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

Concept -

P - test - 

\(\sum \frac{1}{n^p}\) is convergent for p > 1

Explanation -

For option (ii) -

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

If \(\sum_{n=1}^{\infty} a_n^5 = \sum_{n=1}^{\infty} \frac{1}{n^5} \) is convergent by P - test.

but \(\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{1}{n}\) is divergent series 

Hence option(ii) is false.

For option(i) -

If \(\sum_{n=1}^{\infty} a_n \)  is convergent then \(\sum_{n=1}^{\infty} a_n^5 \) is also convergent for any convergent series.

Hence option(i) is true.

For option(iii) -

If \(\sum_{n=1}^{\infty} a_n^{\frac{3}{2}} \) is convergent then \(\sum_{n=1}^{\infty} a_n \) is either cgt or dgt as well

but in both cases, the series \(\sum_{n=1}^{\infty} \frac{ a_n}{n}\) is convergent.

Hence option(iii) is true.

Explanation:

\(X=\left[\begin{array}{rr}1 & -1 \\ 1 & 2 \\ 1 & -1\end{array}\right]\)

Let w₁ = \(\left[\begin{array}{r}1 \\ 1 \\ 1\end{array}\right]\) and w2 = \(\left[\begin{array}{r}-1 \\ 2 \\ -1\end{array}\right]\) and u = \(\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right) \)

then orthogonal projection of u on W is 

\(\frac{}{}\) w1 + \(\frac{}{}\)w2

= \(\frac13\)\(\left[\begin{array}{r}1 \\ 1 \\ 1\end{array}\right]\)+ \(\frac26\)\(\left[\begin{array}{r}-1 \\ 2 \\ -1\end{array}\right]\) = \(\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right) \)

= \(\frac13\)\(\left[\begin{array}{r}1 \\ 1 \\ 1\end{array}\right]\)+ \(\frac13\)\(\left[\begin{array}{r}-1 \\ 2 \\ -1\end{array}\right]\) = \(\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right) \)

(2) correct

Explanation -

Results -

(i) Number of homomorphism from \(\mathbb{Q}_8 \to K_4\) is 16.

(ii) Number of onto homomorphism from \(\mathbb{Q}_8 \to K_4\) is 6.

(iii) Number of 1-1 homomorphism from \(\mathbb{Q}_8 \to K_4\) is 0.

Hence option(2) is correct.

Explanation:

T: ℝ2 →ℝ2 be defined by \(T\left(\begin{array}{l} x \\ y \end{array}\right)=\left(\begin{array}{l} x+y \\ x-2 y \end{array}\right)\)

\(C=\left[\left(\begin{array}{l} 1 \\ 2 \end{array}\right),\left(\begin{array}{l} 2 \\ 1 \end{array}\right)\right]\) be a basis of ℝ2 

So, \(T\left(\begin{array}{l} 1 \\ 2 \end{array}\right)=\left(\begin{array}{l} 3 \\ -3 \end{array}\right)\) = \(-3\left(\begin{array}{l} 1 \\ 2 \end{array}\right)+3\left(\begin{array}{l} 2 \\ 1 \end{array}\right)\)

 \(T\left(\begin{array}{l} 2 \\ 1 \end{array}\right)=\left(\begin{array}{l} 3 \\ 0 \end{array}\right)\) = \(-1\left(\begin{array}{l} 1 \\ 2 \end{array}\right)+2\left(\begin{array}{l} 2 \\ 1 \end{array}\right)\)

So, matrix representation is

\(T[C]=\left[\begin{array}{rr} -3 & -1 \\ 3 & 2 \end{array}\right]\)

Option (3) is true and others are false

Concept:

L’Hospital’s Rule: If \(\lim_{x\to c}f(x)\) = \(\lim_{x\to c}g(x)\) = 0 or ± ∞ and g'(x) ≠ 0 for all x in I with x ≠ c and \(\lim_{x\to c}\frac{f'(x)}{g'(x)}\) exist then \(\lim_{x\to c}\frac{f(x)}{g(x)}\) = \(\lim_{x\to c}\frac{f'(x)}{g'(x)}\)

Explanation:

\(\lim_{x\to0}\frac{x(1-\cos x)-ax\sin x}{x^4}\) (0/0 form so using L'hospital rule)

= \(\lim_{x\to0}\frac{x sin x + 1 - cosx - ax cos x - asinx }{4x^3}\) 

= \(\lim_{x\to0}\frac{1 + (x-a) sin x - (ax + 1) cos x}{4x^3}\)

Again using L'hospital rule

= \(\lim_{x\to0}\frac{(x-a) cos x + sin x + (ax + 1) sin x - acos x}{12x^2}\)

= \(\lim_{x\to0}\frac{(x-2a) cos x + (ax + 2) sin x }{12x^2}\)

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) = \(\frac{siny}{x}\) (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 \(\frac{sin mx}{x}\)
 

We cannot eliminate m from the above function.

Hence limit does not exist.

\(\therefore\) 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 \(\lim _{x \rightarrow 0} g(x) = 1\)

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 \(\frac{f(x+h)-f(x)}{h}= \frac{f(x)[f(h) - 1]}{h}\)

                      = \( \frac{f(x)h. g(h)}{h}=f(x). g(h)\) (using (ii))

Hence, \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} f(x) \cdot g(h)=f(x) \cdot 1\)

Since, by hypothesis \(\lim_{h \rightarrow 0} g(h) = 1\)

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

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

and f'(x) = f(x) 

⇒ \(\frac{d}{dx} f(x) = f(x)\)

(2) is true.

Concept -

If f : [a,b] → \(\mathbb{R}\) and f(a) > 0 and f(b) < 0 then there exist c ∈ (a,b) such that f(c) = 0

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 = \(\frac{9\pm\sqrt{81+ 32}}{8}= \frac{9\pm\sqrt{113}}{8}\)

⇒ we get three critical points of the given polynomial.

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

Now function is decreasing from 0 to 1.

Now f(2) = 16 - 24 - 4 + 4 = -8 < 0

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 

\(e = \sum_{k=0}^n \frac{1}{k!} + \sum_{k=n+1}^{\infty} \frac{1}{k!} \)

⇒ \(2 \pi e n! = 2\pi r + 2\pi n!( \sum_{k=n+1}^{\infty} \frac{1}{k!} )\)

Where r is positive integer. so we have

\(lim_{ n \to \infty } n sin( 2\pi r + 2\pi n!( \sum_{k=n+1}^{\infty} \frac{1}{k!} ))\)

= \(lim_{ n \to \infty } n \ sin( 2\pi n!( \sum_{k=n+1}^{\infty} \frac{1}{k!} ))\)

Further, observe that 

\(\frac{1}{n+1} < ​​n! ( \sum_{k=n+1}^{\infty}\frac{1}{k!}) = \frac{1}{n+1}+ \frac{1}{(n+1)(n+2)}+..... < \frac{1}{n}+ \frac{1}{n^2} + ... = \frac{1}{n-1}\)

By squeeze principle, we have 

\(lim_{n \to \infty }​​n! ( \sum_{k=n+1}^{\infty}\frac{1}{k!}) = lim_{n \to \infty } b_n = 0\) and \(lim_{n \to \infty }n b_n = 1\)

So using the result that \(lim_{x \to 0 } \frac{sinx}{x} = 1\) we get 

\(lim_{n \to \infty } a_n = ​​lim_{n \to \infty } n sin(2 \pi b_n) = ​​lim_{n \to \infty }2 \pi b_n n \frac{sin(2 \pi b_n)}{2 \pi b_n} = 2\pi\)

Hence Option(3) is correct.