At a point MCQ Quiz in मराठी - Objective Question with Answer for At a point - मोफत PDF डाउनलोड करा

Concept:

Definition:

• A function f(x) is said to be continuous at a point x = a in its domain, if \(\rm \displaystyle \lim_{x\to a}f(x)\) exists or if its graph is a single unbroken curve at that point.
• f(x) is continuous at x = a ⇔ \(\rm \displaystyle \lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=\lim_{x\to a}f(x)=f(a)\).

Calculation:

For x ≠ 0, the given function can be re-written as:

\(\rm f(x) = \left\{ \begin{matrix} \rm \dfrac{1-x}{2}; & \rm x \ne 0 \\ \rm K; & \rm x = 0 \end{matrix}\right.\)

Since the equation of the function is same for x < 0 and x > 0, we have:

\(\rm \displaystyle \lim_{x\to 0^+}f(x)=\lim_{x\to 0^-}f(x)=\lim_{x\to 0}\dfrac{1-x}{2}= \dfrac{1-0}{2}=\dfrac{1}{2}\)

For the function to be continuous at x = 0, we must have:

\(\rm \displaystyle \lim_{x\to 0}f(x)=f(0)\)

⇒ K = \(\dfrac{1}{2}\).

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: \(\rm f(x) = \left\{\begin{matrix} bx-3 &if &x \geq 3 \\ 5x-9 & if & x< 3 \end{matrix}\right.\) is continuous at x =3.

AS we know that, if a function f is continuous at point say a then \(\mathop {\lim }\limits_{x \to a} f(x) = l = f\left( a \right)\)

⇒ LHL = lim_x→3^- (5x - 9) = (5 ⋅ 3) - 9 = 6.

⇒ RHL =  limx→3+ (bx - 3) = 3b - 3.

As function is continuous at x = 3 so, LHL = RHL.

⇒ 3b - 3 = 6 ⇒ 3b = 9

⇒ b = 3

Hence, option 2 is correct.

Concept:

A function f(x) is continuous at a certain point x = a only if it follows the following conditions. 

• f(a) exists
• \(\lim\limits_{x \to a} f(x)\) exists
• \(\lim\limits_{x \to a} f(x)\) = f(a)

The second condition can also be written as \(\lim\limits_{x \to a^-} f(x)=​​\lim\limits_{x \to a^+} f(x)\)

All polynomial functions are continuous functions.

Calculation:

Statement I: f(x) is continuous only and only if \(\lim\limits_{x \to a} f(x)\) exists.

We know that function f(x) is continuous at a certain point x = a only if it follows the following conditions. 

• f(a) exists
• \(\lim\limits_{x \to a} f(x)\) exists
• \(\lim\limits_{x \to a} f(x)\) = f(a)

The second condition can also be written as \(\lim\limits_{x \to a^-} f(x)=​​\lim\limits_{x \to a^+} f(x)\)

So, Statement I is incorrect.

Statement II: f(x) = x3 - x2 + 3x + 6 is not continuous at x = 0.

All polynomial functions are continuous functions.

So, Statement II is incorrect.

∴ None of the statements are correct.

Concept:

If f(x) = \({p(x) \over q(x)}\) such that, 

\(\mathop {\lim }\limits_{x \to a}p(x) = 0 \ and \ \mathop {\lim }\limits_{x \to a}q(x) =0\),

{\(\mathop {\lim }\limits_{x \to a}f(x) = {0\over 0}\) form }

then, by L's hospital Rule,

\(\mathop {\lim }\limits_{x \to a}f(x) = \mathop {\lim }\limits_{x \to a}{p'(x) \over q'(x)}\)

Calculation:

Given, \(\mathop {\lim }\limits_{x \to a} \frac{{\left( {{x^{ - 1}} - {a^{ - 1}}} \right)}}{{x - a}} \)

{ \(\mathop {\lim }\limits_{x \to a} \frac{{\left( {{x^{ - 1}} - {a^{ - 1}}} \right)}}{{x - a}} = \frac{{\left( {{a^{ - 1}} - {a^{ - 1}}} \right)}}{{a - a}} ={ 0\over0}\) form}

By L's Hospital Rule,

⇒ \(\mathop {\lim }\limits_{x \to a} \frac{{\left( {{x^{ - 1}} - {a^{ - 1}}} \right)}}{{x - a}} =\mathop {\lim }\limits_{x \to a} \frac{{\left( {{x^{ - 1}} - {a^{ - 1}}} \right)'}}{{(x - a)'}} \)

⇒ \(\mathop {\lim }\limits_{x \to a} \frac{{\left( {{x^{ - 1}} - {a^{ - 1}}} \right)}}{{x - a}} =\mathop {\lim }\limits_{x \to a} \frac{{\left( {(-1){x^{ - 2}} } \right)}}{{1}} \)

⇒ \(\mathop {\lim }\limits_{x \to a} \frac{{\left( {{x^{ - 1}} - {a^{ - 1}}} \right)}}{{x - a}} ={-1\over a^2}\)

∴ The correct answer is option (4).

Concept:

Graph of f(x) = a^-x , a > 1

• Domain: \((-\infty ,\infty )\)
• Range: \((0 ,\infty )\)
• y-intercept: (0, 1)
• Decreasing
• Continuous

Solution:

Statement I: The function f(x) = 2-x continuous at x = 0

Graph of 2^-x is

By graph the function f(x) = 2^-x is continuous at x = 0

∴ Statement I is correct.

Statement II: The function f(x) = \(\frac{5}{x^{4}-16}\) is continuous at all points of its domain.

Given function is f(x) = \(\frac{5}{x^{4}-16}\)

Domain of the function is \((-\infty ,-4)\cup (-4,4)\cup (4,\infty )\)

Given function is continuous at all points of its domain.

