Continuity & Differentiability MCQ Quiz in मल्याळम - Objective Question with Answer for Continuity & Differentiability - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Let D = \(\mathbb Q\)

W√3 = √3 not contained in f(D) because f(D) = f(\(\mathbb Q\)) = \(\mathbb Q\)

Concept: 

If function is continuous and limit exist at the end points of the interval a and b then the function is uniformly continuous at (a, b)

Solution:

Here, a = 0 and b = 1 

Here \(f(x) = x^{\frac{-1}{2}}\) 

At x = 0, \(\lim_{x\rightarrow 0}\frac{1}{\sqrt x} = \infty\) 

so limit does not exist at end points 0

so, \(f(x) = \frac{1}{\sqrt x}\)  is not uniformly continuous at (0, 1)

Therefore, Correct option is Option 2).

Concept:

Let  D be a nonempty subset of  R. A function  f: D→R is called uniformly continuous on  D if for any ε > 0, there exists δ > 0 such that if u, v ∈ D and  |u − v| < δ, then |f(u) − f(v)| < ε

Explanation:

Let δ = ϵ then

||\((x_1, x_2, ..., x_n)-(y_1, y_2, ..., y_n)\)|| < δ

⇒ |x_i - y_i| < ϵ ∀ i ∈ \(\mathbb N\)

⇒ |Max{|xi|} - Max{|yi|}| < ϵ 

So f is uniformly continuous.

Option (4) is correct

Note: Here option (1) is also correct but we have to choose best option

Explanation:

Given 

f(0) = 0 and f(1)(0) = 0. So x = 0 is a repeated root.

Hence x² is factor of given polynomial.

Also f(-1) = 0 so (x + 1) is a factor of f.

let f(x) = x²(x  + 1)(αx + b)

Given f(1) = 1

So 1 = 2(α + b) ⇒ α + b = 1/2...(i)

Hence

f(x) = x2(x  + 1)(αx + b)

     = (x³ + x²)(αx + b)

    = αx⁴ + (α + b)x³ +bx²

   =  αx4 + (1/2)x3 + bx2 (by using (i))

∴ f'(x) = 4αx3 + (3/2)x2 + 2bx

⇒ f''(x) = 12αx² + 3x + 2b

⇒ f⁽³⁾(x) = 24αx + 3

So, f(3)(0) = 3

Option (4) is correct and option (3) is false

For α = -1 f(3)(1/2) = - 12 + 3 = - 9 < 0

Option (2) false.

As f(3)(0) = 3, there exists a ∈ (-1, 1) such that f(3)(a) ≥ 3

Option (1) is correct

Concept:

Continuous image of compact set is compact.

Continuous image of connected set is connected

Calculation:

Let f(x) = e^x

Here f(x) is continuous and f(\(\mathbb R\)) = (0, \(∈fty\))

f(\(\mathbb R\)) is not closed 

Hence option (1) is false

Let f: \(\mathbb R\) →  \(\mathbb R\) defined by f(x) = 1 for all x ∈ \(\mathbb R\)

let A = {1} which is bounded

Now, f^-1(A) = f^-1({1}) = \(\mathbb R\) which is unbounded  

f^-1(A) = R, which is unbounded 

Hence option (2) is false

we known that closed + bounded = compact 

and we know that Continuous image of compact set is compact.

So A is compact implies f(A) is compact

⇒ f(A) is closed and bounded

Hence option (3) is correct

 (4): A is bounded. So there exist m, M such that

m ≤ x ≤ M, ∀ x ∈ A

Now, [m, M] is closed ad bounded so it is compact

Hence f[m , M] is also compact 

f(A) is bounded 

Hence option (4) is correct

Explanation:

\(f(x)=\left\{\begin{array}{c} 1, \text { if } x=0 \\ 0, \text { if } x ∈ \mathbb{R} \backslash \mathbb{Q} \\ 3, \text { if } x=\frac{m}{n} \end{array}\right.\)

\(\lim_{x\to0}f(x)\) ≠ 1

So, f(x) is not continuous at x = 0

Option (2) and (4) false.

