GATE Aptitude-General

GATE Engineering Mathematics

GATE 2011

1. With K as a constant, the possible solution for the first order differential equation  dy/dx = e^-3x  is

a. - 1/3 e^-3x + K (Ans)

b. - 1/3 e^3x + K

c. -3e^-3x + K

d. -3e^-x + K

Ans :     dy/dx = e^-3x

dy = e^-3x dx

Taking integration on both sides,

y = e^-3x /-3 + K = -  1/3 e-^3x + K

2. Roots of the algebraic equation x³ + x² + x  + 1 = 0 are

a. +1, +j - j

b. +1, -1,+1

c. 0,0,0

d. -1, +j, -j (Ans)

Ans :  x³ + x² + x + 1 = 0

(x² + 1) (x +1) = 0

x² + 1 = 0    ⇒  x =  ± j

x + 1 = 0     ⇒  x =  -1

3. The Fourier series expansion f(t) = a₀ + ^∞Σ_{n = 1}   a_n cos nωt + b_n sin nωt of the periodic signal shown below will contain the following non-zero terms

                                                       f (t)                                        

                                                                                                  t

                                                     0

a. a₀ and b_n, n = 1, 3, 5, ... ∞

b. a₀, a_n and b_n, n = 1, 2, 3, ... ∞

c. a₀ and a_n, n = 1, 2, 3, ... ∞

d. a₀ and a_n, n = 1, 3, 5, ... ∞ (Ans)

Ans : The given signal satisfies even symmetry so b_n = 0 and it also satisfies half wave symmetry so, it will contain only odd harmonics.

GATE 2010
 

4. The value of the quantity P, where P =∫¹₀ xe^x dx, is equal to

a. 0

b. 1 (Ans)

c. e

d. 1/e

Ans :      P =∫¹₀ xe^x dx

= [x ∫e^x dx]¹₀ - ∫¹₀ 1.e^x dx

= e - e + e⁰ = 1

5. Divergence of the three-dimensional radial vector field  ^→r  is —

a. 3 (Ans)

b. 1/r

c. Î_i  +  Ä´ +  ^_k

_{d. 3(}Π +  Ä´  +  ^_{k )}

Ans :   The radial vector in 3-D space is^→r  =  xΠ +  yÄ´ +  z^_k

_{Now, div}  ^→r =  ᐁ. ^→r

= (Π ∂/∂x + Ä´∂/∂y + ^_k∂/∂z )(xΠ +  yÄ´ +  z^_k)

= ∂/∂x (x) + ∂/∂y (y) + ∂/∂z (z)

= 1 + 1 + 1 = 3

6. The period of the signal x (t) = 8 sin (0.8 t + /4) is

a. 0.4 s

b. 0.8 s

c. 1.25 s

d. 2.5 s (Ans)

Ans : The given signal  x (t) = 8 sin (0.8t + /4)

We know that, imperiodic function

f (t +T) = f (t),where T = period 

Then,     x (t +T) = x (t)

⇒ 8 sin {0.8 (t +T) + /4}= 8 sin (0.8 t +/4)

⇒ sin {0.8 (t +T) + /4}= sin {2 + (0.8 t + /4)}

{∵ sin (2 + θ) = sin θ, period of sin θ is 2}

⇒ 0.8 (t +T) + /4 = 2 + 0.8 t + /4

⇒ 0.8 t + 0.8 T +2 + 0.8 t

⇒    T = 2 /0.8 ⇒ 2/0.8

⇒    T (period) = 2.5s

GATE 2009

7. The trace and determinant of a 2 * 2 matrix are known to be - 2 and - 35 respectively. Its eigen values are

a. - 30 and - 5

b. - 37 and - 1

c. - 7 and 5  (Ans)

d. 17.5 and - 2
 

Ans :  Given, trace of 2 * 2 matrix = -2 and determinant = -35

Let      A = [a₁₁  a₁₂]   
                 a₂₁   a_{22  2 * 2}

then, trace of matrix

A =  Addition of principal diagonal elements 

- 2 = a₁₁ + a₂₂                                 ....(i)
and ⃒A⃒ =a₁₁ . a₂₂ - a₁₂ . a₂₁ = - 35        .....(ii)
Also the eigen values of [A] = Trace of matrix A
λ₁+ λ₂ = - 2                        ........(iii)  
So, option (c) will satisfy Eq. (iii).

Hence, eigen values are - 7 and 5.

