&\det$A - \lambda I$\\&=&\det\[\begin{array}a-\lambda&b\\c&d-\lambda\end{array}\]\\&=&\lambda^2 - (a+d)\lambda + $ad-bc$\end{array}$

Substituting by matrix $A$ in the polynomial,

$\begin{array}{lll}p(A)&=&A^2 - (a+d)A+$ad-bc$I\\&=&\[ \begin{array}(a^2+bc)&(ab+bd)\\(ac+cd)&(bc+d^2)\end{array}\]-\[ \begin{array}(a^2+ad)&(ab+bd)\\(ac+cd)&(ad+d^2)\end{array}\]+\[ \begin{array}(ad-bc)&0\\0&(ad-bc)\end{array}\]\\&=&\[ \begin{array}0&0\\0&0\end{array}\]\end{array}$ .

Now, applying Cayley Hamilton theorem to the problem at hand,

${A=\[\begin{array}{rr}-5&-3\\2&0\end{array}\]}$ .

The characteristic polynomial is,

$\begin{array}{lll}p(\lambda)&=&\det$A - \lambda I$\\&=&\det\[\begin{array}-5-\lambda&-3\\2&-\lambda\end{array}\]\\&=&\lambda^2 +5\lambda+6\end{array}$ .

Substituting by matrix $A$ in the polynomial,

$\begin{array}p(A)&=&A^2 +5A+6I&=&0\end{array}$ .

Alternatively, $\begin{array}A^2&=&-5A-6I\end{array}$ .

Finding $A^3$ in terms of $A$ by substituting for $A^2$ ,

$\begin{array}{lll}A^3&=&A^2A\\&=&$-5A-6I$A\\&=&-5A^2-6A\\&=&-5$-5A-6I$-6A\\&=&19A+30I\end{array}$

Matlab example

>> A = [-5 -3 ; 2 0];
>> A^3
ans =

  -65  -57
   38   30

>> 19*A + 30*eye(2)
ans =

  -65  -57
   38   30

Based on the above, the right choice is (B) ${19A+30I}$ .

References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

[2] Introduction to Linear Algebra, Glibert Strang (buy from Amazon.com, buy from Flipkart.com)

[3] wiki entry on Cayley Hamilton theorem

GATE-2012 ECE Q36 (math)

Question 36 on math from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q36. A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is

(A) 1/3

(B) 1/2

(D) 3/4

Solution

Let us start by finding the sample space. The possible sample space with odd number of tosses till a head appears for the first time is,

$\begin{array}S=\{H,&TTH,& TTTTH,& TTTTTTH,& ...&\}\end{array}$

Given that the coin is fair,

$p(H)=p(T)=\frac{1}{2}$ .

The total probability is,

$\begin{array}{lll}p(S)&=&\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+...\\&=&\frac{1}{2}$1+\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...$\\&=&\frac{1}{2}$1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...$\end{array}$

Using Taylor series,

$\begin{array}\frac{1}{1-x}&=&\sum_{n=0}^{\infty}x^n,&\mbox{for } |x| < 1\end{array}$ .

Substituting,

$\begin{array}{lll}p(S)&=&\frac{1}{2}*\frac{1}{$1-\frac{1}{4}$}\\&=&\frac{2}{3}\end{array}$

Matlab/Octave example

Felt it would be nice to get similar results using a simple simulation model

% Script to find the probability that odd number of tosses
% are required to get a head for the first time

clear all; close all;
N = 5*10^5;

% HEAD = 0, TAIL = 1;
% definfing the pattern corresponding to odd tosses 
% till the first head - limiting to max of 21 tosses
% converted to integer (for easy comparison)
pattern_v = [ ...
   sum(2.^0         .*  [0 ]);                      ... % H
   sum(2.^(2:-1:0)  .*  [1 1 0 ]);                  ... % TTH
   sum(2.^(4:-1:0)  .*  [1 1 1 1 0 ]);              ... % TTTTH 
   sum(2.^(6:-1:0)  .*  [1 1 1 1 1 1 0 ]);          ... % TTTTTTH 
   sum(2.^(8:-1:0)  .*  [1 1 1 1 1 1 1 1 0 ]);      ... % TTTTTTTTH ...
   sum(2.^(10:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 0 ]);  ...  
   sum(2.^(12:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 1 1 0 ]); ...
   sum(2.^(14:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 ]); ... 
   sum(2.^(16:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 ]); ... 
   sum(2.^(18:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 ]); ...
   sum(2.^(20:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 ]); ...
  ];

% finding the probability for each event in the  sample space
event_v = [1:2:21];
for ii=1:length(event_v)
   kk           = event_v(ii);
   x            = rand(N,kk)>0.5;
   xVal         = sum((ones(N,1)*[2.^([kk-1:-1:0])]).*x,2);
   matchCnt(ii) = size(find(xVal==pattern_v(ii)),1);
end
totalProb = sum(matchCnt./N)

Figure : Probability of odd number of tosses till first head

Based on the above, the right choice is (C) 2/3.

References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

[2] Wiki entry on Taylor series http://en.wikipedia.org/wiki/Taylor_series

GATE-2012 ECE Q46 (math)

Question 46 on math from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q46. The maximum value of ${f(x) = x^3 - 9x^2 + 24x + 5}$ in the interval [1, 6] is

(A) 21

(B) 25

(D) 46

Solution

Let us start by finding the critical points of the function $f(x)$ .

