At a point MCQ Quiz - Objective Question with Answer for At a point - Download Free PDF

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 except -4 & 4 

∴ Statement II is  not correct.

So, The correct option is (1)

Concept:

The limit of a function at exists only when its left hand limit and right hand limit exist and are equal and have a finite value.

Calculation:

Statement I: $\underset{x\to 0}{lim}\frac{1}{x}$ exists.

LHL = $\underset{x\to 0}{lim}-\frac{1}{h}$ = -∞ 

RHL = $\underset{x\to 0}{lim}+\frac{1}{h}$ = +∞ 

LHL ≠ RHL

So,  $\underset{x\to 0}{lim}\frac{1}{x}$ does not exists.

Statement I is incorrect. 

Statement I: $\underset{x\to 0}{lim}{e}^{\frac{1}{x}}$ does not exist.

LHL = $\underset{x\to 0}{lim}{e}^{\frac{1}{x}}$ = e^-∞ = 0

RHL = $\underset{x\to 0}{lim}{e}^{\frac{1}{x}}$ = e^∞ = ∞

LHL ≠ RHL

 $\underset{x\to 0}{lim}{e}^{\frac{1}{x}}$ does not exist.

Statement II is correct.

∴ Only Statement II is correct.

Calculation

$\underset{x\to 0^{-}}{lim}\frac{2}{x}\{\sin (k_1+1)x+\sin (k_2-1)x\}=4$

⇒ 2(k₁ + 1) + 2(k₂ – 1) = 4

⇒ k₁ + k₂ = 2 

⇒ $\underset{x\to 0^{+}}{lim}\frac{2}{x}\ln \left(\frac{2+k_{1}x}{2+k_{2}x}\right)=4$

⇒ $\underset{x\to 0^{+}}{lim}\frac{1}{x}\ln (1+\frac{(k_1-k_2)x}{2+k_{2}x})=2$

⇒ $\frac{{\mathrm{k}}_1-{\mathrm{k}}_2}{2}=2$

⇒ k₁ – k₂ = 4

∴ k₁ = 3, k₂ = – 1

${\mathrm{k}}_1^2+{\mathrm{k}}_2^2=9+\mathrm{l}=10$

Hence option 4 is correct

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:

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 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:

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

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

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

= ${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{2-\frac{{\sin }^{-1}\mathrm{x}}{\mathrm{x}}}{2+\frac{{\tan }^{-1}\mathrm{x}}{\mathrm{x}}}=\frac{2-1}{2+1}=\frac{1}{3}}$

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 = ${\displaystyle \frac{1}{3}}$.

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: $\mathrm{f}(\mathrm{x})=\frac{{\mathrm{x}}^2+\mathrm{x}-6}{{\mathrm{x}}^2+\mathrm{k}\mathrm{x}-3}$ is not continuous at x = 3.

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 = 3 then we can say that f(3) is infinite and limit cannot exist.

Let's find the value of k for which the denominator of f(x) is 0 for x = 3.

So, substitute x = 3 in x2 + kx - 3 = 0

⇒ 3² + 3k - 3 = 0.

⇒ 6 + 3k = 0.

⇒ k = - 2.

Hence, option 1 is correct.

Concept:

For the function to be continuous:

LHL = RHL = f(x)

Where LHL = $\underset{\alpha \to 0}{lim}$f(x - α) and RHL = $\underset{\alpha \to 0}{lim}$f(x + α)

Calculation:

Given that f(x) is continuous function

LHL = f(x) = RHL

$\underset{\alpha \to 0}{lim}$ f(1 - α) = f(1)

$\underset{\alpha \to 0}{lim}$ [a + b(1 - α)] = 5

$\underset{\alpha \to 0}{lim}$ [a + b - bα] = 5

a + b = 5

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: $\mathrm{f}(\mathrm{x})=\{\begin{array}{ccc}\frac{ksin(\pi -x)}{\pi -x}& if& x\ne \pi \\ 1& if& x=\pi \end{array}$ is continuous at x = π.