GATE 2008

8. X is a  uniformly distributed random variable that takes values between O and 1. The value of E{X³}will be

a. 0

b. 1/8

c. 1/4  (Ans)

d. 1/2

Ans :  Since, X is a uniformly distributed random variable between (0,1) = (a, b)

Then, probability density density function

f (x) = 1/(b - a) ⇒  1/(1 - 0)  ⇒ 1

So,  f (x) = {1,      0 < x < 1
                   0,  other value of x

So, E(x³) = ∫¹_{x = 0}  x³ . f (x) dx

=  ∫¹₀  x³ .1 dx

= [x⁴ /4] ¹₀

= 1/4 (1 - 0)

= 1/4

9. The characteristic equation of a (3 * 3)  matrix P is defined as  a (λ) = ⃒λ|  - P⃒ = λ³+  λ² + 2λ + I = 0
If I denotes identity matrix, then the inverse of matrix P will be

a. (P² + P + 2I)

b. (P² + P + I)

c.  - (P² + P + I)

d.  - (P² + P + 2I)  (Ans)

Ans :  Given, the characteristic equation is

 λ³+  λ² + 2λ + I = 0             ..........(i)
We know that, By Caylay - Hamilton theorem Every square matrix satisfies its characteristic equation.
Then, from Eq. (i),putting (λ = P),
P³ + P² + 2P + I = 0

Operating (P^-1) on both sides,

(P^-1 P³) + (P^-1 P²) + 2(P^-1P) + (P^-1 I) = (0.P^-1)

⇒    P² + P + 2I + P^-1 = 0

⇒    P^-1 = - (P² + P +2I)

10. If the rank of a (5 * 6) matrix Q is 4, then which one of the following statements is correct?

a. Q will have four linearly independent rows and four linearly independent columns  (Ans)

b. Q will have four linearly independent rows and five linearly independent columns

c. QQ^T will be invertible

d. Q^T Q will be invertible

Ans :  Given, f ([Q]_{5 *6}) = 4 {/→ denotes Rank of}Then, [Q] must have four linearly independent rows and four linearly independent columns.  Because the matrix Q  of order (4 * 4) will be non- singular matrix i.e., [Q] _4*4
⇒     ⃒Q⃒_4*4  ≠ 0

GATE 2007

11.  x = [x₁ x₂......xn]^T is an n-tuple non-zero vector. The n * n matrix V = xx^T

a. has rank zero (Ans)

b. has rank 1

c. is orthogonal

d. has rank n

Ans :  Given, x = [x₁, x₂, ....x_n]^T

Since, x is an n-tuple non-zero vector, that is, x is a non-singular matrix of order n, so it rank should be n.  

i.e,         I (x) = n

The vector (x^T) also have rank (n) because transpose of any matrix does not altered its rank. 

Then, matrix [v = x T'] must have the rank n, i. e.,I(v) = n, because resultant of the multiplication of two same rank matrices also has the same rank as the rank of multiplicative matrix.

GATE 2005

12. In the matrix equation px = q, which of the following is a necessary condition for the existence of at least one solution for the unknown vector x ?

a.  Augmented matrix [pq] must have the same rank as matrix p  (Ans)

b.  Vector q must have only non - zero elements

c.  Matrix p must be singular

d.  Matrix p must be square 

Ans :  The matrix equation px = q will have solutions if it is consistent.

So, for consistency of matrix equation isRank of [pq]  = Rank of [p]

where, [pq] → Augmented matrix

13. If P and Q are two random events, then which of the following is true?

a. Independence of P and Q implies that probability  (P ⌒ Q) = 0

b. Probability (P ∪Q) ≥ Probability (P) + Probability (Q)

c. If P and Q are mutually exclusive, then they must be independent

d. Probability (P ⌒ Q) ≤ Probability (P) (Ans)

Ans :  Given, P and Q are two random events, then

(a)  Independence of P and Q, Prob (P ⌒ Q) = Prob (P) . Prob (Q)

If two events P and Q are mutually disjoint, then Prob (P ⌒ Q) = 0

So, option (a) is incorrect.

(b) For two events P and Q,

Prob (P ∪Q) = Prob (P) + Prob (Q) - Prob (P ⌒ Q)

Here, Prob (P ∪Q) ≤ Prob (P) + Prob (Q)

So, option (b) is incorrect.

(c) There is no relation between mutually exclusive and independent for two random variables P and Q.

So, option (c) is incorrect.

(d) And Prob (P ⌒ Q) ≤ Prob (P) which is true for every events.

So, option (d) is correct.

