Continuity of a function MCQ Quiz in मल्याळम - Objective Question with Answer for Continuity of a function - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

For a function say f, $\underset{x\to a}{lim}f(x)$ exists

$\Rightarrow \underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)=l=\underset{x\to a}{lim}f(x)$, where l is a finite value.

Any function say f is said to be continuous at point say a if and only if $\underset{x\to a}{lim}f(x)=l=f\left(a\right)$, where l is a finite value.

Calculation:

Given: $\mathrm{f}(\mathrm{x})=\frac{{\mathrm{x}}^2-5\mathrm{x}-6}{{\mathrm{x}}^2+5\mathrm{x}-6}$

Here, we have to find the value of x for which f(x) is not continuous.

So, if any function is not continuous at x = a then $\underset{x\to a}{lim}f(x)=l\ne f\left(a\right)$

So, for the function f(x) if denominator is 0 at x = a then we can say that f(a) is infinite and limit cannot exist.

Let's find the value of x for which the denominator of f(x) is 0.

⇒ x² + 5x - 6 = 0

⇒ (x + 6) (x - 1) = 0.

⇒ x = -6, 1.

Hence, option 3 is correct.

Concept:

Continuity:

For a function say f, $\underset{x\to a}{lim}f(x)$ exists

 $\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)=l=\underset{x\to a}{lim}f(x)$

where l is a finite value.

Any function say f is said to be continuous at a point say 'a', if and only if:

$\underset{x\to a}{lim}f(x)=l=f\left(a\right)$

where l is a finite value.

Calculation:

$\underset{x\to 2}{lim}\frac{\sin \left(e^{x-2}-1\right)}{\log \left(x-1\right)}$

On substituting h = x – 2, we get:

$=\underset{h\to 0}{lim}\frac{\sin \left(e^h-1\right)}{\log \left(1+h\right)}$

This can be rearranged as:

$=\underset{h\to 0}{lim}\frac{\sin \left(e^h-1\right)}{e^h-1}\cdot \frac{e^h-1}{h}\cdot \frac{h}{\log \left(1+h\right)}$

= 1 ⋅ 1 ⋅ 1

= 1 and f(2) = k

∴ For the function to be continuous the value of the function f(x) at x = 2 must equal the limiting value of 1, i.e. k = 1

Concept:

Limit exists if LHL = RHL ⇔  $\underset{\mathrm{x}\to {\mathrm{a}}^{-}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to {\mathrm{a}}^{+}}{lim}\mathrm{f}(\mathrm{x})$

Calculation:

1. ${\displaystyle \underset{\mathrm{x}\to 0}{lim}\sin {\displaystyle \frac{1}{\mathrm{x}}}}$

⇒ -1 ≤ sin $\frac{1}{\mathrm{x}}$ ≤ 1

⇒ -1 ≤ ${\displaystyle \underset{\mathrm{x}\to 0}{lim}\sin {\displaystyle \frac{1}{\mathrm{x}}}}$ ≤ 1

Here, LHL ≠ RHL, so limit doesn't exist.

2. We have, ${\displaystyle \underset{\mathrm{x}\to 0}{lim}\mathrm{x}\cdot \mathrm{s}\mathrm{i}\mathrm{n}{\displaystyle \frac{1}{\mathrm{x}}}}$

Let, x = 1/t, if x → 0, t → 1/x = ∞ 

 $\therefore {\displaystyle \underset{\mathrm{t}\to \infty }{lim}\frac{1}{\mathrm{t}}\cdot \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{t}=0}$       (∵ something divide by ∞ = 0)

Here, LHL = RHL, so the limit exists

Hence, option (3) is correct.

The correct answer is option 4.

Given: $f(x)={\displaystyle \frac{x-|x|}{x}}$

Calculation:

⇒ $f(x)={\displaystyle \frac{x-|x|}{x}}$

For the value of x = 2

The function f(2) = ${\displaystyle \frac{2-|2|}{2}}$ = 0

For the value of x = 0; f(0) = ${\displaystyle \frac{0-|0|}{0}}$ = Impossible value

For the value of x = -2; f(-2) = ${\displaystyle \frac{-2-|-2|}{-2}}$ = 2

So, the function has some definite solution for all the values of x except x = 0.