For, any irrational number f(x) = 0

so, continuous at for irrational number.

Option (1) is true.

Explanation:

 Given f(x) = ln(x) 

now, we have to find the value of 

\(lim_{m\rightarrow0}\) \(\left[\displaystyle\lim _{n \rightarrow 0} \frac{f(2+m+n)-f(2+m)-f(2+n)+f(2)}{m n}\right]\)

\(lim_{m\rightarrow0} \frac{1}{m}[lim_{n\rightarrow0} \frac{f(2+m+n) - f(2+m)}{n} - lim_{n\rightarrow0}\frac{f(2+n)-f(n)}{n}]\)

\(lim_{m\rightarrow 0}\frac{f'(2+m)-f'(2)}{m} = f''\)

Given f(x) = ln x

when we differentiate the given function we get,

f'(x) = \(\frac{1}{x}\)

Again differentiation ,

\(f"(x) = \frac{-1}{x^2}\)

At x = 2    ,  f" (2) = \(\frac{-1}{4}\) 

Therefore, correct answer is option 4 .

Explanation -

For option(1) -

We have  f be continuously differentiable 2 π periodic real valued function,

\(f'(a+ 2π ) = lim_{h → 0} \frac{f(a+2π+h) - f(a+ 2π )}{h}\)

= \(lim_{h → 0} \frac{f(a+h) - f(a)}{h} = f'(a)\)

which shows that f' is also a 2π periodic.

Hence Option(1) is true and Option(2) is false.

For option(3) -

Since f is continuous and 2 π periodic means it is bounded, being uniformly continuous. Let M  be bound on f, then

\(|a_n| \le \int_{-\pi}^{\pi} |f(t)cos(nt)| dt \le 2M \int_0^{\pi} |cos(nt)|dt =\frac{2M}{n}\int_0^{n \pi} |cosp| dp\)

Observe that \(\int_0^{n \pi} |cosp| dp\)  depends on n which means that no required C exists.

Hence Option(3) is false.

For Option(4) -

By Riemann Lebesgue lemma, If f be continuously differentiable 2 π periodic real valued function on the real line and \(a_n = \int_{-\pi}^{\pi} f(t) cos (nt) dt\)

then a_n → 0 as n → ∞ 

Hence Option(4) is false.

Concept:

(i) A function f(x) is uniformly continuous then it is continuous

(ii) If |f'(x)| ≤ K then f is uniformly continuous

Explanation:

(1): f(x) = sin\(\frac1x\) does not exist at x = 0 so not continuous at x = 0 so not uniformly continuous

Option (1) is false

(2): f(x) = (sin x)2

f'(x) = 2 sin x cos x = sin 2x

Here |f'(x)| = |sin 2x| < 1

⇒ f(x) =  (sin x)2 is uniformly continuous.

option (2) is true

(3) f(x) = x2 is uniformly continuous only on [-M, M] for fixed M > 0

Option (3) is false 

Explanation:

 Given f(x) = ln(x) 

now, we have to find the value of 

\(lim_{m\rightarrow0}\) \(\left[\displaystyle\lim _{n \rightarrow 0} \frac{f(2+m+n)-f(2+m)-f(2+n)+f(2)}{m n}\right]\)

\(lim_{m\rightarrow0} \frac{1}{m}[lim_{n\rightarrow0} \frac{f(2+m+n) - f(2+m)}{n} - lim_{n\rightarrow0}\frac{f(2+n)-f(n)}{n}]\)

\(lim_{m\rightarrow 0}\frac{f'(2+m)-f'(2)}{m} = f''\)

Given f(x) = ln x

when we differentiate the given function we get,

f'(x) = \(\frac{1}{x}\)

Again differentiation ,

\(f"(x) = \frac{-1}{x^2}\)

At x = 2    ,  f" (2) = \(\frac{-1}{4}\) 

Therefore, correct answer is option 4 .