Mathematical Science MCQ Quiz in தமிழ் - Objective Question with Answer for Mathematical Science - இலவச PDF ஐப் பதிவிறக்கவும்

Concept -

P - test - 

 is convergent for p > 1

Explanation -

For option (ii) -

If a_n = 1/n be a sequence of non - negative real number.

If  is convergent by P - test.

but  is divergent series 

Hence option(ii) is false.

For option(i) -

If   is convergent then  is also convergent for any convergent series.

Hence option(i) is true.

For option(iii) -

If  is convergent then  is either cgt or dgt as well

but in both cases, the series  is convergent.

Hence option(iii) is true.

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

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.

Explanation:

Here, it is given that

(i) f(x + y) = f(x).f(y) and

(ii) f(x) = 1 + x g(x), where 

Now, writing for y in the given condition. We have

f(x + h) = f(x).f(h)

Then, f(x + h) - f(x) = f(x)f(h) - f(x)

Or 

                      =  (using (ii))

Hence, 

Since, by hypothesis 

It follows that f'(x) = f(x)

Since, f(x) exists, f'(x) also exists

and f'(x) = f(x) 

⇒ 

(2) is true.

Concept -

If f : [a,b] →  and f(a) > 0 and f(b)

Explanation -

We have the polynomial f(x) = x4 - 3x3 - x2 + 4

Now f'(x) = 4x³ - 9x² - 2x = x( 4x² - 9x - 2) 

Now for the critical points 

f'(x) = 0

⇒  x( 4x2 - 9x - 2) = 0

⇒ x = 0 or 4x2 - 9x - 2 = 0

Now for 4x2 - 9x - 2 = 0 ⇒ x = 

⇒ we get three critical points of the given polynomial.

Now f(0) = 4 and f(1/2) = 1/16 - 3/8 -1/4 + 4

Now function is decreasing from 0 to 1.

Now f(2) = 16 - 24 - 4 + 4 = -8

Hence we get a one real roots in between 1 & 2.

Now f(3) > 0 and f(4) > f(3) 

Hence we get a one real roots in between 2 & 3.

Therefore we get two real roots in between  [1,4].

Hence option(3) is correct.