Hence, the function is a continuous function for all the values of x except x = 0. 

Concept:

Differentiable Functions:

• If a graph has a sharp corner at a point, then the function is not differentiable at that point.
• If a graph has a break at a point, then the function is not differentiable at that point.
• If a graph has a vertical tangent line at a point, then the function is not differentiable at that point.

Calculation:

Givne that,

f(x) = |x| - 1     ----(1)

Step: 1

Step: 2 

Step: 3 

Clearly, we can see that,

f(x) is continuous at x = 1 and

f(x) is not differentiable at x = 0 because there is a corner at x = 0.

∴ Only statement 1 is correct.

Additional Information 

• Differentiable functions are those functions whose derivatives exist.
• If a function is differentiable, then it is continuous.
• If a function is continuous, then it is not necessarily differentiable.
• The graph of a differentiable function does not have breaks, corners, or cusps.

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I. $\underset{\mathrm{x}\to \mathrm{a}}{lim}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\frac{0}{0}$

II. $\underset{\mathrm{x}\to \mathrm{a}}{lim}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\frac{\mathrm{\infty }}{\mathrm{\infty }}$

Then we can apply L-Hospital Rule ⇔ $\underset{\mathbf{x}\to \mathbf{a}}{lim}\frac{\mathbf{f}\left(\mathbf{x}\right)}{\mathbf{g}\left(\mathbf{x}\right)}=\underset{\mathbf{x}\to \mathbf{a}}{lim}\frac{{\mathbf{f}}^{\prime }\left(\mathbf{x}\right)}{{\mathbf{g}}^{\prime }\left(\mathbf{x}\right)}$ 

Note: We have to differentiate both the numerator and denominator with respect to x unless and until $\underset{\mathrm{x}\to \mathrm{a}}{lim}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\mathrm{l}\ne \frac{0}{0}$ where l is a finite value.

A function f(x) is said to be continuous at a point x = a, in its domain if $\underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{a}\right)$ exists 

Calculation:

Given: $\mathrm{f}(\mathrm{x})={\displaystyle \frac{\sin \mathrm{x}}{\mathrm{x}}}$

To Find: f(0)

Function is continuous at x = 0

Therefore, f(0) = $\underset{\mathrm{x}\to 0}{lim}\mathrm{f}(\mathrm{x})$

$=\underset{\mathrm{x}\to 0}{lim}\frac{\sin \mathrm{x}}{\mathrm{x}}$          (Form 0/0)

Apply L-Hospital Rule,

$$\begin{gathered} =\underset{\mathrm{x}\to 0}{lim}\frac{\cos \mathrm{x}}{1} \\ =\frac{\cos 0}{1}=1 \end{gathered}$$

Concept:

f(x) is Continuous at x = a, if $\underset{\mathrm{x}\to {\mathrm{a}}^{-}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to {\mathrm{a}}^{+}}{lim}\mathrm{f}(\mathrm{x})=\mathrm{l}\mathrm{i}{\mathrm{m}}_{\mathrm{x}\to \mathrm{a}}\mathrm{f}(\mathrm{x})$ OR if derivative of function is defined 

If f(x) = |x| ⇔ f(x) = -x, for x < 0, and f(x) = x, for x > 0,

Calculation:

1 f(x) = ex

The derivative of the function is ex and it is defined everywhere from negative infinity to positive infinity and it doesn't take zero value at any point. So, ex is a continuous function.

2.

$\begin{array}{l}\mathrm{L}\mathrm{H}\mathrm{L}=\underset{\mathrm{x}\to 3^{-}}{lim}\mathrm{g}(\mathrm{x})=\underset{\mathrm{x}\to 3^{-}}{lim}(-(\mathrm{x}-3))=0\\ \mathrm{R}\mathrm{H}\mathrm{L}=\underset{\mathrm{x}\to 3^{+}}{lim}\mathrm{g}(\mathrm{x})=\underset{\mathrm{x}\to 3^{+}}{lim}\mathrm{x}-3=0\\ \underset{\mathrm{x}\to 3^{-}}{lim}\mathrm{g}(\mathrm{x})=\underset{\mathrm{x}\to 3^{+}}{lim}\mathrm{g}(\mathrm{x})\end{array}$

So, g(x) is continuous

Hence, option Both 1 and 2 correct.

