Continuity of a function MCQ Quiz in मराठी - Objective Question with Answer for Continuity of a function - मोफत PDF डाउनलोड करा

Concept:

For a function say f, \(\mathop {\lim }\limits_{x \to a} f(x)\) exists

\( ⇒ \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \;\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = l = \;\mathop {\lim }\limits_{x \to a} f(x)\), where l is a finite value.

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

Calculation:

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

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 \(\mathop {\lim }\limits_{x \to a} f(x) = l \neq f\left( a \right)\)

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, \(\mathop {\lim }\limits_{x \to a} f(x)\) exists

 \( \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \;\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = l = \;\mathop {\lim }\limits_{x \to a} f(x)\)

where l is a finite value.

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

\(\mathop {\lim }\limits_{x \to a} f(x) = l = f\left( a \right)\)

where l is a finite value.

Calculation:

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

On substituting h = x – 2, we get:

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

This can be rearranged as:

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

= 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 ⇔  \(\rm\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)\)

Calculation:

1. \(\displaystyle\lim_{\rm x\rightarrow 0} \sin \dfrac{1}{\rm x}\)

⇒ -1 ≤ sin \(\rm \frac{1}{x}\) ≤ 1

⇒ -1 ≤ \(\displaystyle\lim_{\rm x\rightarrow 0} \sin \dfrac{1}{\rm x}\) ≤ 1

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

2. We have, \(\displaystyle\lim_{\rm x→ 0} \rm x \cdot sin \dfrac{1}{\rm x} \)

Let, x = 1/t, if x → 0, t → 1/x = ∞ 

 \(\therefore \displaystyle\lim_{\rm t→ ∞} \rm \frac{1}{t} \cdot sin \ t =0\)       (∵ something divide by ∞ = 0)

Here, LHL = RHL, so the limit exists

Hence, option (3) is correct.

The correct answer is option 4.

Given: \(f(x)=\dfrac{x-|x|}{x}\)

Calculation:

⇒ \(f(x)=\dfrac{x-|x|}{x}\)

For the value of x = 2

The function f(2) = \(\dfrac{2-|2|}{2}\) = 0

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

For the value of x = -2; f(-2) = \(\dfrac{-2-|-2|}{-2}\) = 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. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{0}{0}\)

II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule ⇔ \(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \;\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\) 

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

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

Calculation:

Given: \(\rm f(x)=\dfrac{\sin x}{x}\)

To Find: f(0)

Function is continuous at x = 0

Therefore, f(0) = \( \rm \lim_{x\rightarrow 0} f(x)\)

\(= \rm \lim_{x\rightarrow 0} \frac{\sin x}{x}\)          (Form 0/0)

Apply L-Hospital Rule,

\(= \rm \lim_{x\rightarrow 0} \frac{\cos x}{1}\\=\frac{\cos 0}{1} = 1\)

Concept:

f(x) is Continuous at x = a, if \(\rm\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=lim _{x \rightarrow a} f(x)\) 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.

\(\begin{array}{l} \rm LHL=\lim _{x \rightarrow 3^{-}} g(x)=\lim _{x \rightarrow 3^{-}}(-(x-3))=0 \\ \rm RHL=\lim _{x \rightarrow 3^{+}} g(x)=\lim _{x \rightarrow 3^{+}} x-3=0 \\ \rm \lim _{x \rightarrow 3^{-}} g(x) =\lim _{x \rightarrow 3^{+}} g(x) \end{array}\)

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 = \(\lim _{x \rightarrow a^{-}}\) f(x) = \(\lim _{h \rightarrow 0} \) f(a - h) = \(\lim _{h \rightarrow 0}\) \(\frac{(a-h)^2}{a}-a\)

\(=\lim _{h \rightarrow 0} \frac{(a-h)^2-a^2}{a}=\frac{a^2-a^2}{a}=0\)

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

\(=\lim _{h \rightarrow 0} \frac{a(a+h)^2-a^3}{(a+h)^2}=\frac{a^3-a^3}{a^2}=0\)

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) = \(\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}\) 

\(=\lim _{h \rightarrow 0}\frac{ \frac{(a-h)^2}{a}-a-0 }{-h}\)

\(=\lim _{h \rightarrow 0} \frac{(a-h)^2-a^2}{-a h}=\lim _{h \rightarrow 0} \frac{a^2+h^2-2 a h-a^2}{-a h} \)

\(=\lim _{h \rightarrow 0} \frac{h(h-2 a)}{-a h}=\frac{-2 a}{-a}=2\)

and \(R f^{\prime}(a) =\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\)

\( =\lim _{h \rightarrow 0} \frac{a-\frac{a^3}{(a+h)^2}-0}{h}\)