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

$\Rightarrow \underset{\mathrm{x}\to \pi }{lim}\frac{\mathrm{k}\mathrm{s}\mathrm{i}\mathrm{n}(\pi -\mathrm{x})}{\pi -\mathrm{x}}=\mathrm{f}(\pi )=1$

∵ $\underset{\mathrm{x}\to \pi }{lim}\frac{\mathrm{k}\mathrm{s}\mathrm{i}\mathrm{n}(\pi -\mathrm{x})}{\pi -\mathrm{x}}=\frac{0}{0}$, we can use L'Hospitals rule.

⇒ $\underset{\mathrm{x}\to \pi }{lim}\frac{-\mathrm{k}\mathrm{c}\mathrm{o}\mathrm{s}(\pi -\mathrm{x})}{-1}=1$

⇒ $\frac{\mathrm{k}\mathrm{c}\mathrm{o}\mathrm{s}(\pi -\pi )}{1}=1$ 

⇒ k cos(0) = 1

⇒ k = 1.

Hence, option 2 is correct.

Concept:

Greatest Integer Function:

Greatest Integer Function [x] indicates an integral part of the real number x which is a nearest and smaller integer to x. It is also known as floor of x

• In general, If n≤ x ≤ n+1 Then [x] = n     (n ∈ Integer)
• Means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

Calculation:

Given: 

f(x) =$\{\begin{array}{c}{\displaystyle \frac{\sin [\mathrm{x}]}{[\mathrm{x}]}},[\mathrm{x}]\ne 0\\ 0,[\mathrm{x}]=0\end{array}$

f(x) =$\{\begin{array}{c}{\displaystyle \frac{\sin (-1)}{-1}}=\sin 1,-1\le \mathrm{x}<0\\ 0,0\le \mathrm{x}<1\end{array}$

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

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

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

So, ${\displaystyle \underset{\mathrm{x}\to 0}{lim}\mathrm{f}(\mathrm{x})}$ doesn't exists

Concept:

Continuity of a Function:

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

Differentiability of a Function:

• A function f(x) is differentiable at a point x = a in its domain if its derivative is continuous at a.

  This means that f'(a) must exist, or equivalently: ${\displaystyle \underset{\mathrm{x}\to {\mathrm{a}}^{+}}{lim}{\mathrm{f}}^{\prime }(\mathrm{x})=\underset{\mathrm{x}\to {\mathrm{a}}^{-}}{lim}{\mathrm{f}}^{\prime }(\mathrm{x})=\underset{\mathrm{x}\to \mathrm{a}}{lim}{\mathrm{f}}^{\prime }(\mathrm{x})={\mathrm{f}}^{\prime }(\mathrm{a})}$.

Maxima/Minima:

• If f(x) has a local maximum or a local minimum at a point x = a, then it must be either a critical point [f'(a) = 0] or a point of non-differentiability.

​

Calculation:

Let us check for the continuity and differentiability (maxima/minima) of the function at x = 0.

Continuity:

$\mathrm{f}(\mathrm{x})=\{\begin{array}{cc}{\mathrm{x}}^2;& \mathrm{x}\le 0\\ 2\sin \mathrm{x};& \mathrm{x}>0\end{array}$

${\displaystyle \underset{\mathrm{x}\to 0^{-}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to 0^{-}}{lim}{\mathrm{x}}^2}$ = 0² = 0.

${\displaystyle \underset{\mathrm{x}\to 0^{+}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to 0^{+}}{lim}2\sin \mathrm{x}}$ = 2 sin 0 = 0.

f(0) = 0² = 0.

∵ ${\displaystyle \underset{\mathrm{x}\to 0^{-}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to 0^{+}}{lim}\mathrm{f}(\mathrm{x})}$ = f(0), the function f(x) is continuous at x = 0.

Differentiability:

${\mathrm{f}}^{\prime }(\mathrm{x})=\{\begin{array}{cc}2\mathrm{x};& \mathrm{x}\le 0\\ 2\cos \mathrm{x};& \mathrm{x}>0\end{array}$

${\displaystyle \underset{\mathrm{x}\to 0^{-}}{lim}{\mathrm{f}}^{\prime }(\mathrm{x})=\underset{\mathrm{x}\to 0^{-}}{lim}2\mathrm{x}}$ = 2×0 = 0.