The first derivative is,

$f'(x) = 3x^2 - 18x + 24$ .

Solving by setting $f'(x) =0$ .

$\begin{array}{ll}3x^2 - 18x + 24 & = 0 \\ (x-4)(x-2)&=0\end{array}$ .

Taking second derivative, $f''(x) = 6x - 18$ .

At $\begin{array}{lll}x=4, &f''(x) &= +6 & (+ve)\end{array}$ and this corresponds to a local minimum.

At $\begin{array}{lll}x=2, &f''(x) &= -6 & (-ve)\end{array}$ and this corresponds to a local maximum.

The value of the function at these critical points are,

$\begin{array}f(2)&=&2^3 - 9*2^2 + 24*2 + 5&=&25 &\mbox{(local maximum)}\\f(4)&=&4^3 - 9*4^2 + 24*4 + 5&=&21&\mbox{(local minimum)}\end{array}$ .

Hold on, we are not done yet… 🙂

From the wiki entry on Extereme Value Theorem,

In calculus, the extreme value theorem states that if a real-valued function f is continuous in the closed and bounded interval [a,b], then f must attain its maximum and minimum value, each at least once. That is, there exist numbers c and d in [a,b] such that:

$\begin{array}f(c) &\ge & f(x)& \ge& f(d), & \mbox{for all x} & \in & \[a,b\]\end{array}$ .

Further, from wiki entry on finding maxima and minima,

If a function is continuous on a closed interval, then by the extreme value theorem global maxima and minima exist. Furthermore, a global maximum (or minimum) either must be a local maximum (or minimum) in the interior of the domain, or must lie on the boundary of the domain. So a method of finding a global maximum (or minimum) is to look at all the local maxima (or minima) in the interior, and also look at the maxima (or minima) of the points on the boundary; and take the biggest (or smallest) one.

For our example, the function $f(x) = x^3 - 9x^2 + 24x + 5$ is real valued and is continuous in the closed and bounded interval [1,6].

Finding out the value of the function at the boundaries,

$\begin{array}f(1)&=&1^3 - 9*1^2 + 24*1 + 5 &= & 21\\f(6)&= & 6^3 - 9*6^2 + 24*6 + 5 & =& 41\end{array}$ .

So the global maximum of the function $f(x) = x^3 - 9x^2 + 24x + 5$ in the interval [1, 6] is $\max{$25, 41$} = 41$ and it occurs at the boundary i.e. x = 6.

Matlab example

clear all; close all;
x = [1:.1:6];
fx = x.^3 - 9*x.^2 + 24*x + 5;
plot(x,fx);
xlabel('x'); ylabel('fx');
grid on; axis([1 6 20 45]);
title('fx  =  x^3 - 9x^2 + 24x + 5');

Figure : Plot of the function $f(x) = x^3 - 9x^2 + 24x + 5$ in the interval [1,6]

From the figure, it is relatively easy to observe that the maximum is hit at x=6 (and not at the local maximum point of x=2).

Based on the above, the right choice is (C) 41.

References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

[2] Wiki entry on Extreme Value Theorem

http://en.wikipedia.org/wiki/Extreme_value_theorem

[3] Wiki entry on Maxima and Minima

http://en.wikipedia.org/wiki/Maxima_and_minima

GATE-2012 ECE Q24 (math)

Question 24 on math from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q24. Two independent random variables X and Y are uniformly distributed in the interval [-1, 1]. The probability that max[X,Y] is less than 1/2 is

(A) 3/4

(B) 9/16

(D) 2/3

Continue reading “GATE-2012 ECE Q24 (math)”

GATE-2012 ECE Q25 (math)

Question 25 on math from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q25. If ${x=\sqrt{-1}}$ , then the value of ${x^x}$ is,

(a) $e^{-\pi/2}$

(b) $e^{\pi/2}$

(d) 1

Continue reading “GATE-2012 ECE Q25 (math)”

Q12. With initial condition the solution of the differential equation,

is

(A)

(B)

(C)

(D)

Solution

References

Q47. Given that

and , the value of is

(A)

(B)

(C)

(D)

Solution

References

Q36. A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is

Solution

References

Q46. The maximum value of in the interval [1, 6] is

Solution

Based on the above, the right choice is (C) 41.

References

Q24. Two independent random variables X and Y are uniformly distributed in the interval [-1, 1]. The probability that max[X,Y] is less than 1/2 is

Q25. If , then the value of is,

Q12. With initial condition ${x(1)=0.5}$ the solution of the differential equation,

${t\frac{dx}{dt}+x= t}$ is

(A) ${x=t-\frac{1}{2}}$

(B) ${x=t^2-\frac{1}{2}}$

(C) ${x=\frac{t^2}{2}}$

(D) ${x=\frac{t}{2}}$

${A=\[\begin{array}{rr}-5&-3\\2&0\end{array}\]}$ and ${I=\[\begin{array}{rr}1&0\\0&1\end{array}\]}$ , the value of ${A^3}$ is

(A) ${15A+12I}$

(B) ${19A+30I}$

(C) ${17A+15I}$

(D) ${17A+21I}$

Q46. The maximum value of ${f(x) = x^3 - 9x^2 + 24x + 5}$ in the interval [1, 6] is

Q25. If ${x=\sqrt{-1}}$ , then the value of ${x^x}$ is,