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

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

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

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. 

Explanation -

Let a_n = n sin(2 π en!) we have 

⇒ 

Where r is positive integer. so we have

= 

Further, observe that 

By squeeze principle, we have 

 and 

So using the result that  we get 

Hence Option(3) is correct.

Concept -

Reimann Lebesgue Lemma -

If the Lebesgue Integral of |f| is finite then the fourier transform of |f| vanishes as its argument does to infinity.

Explanation -

We have the sequence  

Note that f(x) being continuous on a compact set is bounded and |sin t | ≤ 1

Therefore     ∀ n  where M is bound on f(x).

Thus the sequence {b_n} is bounded.

integration by parts, we get 

b_n  = 

Since f'(t) is continuous then by Reimann Lebesgue Lemma, which is " If the Lebesgue Integral of |f| is finite then the fourier transform of |f| vanishes as its argument does to infinity. "

Thus in particular, b_n and n b_n → 0 as n → ∞ 

Hence option (1) and (2) is correct.

For option(iii) -

 is also absolutely convergent because b_n being bounded and and cgs to 0.

Hence the option (3) is correct. 

Hence option(4) is the correct option.

Concept -

(i) ∑ |a_n | is convergent then ∑ an is absolutely convergent.

(ii) Ratio Test - 

If  then the series  ∑ an is convergent.

Explanation -

We have the series  

Now for Absolutely convergent -

Now using Ratio Test -

Hence the series  is convergent and the given series is absolutely convergent.

Hence Option (i) is true.