Mathematical Science MCQ Quiz in मल्याळम - Objective Question with Answer for Mathematical Science - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept -

(1) The collection of all the sequences on two symbols or more than two symbols is uncountable.

Explanation -

The set T= {(x1, x2,..., xn,...): x1, x2,..., xn, ... ∈ {1, 3, 5, 7, 9}} 

Now we have 5 symbols and T represents the collection of all the sequences.

Hence the set T is Uncountable.

Concept use:

Bounded set : A set S is bounded if it has both upper and lower bounds. 

Closed set: If a set contain each of its limit point in the set 

Calculations:

S = {x5 - x4

T = { x2 - 2x

Hence the Intersection of the Closed set and Open Set need not be closed set, but it is bounded also.

So, The Correct option is 2.

Explanation:

Let w₁ =  and w2 =  and u = 

then orthogonal projection of u on W is 

}{}\) w1 + }{}\)w2

= +  = 

= +  = 

(2) correct

Concept -

(i)  If n is even then (-1)n = 1 

(ii)  If n is odd then (-1)n = -1

(iii)   then 

Explanation -

We have the sequence 

Now as n →  ∞ ,

a_n = 0 + (-1)ⁿ cos³(0) + (-1)ⁿ

Now we make the cases -

Case - I - If n is even then put (-1)ⁿ = 1 in the above equation we get

an = 0 + 1 x cos3(0) + 1 x  = 1 + 1 = 2

Case - II - If n is odd then put (-1)n = -1 in the above equation, we get

an = 0 - 1 x cos3(0) - 1 x  = -1 + 1 = 0

Hence largest and smallest limit points are 2 & 0.

So Options (i) & (iv) are wrong.

And we know that limit of the sequence is different in both the cases so not convergent.

Hence option (iii) is correct and (ii) is wrong.

Explanation -

Results -

(i) Number of homomorphism from  is 16.

(ii) Number of onto homomorphism from  is 6.

(iii) Number of 1-1 homomorphism from  is 0.

Hence option(2) is correct.

Explanation:

T: ℝ2 →ℝ2 be defined by 

 be a basis of ℝ2 

So,  = 

  = 

So, matrix representation is

Option (3) is true and others are false

Concept:

L’Hospital’s Rule: If  =  = 0 or ± ∞ and g'(x) ≠ 0 for all x in I with x ≠ c and  exist then  = 

Explanation:

 (0/0 form so using L'hospital rule)

=  

= 

Again using L'hospital rule

= 

= 

It will be 0/0 form if

x - 2a = 0

⇒ a = 0

Option (1) is correct

Concept:

If f(z) is an analytic function within and on a simple closed curve C and if a is any point within C, then 

f(a) = dz

Here, the integral should be taken in the positive sense around C.

Solution - Given , function

f(z) = 

and the function has singularity at z = i, z = - i

C : {z : |z - i|

So z = - i does not lie on the curve and z = i lies inside the curve

Hence

I = 2πi ×  = 2πi  = 2πi  = π (log i)³

Now, log (i) = log 1 + i tan^-1(1/0) = 0 + i = i

hence I = π = 

Therefore, Correct Option is Option 4).

Solution:

Given function is:

F(x, y) = x3 + y3 – 3xy – 4 = 0

And x = 2 and g(2) = 2

Now,

F(2, 2) = (2)3 + (2)3 – 3(2)(2) – 4

= 8 + 8 – 12 – 4

= 0

So, F(2, 2) = 0

∂F/∂x = ∂/∂x (x3 + y3 – 3xy – 4) = 3x2 – 3y

∂F/∂y = ∂/∂y (x3 + y3 – 3xy – 4) = 3y2 – 3x

Let us calculate the value of ∂F/∂y at (2, 2).

That means, ∂F(2, 2)/∂y = 3(2)2 – 3(2) = 12 – 6 = 6 ≠ 0.

Thus, ∂F/∂y is continuous everywhere.

Hence, by the implicit function theorem, we can say that there exists a unique function g defined in the neighborhood of x = 2 by g(x) = y, where F(x, y) = 0 such that g(2) = 2.

Also, we know that ∂F/∂x is continuous.

Now, by implicit function theorem, we get;

g’(x) = -[∂F(x, y)/∂x]/ [∂F(x, y)/ ∂y]

= -(3x2 – 3y)/(3y2 – 3x)

= -3(x2 – y)/ 3(y2 – x)

= -(x2 – y)/(y2 – x)

Hence, option 3 is correct

Given:

f(x, y) =  (x, y) → (0, 0)

Concept Used:

Putting y = mx in the function and checking whether the function is free from m then limit will exist if not then limit will not exist.

Solution:

We have,

f(x, y) = \(\frac{siny}{x}\) (x, y) → (0, 0)

Put y = mx

So, 

lim (x, y) → (0, 0) \(\frac{siny}{x}\)

⇒ lim x → 0 
 

We cannot eliminate m from the above function.

Hence limit does not exist.

 Option 4 is correct.