14. If S = ∫^∞₁ x^-3dx, then S has the value

a. -1/3

b. 1/4

c. 1/2 (Ans)

d. 1

Ans :      S = ∫^∞₁ x^-3dx

= [x^-2 /-2]^∞₁  = - 1/2[1/x²]^∞₁

= 1/2 [1/∞ -  1/1] = - 1/2(0-1)

= 1/2

15. The solution of the first order differential equation  x '(t) = -3x(t),x(0) = x₀  is

a. x (t) = x₀e^-3t (Ans)

b. x (t) = x₀e^-3

c. x (t) = x₀e^-1/3

d. x (t) = x₀e^-t

Ans :       x '(t) = -3x (t)  ⇒  dx/dt = -3x

⇒      ∫dx/x = -∫3 dt             (on integrating)

⇒    log x = -3t + log c

⇒    x = ce^-3t                .........(i)

But at (t = 0) and (x = x₀),

⇒    x₀ = ce⁰     ⇒ (c = x₀)

Then, from Eq. (i),  x = x₀ e^-3t

or          x (t) =  x₀ e^-3t

GATE 2011

16. The two vectors [1,1,1] and [1,a,a²], where a = (-  1/2 + j√3/2) are

a. orthonormal

b. orthogonal (Ans)

c. parallel

d. collinear

Ans :  These two vectors is orthogonal because the dot product of these two vectors is zero.      

17. The matrix [A] =   2     1  is decomposed into a product of a lower triangular matrix [L] and an upper triangular matrix [U].
                               4    -1      

The properly decomposed [L] and [U] matrices respectively are

a.       1    0    and   1     1   
         4    -1            0   -2  

b.       2    0    and    1    1
          4   -1            0     1

c.       1     0    and   2     1
         4     1            0    -1

d.       2    0             1   0.5       (Ans)
          4   -3    and   0      1

Ans :   [A] = [L] [U]

=     2    0   1  0.5    =     2+0   1+0
        4  -3   0    1            4+0   2-3

=     2    1
        4  -1

18. The function f (x) = 2x - x² + 3  has

a. a maximum at x = 1 and a minimum at x = 5

b. a maximum at x = 1 and a minimum at x = -5

c. only a maximum at x = 1   (Ans)

d. only a minimum at x = 1

Ans :  f (x) = 2x - x² + 3

f' (x) = 2 - 2x = 0

∴    x = 1

at     x = 1, f" (x) = - 2 < 0  ⇒    f" (x) < 0

So,  f (x) having only a maximum at x = 1.

GATE 2010

19. At t = 0, the function f (t) = sin t/ t  has

a. a minimum

b. a discontinuity

c. a point of inflection

d. a maximum  (Ans)

Ans :  Given,  f (t) = sin t/ t ;

At t = 0, first we will check continuity of the function.

LHL f (0 - h)  =   lim      sin (0 - h)   
                       h→ 0     (0 - h)

                    =  lim      - sin h                 [ ∵  sin (-θ) = - sin θ]
                      h→ 0       -h
 

              = 1                                   [ ∵   lim    sin θ/θ = 1]
                                                            h→ 0

RHL   f (0 + h) =   lim     sin (0 + h)/(0 + h)
                         h→ 0

                     =   lim      sin h/h  = 1
                         h→ 0

and          f(0) = 1

since,       LHL = RHL = f(0)

So, the function is continuous at t = 0.

Now, we check the function is maximum or minimum.

            f '(t)  =  1/t cos t -1/ t² sin t

and       f "(t) = 1/t sin t -1/t² cos t - 1/t² cos t + 2sin t/ t³

                    =  - sin t /t -  2 cos t/ t² + 2sin t/t³
 

For max or min value of f (x),

                              f '(x) = 0

            cos t /t - sin t/ t²= 0

⇒                        tan t/t = 1

Now,  lim  f "(t) = - lim      sin t/t 
        t→ 0            t→ 0

         +  lim      (2 sin t - 2 t cos t)/t³        [∴(0/0) form]
            t→ 0

        = - 1 +  lim      (2 cost - 2 cos t + 2 t sin t)/3 t²         [∵ Using L hospital's rule]
                   t→ 0

        =  - 1 +  lim        2 t sin t/3 t
                    t→ 0

        =  - 1 + 2/3    lim        sin t/t
                           t→ 0

        = -1 + 2/3 * 1 = - 1/3 < 0    (Maxima)

    So, function   f (t) is maximum at t = 0.

20. A box contains 4 white balls and 3 red balls. In succession, two balls are randomly selected and removed from the box. Given that the first removed ball is white, the probability that the second removed ball is red is

a. 1/3

b. 3/7

c. 1/2   (Ans)

d. 4/7

Ans : Probability (IInd ball is red/lst ball is white)

=  P (IInd is red and Ist is white)
            P (Ist ball is white)

= P (Ist ball is white and IInd ball is red)
            P (Ist ball is white)

    (Probability of Ist ball is white)

=     x (Probability of II nd ball is red)    
        (Probability of Ist ball is white)

= Probability of IInd ball is red

= 3/6 = 1/2