\(=\lim _{h \rightarrow 0} \frac{a(a+h)^2-a^3}{h(a+h)^2}\)

\(=\lim _{h \rightarrow 0} \frac{a^3+a h^2+2 a^2 h-a^3}{h(a+h)^2}\)

\(=\lim _{h \rightarrow 0} \frac{a h(h+2 a)}{h(a+h)^2}=\lim _{h \rightarrow 0} \frac{a h+2 a^2}{(a+h)^2}=\frac{2 a^2}{a^2}=2\)

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 \(\lim_{x\rightarrow a^-}\)f(x)= \(\lim_{x\rightarrow a^+}\)f(x)= f(a)
•  \(\lim_{x\rightarrow 0} \)\(\frac{sinx}{x} \)=1

Explanation:

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

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

 \(\lim_{x\rightarrow 0^-}\)f(x) = \(\lim_{x\rightarrow 0^+}\)f(x) = f(0)  ---- (1)

Now \(\lim_{x\rightarrow 0^-}\)f(x)= \(\lim_{x\rightarrow 0^-}\) \(\frac{ sin (a+1)× x + sin 2x}{2x}\)

 = \(\lim_{x\rightarrow 0^-}\) \(\frac{ sin (a+1)× x}{(a+1)x}× \frac{ (a+1)}{2}\)+ \(\lim_{x\rightarrow 0^-}\) \(\frac{sin2x}{2x} \)

= \(\frac{a+1}{2} \) + 1   ---- (2)

\(\lim_{x\rightarrow 0^+}\)f(x)= \(\lim_{x\rightarrow 0^+}\) \(\frac{\sqrt{x+bx^{3}}-\sqrt{x}}{bx^{\frac{5}{2}}}\)

= \(\lim_{x\rightarrow 0^+}\)\(\frac{\sqrt{x+bx^{3}}-\sqrt{x}}{bx^{\frac{5}{2}}}\)× \(\frac{\sqrt{x+bx^{3}}+\sqrt{x}}{\sqrt{x+bx^{3}}+\sqrt{x}}\)

= \(\lim_{x\rightarrow 0^+}\) \(\frac{x+bx^{3}-x}{bx^{\frac{5}{2}}×\sqrt{x}×(\sqrt{1+bx^{2}}+1)} \)   ----- [∵ (a+b) (a- b) = a² - b²)]

= \(\lim_{x\rightarrow 0^+}\)\(\frac{1}{\sqrt{1+bx^{2}}+1}\) = \(\frac{1}{\sqrt{1+b\times0^{2}}+1} \) =\(\frac{1}{2} \)  ----- (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. \(\rm \lim_{\alpha \to 0}f(a-\alpha)\text{ and }\lim_{\alpha \to 0}f(a+\alpha)\) are real and finite.

3. \(\rm \lim_{\alpha \to 0}f(a-\alpha)=\lim_{\alpha \to 0}f(a+\alpha)=f(a)\)

Calculation:

f(x) = \(\left\{ {\begin{array}{*{20}{c}} {\frac{5}{2} - \rm x\;\;\;\;\;\;x < 2}\\ {1\;\;\;\;\;\;\;\;\;\;\;\;\;\rm x = 2}\\ {\rm x - \;\frac{3}{2}\;\;\;\;\;x > 2} \end{array}} \right.\) 

\(\rm \lim_{\alpha \to 0}f(x-\alpha)={\frac{5}{2} - \rm x}\)

For x = 2,

⇒ \(\rm \lim_{\alpha \to 0}f(2-\alpha)=\lim_{\alpha \to 0}\left[\frac{5}{2} - (2-\alpha)\right]\)

⇒ \(\rm \lim_{\alpha \to 0}f(2-\alpha)=\rm \lim_{\alpha \to 0}\left(\frac{1}{2}+\alpha\right) =\boldsymbol{\rm \frac{1}{2}}\)

\(\rm \lim_{\alpha \to 0}f(x+\alpha)=x-\frac{3}{2}\)

For x = 2,

⇒ \(\rm \lim_{\alpha \to 0}f(2+\alpha)=\lim_{\alpha \to 0}\left[(2+\alpha)-\frac{3}{2}\right]\)

⇒ \(\rm \lim_{\alpha \to 0}f(2+\alpha)=\rm \lim_{\alpha \to 0}\left(\frac{1}{2}+\alpha\right) = \boldsymbol{\rm \frac{1}{2}}\)

f(x) = f(2) = 1

\(\rm \lim_{\alpha \to 0}f(2-\alpha)=\lim_{\alpha \to 0}f(2+\alpha)\neq f(2)\)

∴ The function is not continuous at x = 2