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<=100 where x ∈ R} is unbounded and Closed 

T = { x2 - 2x <67 where x ∈ (0, ∞)} is Open and bounded

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:

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

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

= $\frac{1}{3}$$\left[\begin{array}{r}1\\ 1\\ 1\end{array}\right]$+ $\frac{2}{6}$$\left[\begin{array}{r}-1\\ 2\\ -1\end{array}\right]$ = $\left(\begin{array}{l}0\\ 1\\ 0\end{array}\right)$

= $\frac{1}{3}$$\left[\begin{array}{r}1\\ 1\\ 1\end{array}\right]$+ $\frac{1}{3}$$\left[\begin{array}{r}-1\\ 2\\ -1\end{array}\right]$ = $\left(\begin{array}{l}0\\ 1\\ 0\end{array}\right)$

(2) correct

Concept -

(i)  If n is even then (-1)n = 1 

(ii)  If n is odd then (-1)n = -1

(iii)  $\frac{19}{e^3}<1$ then $(\frac{19}{e^3}{)}^n\to 0asn\to \mathrm{\infty }$

Explanation -

We have the sequence $a_n=e^{-n}+(-1{)}^{n}co{s}^3(\frac{19}{e^3}{)}^n+(-1{)}^n(sin(\frac{1}{n^2}+\frac{(-1{)}^n\pi }{2}))$

Now as n →  ∞ ,

a_n = 0 + (-1)ⁿ cos³(0) + (-1)ⁿ$sin(\frac{(-1{)}^n\pi }{2})$

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 $sin(\frac{\pi }{2})$ = 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 $sin(\frac{-\pi }{2})$ = -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 ${\mathbb{Q}}_8\to K_4$ 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 $T\left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}x+y\\ x-2y\end{array}\right)$

$C=[\left(\begin{array}{l}1\\ 2\end{array}\right),\left(\begin{array}{l}2\\ 1\end{array}\right)]$ 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 $\underset{x\to c}{lim}f(x)$ = $\underset{x\to c}{lim}g(x)$ = 0 or ± ∞ and g'(x) ≠ 0 for all x in I with x ≠ c and $\underset{x\to c}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$ exist then $\underset{x\to c}{lim}\frac{f(x)}{g(x)}$ = $\underset{x\to c}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Explanation:

$\underset{x\to 0}{lim}\frac{x(1-\cos x)-ax\sin x}{x^4}$ (0/0 form so using L'hospital rule)

= $\underset{x\to 0}{lim}\frac{xsinx+1-cosx-axcosx-asinx}{4x^3}$ 

= $\underset{x\to 0}{lim}\frac{1+(x-a)sinx-(ax+1)cosx}{4x^3}$

Again using L'hospital rule

= $\underset{x\to 0}{lim}\frac{(x-a)cosx+sinx+(ax+1)sinx-acosx}{12x^2}$

= $\underset{x\to 0}{lim}\frac{(x-2a)cosx+(ax+2)sinx}{12x^2}$

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 $\frac{sinmx}{x}$
 

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 $\underset{x\to 0}{lim}g(x)=1$

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, $\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}=\underset{h\to 0}{lim}f(x)\cdot g(h)=f(x)\cdot 1$

Since, by hypothesis $\underset{h\to 0}{lim}g(h)=1$

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.