Explanation:

Here, it is given that

\(f(x)= \begin{cases}\frac{x^2}{a}-a, & (0

Continuity at x = a, f(a) = 0

Hence, LHL = $\underset{x\to a^{-}}{lim}$ f(x) = $\underset{h\to 0}{lim}$ f(a - h) = $\underset{h\to 0}{lim}$ $\frac{(a-h{)}^2}{a}-a$

$=\underset{h\to 0}{lim}\frac{(a-h{)}^2-a^2}{a}=\frac{a^2-a^2}{a}=0$

and RHL = $\underset{x\to a^{+}}{lim}$ f(x) = $\underset{h\to 0}{lim}$ $a-\frac{a^3}{(a+h{)}^2}$

$=\underset{h\to 0}{lim}\frac{a(a+h{)}^2-a^3}{(a+h{)}^2}=\frac{a^3-a^3}{a^2}=0$

Hence, LHL = f(a) = RHL

Hence, f(x) is continuous.

Hence, f(x) is continuous on ] 0, ∞[

Now, we can check the differentiability at x = a

Lf'(a) = $\underset{h\to 0}{lim}\frac{f(a-h)-f(a)}{-h}$ 

$=\underset{h\to 0}{lim}\frac{\frac{(a-h{)}^2}{a}-a-0}{-h}$

$=\underset{h\to 0}{lim}\frac{(a-h{)}^2-a^2}{-ah}=\underset{h\to 0}{lim}\frac{a^2+h^2-2ah-a^2}{-ah}$

$=\underset{h\to 0}{lim}\frac{h(h-2a)}{-ah}=\frac{-2a}{-a}=2$

and $R{f}^{\prime }(a)=\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}$

$=\underset{h\to 0}{lim}\frac{a-\frac{a^3}{(a+h{)}^2}-0}{h}$

$=\underset{h\to 0}{lim}\frac{a(a+h{)}^2-a^3}{h(a+h{)}^2}$

$=\underset{h\to 0}{lim}\frac{a^3+a{h}^2+2a^{2}h-a^3}{h(a+h{)}^2}$

$=\underset{h\to 0}{lim}\frac{ah(h+2a)}{h(a+h{)}^2}=\underset{h\to 0}{lim}\frac{ah+2a^2}{(a+h{)}^2}=\frac{2a^2}{a^2}=2$

Hence, Lf' (a) = Rf' (a)

Hence, f(x) is differentiable at x = a.

Therefore, f(x) is differentiable at (0, ∞).

Concept:

• A function f(x) is continuous at x = a if $\underset{x\to a^{-}}{lim}$f(x)= $\underset{x\to a^{+}}{lim}$f(x)= f(a)
•  $\underset{x\to 0}{lim}$$\frac{sinx}{x}$=1

Explanation:

f(x)=$\{\begin{array}{ll}\frac{\sin (a+1)x+\sin 2x}{2x}& ,\text{if }x<0\\ b& ,\text{if }x=0\\ \frac{\sqrt{x+b{x}^3}-\sqrt{x}}{b{x}^{5/2}}& ,\text{if }x>0\end{array}$

Function f(x) is continuous at x= 0 if 

 $\underset{x\to 0^{-}}{lim}$f(x) = $\underset{x\to 0^{+}}{lim}$f(x) = f(0)  ---- (1)

Now $\underset{x\to 0^{-}}{lim}$f(x)= $\underset{x\to 0^{-}}{lim}$ $\frac{sin(a+1)\times x+sin2x}{2x}$

 = $\underset{x\to 0^{-}}{lim}$ $\frac{sin(a+1)\times x}{(a+1)x}\times \frac{(a+1)}{2}$+ $\underset{x\to 0^{-}}{lim}$ $\frac{sin2x}{2x}$