${\displaystyle \underset{\mathrm{x}\to 0^{+}}{lim}{\mathrm{f}}^{\prime }(\mathrm{x})=\underset{\mathrm{x}\to 0^{-}}{lim}2\cos \mathrm{x}}$ = 2 cos 0 = 2.

∵ ${\displaystyle \underset{\mathrm{x}\to 0^{+}}{lim}{\mathrm{f}}^{\prime }(\mathrm{x})\ne \underset{\mathrm{x}\to 0^{-}}{lim}{\mathrm{f}}^{\prime }(\mathrm{x})}$, the function is not differentiable at x = 0.

Since, the function is not differentiable at x = 0, let us examine the possibility of maximum/minimum at the point.

The function f(x) = x² is strictly decreasing in (-∞, 0] and its minimum is 0² = 0 at x = 0.

The function f(x) = 2sin x is strictly increasing in $(0,{\displaystyle \frac{\pi }{2}}]$ and its minimum is 2 sin 0 = 0 at x = 0.

∴ The function f(x) has a local minimum at x = 0.

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:

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:

Greatest Integer Function: Greatest Integer Function [x] indicates an integral part of the real number x which is the nearest and smaller integer to x. It is also known as the floor of x

• In general, If n ≤ x ≤ n+1 Then [x] = n     (n ∈ Integer)
• Means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

Example:

Continuity: 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 or its graph is a single unbroken curve.

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

Calculation:

Given

⇒ $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{2}-1$   for 0 ≤  x ≤  π

$\Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}-1& 0\le x<2\\ 0& 2\le x\le \pi \end{array}\right\}$

$\Rightarrow \mathrm{t}\mathrm{a}\mathrm{n}\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}\mathrm{t}\mathrm{a}\mathrm{n}(-1)& 0\le \mathrm{x}<2\\ \mathrm{t}\mathrm{a}\mathrm{n}(0)& 2\le \mathrm{x}\le \pi \end{array}\right\}$

$\Rightarrow \underset{\mathrm{x}\to 2^{-}}{lim}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{f}(\mathrm{x})=-\mathrm{t}\mathrm{a}\mathrm{n}(1)$

$\Rightarrow \underset{\mathrm{x}\to 2^{+}}{lim}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{f}(\mathrm{x})=0$

$\underset{\mathrm{x}\to 2^{+}}{lim}\mathrm{f}\left(\mathrm{x}\right)\ne \underset{\mathrm{x}\to 2^{-}}{lim}\mathrm{f}\left(\mathrm{x}\right)$

So, tan f(x) is not continuous at x = 2

Concept:

• We say f(x) is continuous at x = c if

LHL = RHL = value of f(c) i.e., $\underset{\mathrm{x}\to {\mathrm{c}}^{-}}{lim}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {\mathrm{c}}^{+}}{lim}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{c}\right)$

• $\sqrt{{\mathrm{x}}^2}=\left|\mathrm{x}\right|$

• $\mathrm{f}\left(\mathrm{x}\right)=\left|\mathrm{x}\right|\{\begin{array}{c}-x,x<0\\ x,x\ge 0\end{array}$

Calculation:

Given: $\mathrm{f}\left(\mathrm{x}\right):\{\begin{array}{c}-\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2}},\mathrm{x}\ne 0\\ 0.\mathrm{x}=0\end{array}$ 

$\mathrm{f}\left(\mathrm{x}\right)=-\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2}}$

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=-\frac{\mathrm{x}}{\left|\mathrm{x}\right|}$

$\mathrm{L}\mathrm{H}\mathrm{L}=\underset{\mathrm{x}\to 0^{-}}{lim}\mathrm{f}\left(\mathrm{x}\right)=-\frac{\mathrm{x}}{-\mathrm{x}}=1$                      (∵ |x| = -x, if x < 0)

$\mathrm{R}\mathrm{H}\mathrm{L}=\underset{\mathrm{x}\to 0^{+}}{lim}\mathrm{f}\left(\mathrm{x}\right)=-\frac{\mathrm{x}}{\mathrm{x}}=-1$                    (∵ |x| = x, if x > 0)

Here, LHL ≠ RHL, so f(x) is discontinuous.