∴ Statement II is correct.

So, The correct option is (3)

Concept:

The function f(x) is continuous at x = a if

 f(a^-) = f(a) = f(a⁺)

Calculation:

Given, \(f\left( x \right) = \left\{ \begin{array}{l} 2x,x < 0\\ 2x + 1,x \ge 0.\, \end{array} \right.\)

⇒ f(0^-) = 2(0) = 0

f(0) = 2(0) + 1 = 1

and f(0⁺) = 2(0) + 1 = 1

So, f(0-) ≠ f(0) = f(0+)

⇒ f(x) is discontinuous at x = 0.

∴ The correct answer is option (3).

Concept:

• If, \(\lim_{x\rightarrow a^{-}}f(x)=\lim_{x\rightarrow a^{+}}f(x)\), then, \(\lim_{x\rightarrow a}f(x)\) exists or the function is continuous at x = a.

• If, \(\lim_{x\rightarrow a^{-}}f(x)≠\lim_{x\rightarrow a^{+}}f(x)\), then, \(\lim_{x\rightarrow a}f(x)\) does not exist or the function is discontinuous at x = a.

Calculation:

LHL of f(x) at  x = 0 = \(\lim_{x\rightarrow 0^{-}}f(x)\)

\( ⇒\lim_{h\rightarrow 0}f(0-h)\)

\(⇒\lim_{h\rightarrow 0}f(\frac{\left | -h \right |}{3(-h)^{2}-5(-h)})\)

\(⇒\lim_{h\rightarrow 0}f(\frac{h}{3h^{2}+5h})\)

\(⇒\lim_{h\rightarrow 0}f(\frac{1}{3h+5})\)

\(⇒\frac{1}{3(0)+5}\)

⇒ 1/5

RHL of f(x) at  x = 0 = \(\lim_{x\rightarrow 0^{+}}f(x)\)

\( ⇒\lim_{h\rightarrow 0}f(0+h)\)

\(⇒\lim_{h\rightarrow 0}f(\frac{\left | h \right |}{3h^{2}-5h})\)

\(⇒\lim_{h\rightarrow 0}f(\frac{1}{3h-5})\)

⇒ -1/5 

Hence, LHL ≠ RHL

\(⇒\lim_{x\rightarrow 0^{-}}f(x)≠\lim_{x\rightarrow 0^{+}}f(x)\)

⇒ \(\lim_{x\rightarrow 0}f(x)\) does not exist.

Hence, f(x) is not continuous at x = 0 because \(\lim_{x\rightarrow 0}f(x)\) does not exist.

Concept:

Let y = f(x) be a function. Then,

The function is continuous if it satisfies the following conditions:

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

Calculation:

 f(x) = \(\rm \left\{\begin{matrix} \rm 12x + 3 λ, \, x\ne1 \\ \rm 0, \, x=1 \end {matrix} \right.\)

Since, f(x) is continuous at x  =1 

⇒  lim_x→112x + 3λ = 0

⇒ 12(1) + 3λ = 0 ⇒ 12 + 3λ = 0 

⇒ λ = -4

∴ Option 1 is correctlimx→a−⁡f(x)=limx→a+⁡f(x)=f(a)" id="MathJax-Element-3-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">

Concept:

Let y = f(x) be a function. Then for a function, we say,

The function is continuous if it satisfies the following conditions.

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

Calculation:

Given that,

\(\rm f(x) = sin \left(\frac{1}{x^2}\right)\)           ----(1)

Statement: 1

\(LHL=\rm \displaystyle\lim_{x→0^-}sin(\frac{1}{x^2})\)

⇒ \(LHL=\rm \displaystyle\lim_{h→ 0}sin[\frac{1}{(0-h)^2}]=sin\ ∞\)     ----(2)

We know that, 

sin θ ∈  [ -1 , 1] 

⇒ sin (∞) is a definite value.

Similarly,

\(RHL=\rm \displaystyle\lim_{h→ 0}sin[\frac{1}{(0+h)^2}]=sin\ ∞\)   ----(3)

According to the question,

f(0) = 0        ----(4)

From equations (2), (3), (4),

Statement 1 is incorrect.

Statement: 2

\(LHL=\rm \displaystyle\lim_{h→ 0}sin[\frac{1}{(0-\frac{2}{\sqrt \pi})^2}]=sin\ \frac{\pi}{4}\)

⇒ LHL = \(\frac{1}{\sqrt 2}\)    ----(5)

Similarly, 

RHL =  \(\frac{1}{\sqrt 2}\)       -----(6)

And,

f(\(\frac{2}{\sqrt{\pi}}\)) = sin \(\rm \left ( \frac{1}{(\frac{2}{\sqrt{\pi}})^{2}} \right ) \)

⇒ f(\(\frac{2}{\sqrt{\pi}}\)) = \(\rm \frac{1}{\sqrt{2}} \)       ----(7)

From equation (5), (6), (7)

We can say that f(x) is continuous at x = \(\frac{2}{\sqrt{\pi}}\)

∴  Statement 2 is incorrect.

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: y = |2x - 4| 

⇒ \(\rm y=\left\{\begin{matrix} 2x-4 & , &x\geq 2 \\ 4-2x& , & x< 2 \end{matrix}\right.\)

⇒ LHL=  lim_x→2^- (4 - 2x) = 4 - 2 ⋅ 2 = 0.

⇒ RHL =  limx→2+ (2x - 4) = 2 ⋅ 2 - 4 = 0.

Hence limit exists at x = 2

y(2) = 2x - 4 = 2 ⋅ 2 - 4 = 0

⇒ LHL = RHL = y(2)

Hence function is continuous at x = 2.

Hence, option 1 is correct.