= $\frac{a+1}{2}$ + 1   ---- (2)

$\underset{x\to 0^{+}}{lim}$f(x)= $\underset{x\to 0^{+}}{lim}$ $\frac{\sqrt{x+b{x}^3}-\sqrt{x}}{b{x}^{\frac{5}{2}}}$

= $\underset{x\to 0^{+}}{lim}$$\frac{\sqrt{x+b{x}^3}-\sqrt{x}}{b{x}^{\frac{5}{2}}}$× $\frac{\sqrt{x+b{x}^3}+\sqrt{x}}{\sqrt{x+b{x}^3}+\sqrt{x}}$

= $\underset{x\to 0^{+}}{lim}$ $\frac{x+b{x}^3-x}{b{x}^{\frac{5}{2}}\times \sqrt{x}\times (\sqrt{1+b{x}^2}+1)}$   ----- [∵ (a+b) (a- b) = a² - b²)]

= $\underset{x\to 0^{+}}{lim}$$\frac{1}{\sqrt{1+b{x}^2}+1}$ = $\frac{1}{\sqrt{1+b\times 0^2}+1}$ =$\frac{1}{2}$  ----- (3)

f(0) = b  ----- (4)

From equations 1, 2, 3, and 4 we get 

$\frac{1}{2}=\frac{a+1}{2}+1=b$

⇒ b =$\frac{1}{2}$  and  

⇒ $\frac{1}{2}$ =$\frac{a+1}{2}+1$ ⇒ a = -2

∴ a = -2 and b = $\frac{1}{2}$

so a +  b = $\frac{-3}{2}$

The correct option is option (4).

Concept:

The function f(x) is continuous at an if all the below condition follows:

1. f(a) is real and finite.

2. $\underset{\alpha \to 0}{lim}\mathrm{f}(\mathrm{a}-\alpha )\text{ and }\underset{\alpha \to 0}{lim}\mathrm{f}(\mathrm{a}+\alpha )$ are real and finite.

3. $\underset{\alpha \to 0}{lim}\mathrm{f}(\mathrm{a}-\alpha )=\underset{\alpha \to 0}{lim}\mathrm{f}(\mathrm{a}+\alpha )=\mathrm{f}(\mathrm{a})$

Calculation:

f(x) = $\{\begin{array}{c}\frac{5}{2}-\mathrm{x}\mathrm{x}<2\\ 1\mathrm{x}=2\\ \mathrm{x}-\frac{3}{2}\mathrm{x}>2\end{array}$ 

$\underset{\alpha \to 0}{lim}\mathrm{f}(\mathrm{x}-\alpha )=\frac{5}{2}-\mathrm{x}$

For x = 2,

⇒ $\underset{\alpha \to 0}{lim}\mathrm{f}(2-\alpha )=\underset{\alpha \to 0}{lim}[\frac{5}{2}-(2-\alpha )]$

⇒ $\underset{\alpha \to 0}{lim}\mathrm{f}(2-\alpha )=\underset{\alpha \to 0}{lim}(\frac{1}{2}+\alpha )=\text{boldsymbol}\frac{1}{2}$

$\underset{\alpha \to 0}{lim}\mathrm{f}(\mathrm{x}+\alpha )=\mathrm{x}-\frac{3}{2}$

For x = 2,

⇒ $\underset{\alpha \to 0}{lim}\mathrm{f}(2+\alpha )=\underset{\alpha \to 0}{lim}[(2+\alpha )-\frac{3}{2}]$

⇒ $\underset{\alpha \to 0}{lim}\mathrm{f}(2+\alpha )=\underset{\alpha \to 0}{lim}(\frac{1}{2}+\alpha )=\text{boldsymbol}\frac{1}{2}$

f(x) = f(2) = 1

$\underset{\alpha \to 0}{lim}\mathrm{f}(2-\alpha )=\underset{\alpha \to 0}{lim}\mathrm{f}(2+\alpha )\ne \mathrm{f}(2)$

∴ The function is not continuous at x = 2