At a point MCQ Quiz in मल्याळम - Objective Question with Answer for At a point - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Definition:

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

Calculation:

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

$\mathrm{f}(\mathrm{x})=\{\begin{array}{cc}{\displaystyle \frac{1-\mathrm{x}}{2}};& \mathrm{x}\ne 0\\ \mathrm{K};& \mathrm{x}=0\end{array}$

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

${\displaystyle \underset{\mathrm{x}\to 0^{+}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to 0^{-}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{1-\mathrm{x}}{2}}={\displaystyle \frac{1-0}{2}}={\displaystyle \frac{1}{2}}}$

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

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

⇒ K = ${\displaystyle \frac{1}{2}}$.

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: $\mathrm{f}(\mathrm{x})=\{\begin{array}{ccc}bx-3& if& x\ge 3\\ 5x-9& if& x<3\end{array}$ is continuous at x =3.

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

⇒ 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
• $\underset{x\to a}{lim}f(x)$ exists
• $\underset{x\to a}{lim}f(x)$ = f(a)

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

All polynomial functions are continuous functions.

Calculation:

Statement I: f(x) is continuous only and only if $\underset{x\to a}{lim}f(x)$ 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
• $\underset{x\to a}{lim}f(x)$ exists
• $\underset{x\to a}{lim}f(x)$ = f(a)

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

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) = $\frac{p(x)}{q(x)}$ such that, 

$\underset{x\to a}{lim}p(x)=0and\underset{x\to a}{lim}q(x)=0$,

{$\underset{x\to a}{lim}f(x)=\frac{0}{0}$ form }

then, by L's hospital Rule,

$\underset{x\to a}{lim}f(x)=\underset{x\to a}{lim}\frac{p^{\prime }(x)}{q^{\prime }(x)}$

Calculation:

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

{ $\underset{x\to a}{lim}\frac{\left(x^{-1}-a^{-1}\right)}{x-a}=\frac{\left(a^{-1}-a^{-1}\right)}{a-a}=\frac{0}{0}$ form}

By L's Hospital Rule,

⇒ $\underset{x\to a}{lim}\frac{\left(x^{-1}-a^{-1}\right)}{x-a}=\underset{x\to a}{lim}\frac{{\left(x^{-1}-a^{-1}\right)}^{\prime }}{(x-a{)}^{\prime }}$

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

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

∴ The correct answer is option (4).

Concept:

Graph of f(x) = a^-x , a > 1

• Domain: $(-\mathrm{\infty },\mathrm{\infty })$
• Range: $(0,\mathrm{\infty })$
• 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 $(-\mathrm{\infty },-4)\cup (-4,4)\cup (4,\mathrm{\infty })$

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

⇒ 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, $\underset{x\to a^{-}}{lim}f(x)=\underset{x\to a^{+}}{lim}f(x)$, then, $\underset{x\to a}{lim}f(x)$ exists or the function is continuous at x = a.

• If, $\underset{x\to a^{-}}{lim}f(x)\ne \underset{x\to a^{+}}{lim}f(x)$, then, $\underset{x\to a}{lim}f(x)$ does not exist or the function is discontinuous at x = a.

Calculation:

LHL of f(x) at  x = 0 = $\underset{x\to 0^{-}}{lim}f(x)$

$\Rightarrow \underset{h\to 0}{lim}f(0-h)$

$\Rightarrow \underset{h\to 0}{lim}f(\frac{|-h|}{3(-h{)}^2-5(-h)})$

$\Rightarrow \underset{h\to 0}{lim}f(\frac{h}{3h^2+5h})$

$\Rightarrow \underset{h\to 0}{lim}f(\frac{1}{3h+5})$

$\Rightarrow \frac{1}{3(0)+5}$

⇒ 1/5

RHL of f(x) at  x = 0 = $\underset{x\to 0^{+}}{lim}f(x)$

$\Rightarrow \underset{h\to 0}{lim}f(0+h)$

$\Rightarrow \underset{h\to 0}{lim}f(\frac{\left|h\right|}{3h^2-5h})$

$\Rightarrow \underset{h\to 0}{lim}f(\frac{1}{3h-5})$

⇒ -1/5 

Hence, LHL ≠ RHL

$\Rightarrow \underset{x\to 0^{-}}{lim}f(x)\ne \underset{x\to 0^{+}}{lim}f(x)$

⇒ $\underset{x\to 0}{lim}f(x)$ does not exist.

Hence, f(x) is not continuous at x = 0 because $\underset{x\to 0}{lim}f(x)$ does not exist.

Concept:

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

The function is continuous if it satisfies the following conditions:

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

Calculation:

 f(x) = $\{\begin{array}{c}12\mathrm{x}+3\lambda ,\mathrm{x}\ne 1\\ 0,\mathrm{x}=1\end{array}$

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.

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

Calculation:

Given that,

$\mathrm{f}(\mathrm{x})=\mathrm{s}\mathrm{i}\mathrm{n}\left(\frac{1}{{\mathrm{x}}^2}\right)$           ----(1)

Statement: 1

$LHL={\displaystyle \underset{\mathrm{x}\to 0^{-}}{lim}\mathrm{s}\mathrm{i}\mathrm{n}(\frac{1}{{\mathrm{x}}^2})}$

⇒ $LHL={\displaystyle \underset{\mathrm{h}\to 0}{lim}\mathrm{s}\mathrm{i}\mathrm{n}[\frac{1}{(0-\mathrm{h}{)}^2}]=\mathrm{s}\mathrm{i}\mathrm{n}\infty }$     ----(2)

We know that, 

sin θ ∈  [ -1 , 1] 

⇒ sin (∞) is a definite value.

Similarly,

$RHL={\displaystyle \underset{\mathrm{h}\to 0}{lim}\mathrm{s}\mathrm{i}\mathrm{n}[\frac{1}{(0+\mathrm{h}{)}^2}]=\mathrm{s}\mathrm{i}\mathrm{n}\infty }$   ----(3)

According to the question,

f(0) = 0        ----(4)

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

Statement 1 is incorrect.

Statement: 2

$LHL={\displaystyle \underset{\mathrm{h}\to 0}{lim}\mathrm{s}\mathrm{i}\mathrm{n}[\frac{1}{(0-\frac{2}{\sqrt{\pi }}{)}^2}]=\mathrm{s}\mathrm{i}\mathrm{n}\frac{\pi }{4}}$

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

Similarly, 

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

And,

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

⇒ f($\frac{2}{\sqrt{\pi }}$) = $\frac{1}{\sqrt{2}}$       ----(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, $\underset{x\to a}{lim}f(x)$ 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: y = |2x - 4| 

⇒ $\mathrm{y}=\{\begin{array}{ccc}2x-4& ,& x\ge 2\\ 4-2x& ,& x<2\end{array}$

⇒ 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.