Continuity Sums for 12th Students (Problems 1 to 20)
Comprehensive guide covering all sections of continuity problems. Below are the detailed solutions for questions 1 through 20.
Problem 1: Check the continuity of the function $f(x) = 2x^2 - 1$ at $x = 3$.
- A) Discontinuous
- B) Continuous at $x = 3$ ✓ Correct
- C) Limit does not exist
- D) None of the above
Solution:
First, find the value of the function at $x = 3$:
$$ f(3) = 2(3)^2 - 1 = 2(9) - 1 = 17 $$
Next, evaluate the limit as $x \to 3$:
$$ \lim_{x \to 3} f(x) = \lim_{x \to 3} (2x^2 - 1) = 2(3)^2 - 1 = 17 $$
Since $\lim_{x \to 3} f(x) = f(3)$, the function is continuous at $x = 3$.
Problem 2: Examine the continuity of the function $f(x) = 5x - 3$ at $x = 0$.
- A) Discontinuous at $x = 0$
- B) Continuous at $x = 0$ ✓ Correct
- C) Continuous everywhere except $x = 0$
- D) Limit approaches infinity
Solution:
Find the value of the function at $x = 0$:
$$ f(0) = 5(0) - 3 = -3 $$
Now, find the limit of the function as $x \to 0$:
$$ \lim_{x \to 0} (5x - 3) = 5(0) - 3 = -3 $$
Because $\lim_{x \to 0} f(x) = f(0)$, the function is continuous at $x = 0$. Since it is a polynomial, it is continuous for all real numbers.
Problem 3: Discuss the continuity of the modulus function $f(x) = |x|$ at $x = 0$.
- A) Continuous at $x = 0$ ✓ Correct
- B) Discontinuous at $x = 0$
- C) Undefined at $x = 0$
- D) Continuous only for $x > 0$
Solution:
The modulus function is defined as:
$$ f(x) = \begin{cases} -x, & x < 0 \\ x, & x \ge 0 \end{cases} $$
Value at $x = 0$: $f(0) = 0$.
Left Hand Limit (LHL): $\lim_{x \to 0^-} (-x) = 0$
Right Hand Limit (RHL): $\lim_{x \to 0^+} (x) = 0$
Since LHL = RHL = $f(0)$, $f(x)$ is continuous at $x = 0$.
Problem 4: Check the continuity of the Greatest Integer Function $f(x) = [x]$ at $x = 1$.
- A) Continuous at $x = 1$
- B) Discontinuous at $x = 1$ ✓ Correct
- C) Limit exists but $f(1)$ is undefined
- D) None of the above
Solution:
Value of the function at $x = 1$: $f(1) = [1] = 1$.
Calculate LHL at $x = 1$ ($x \to 1^-$). For values slightly less than 1 (e.g., 0.99), the greatest integer is 0.
$$ \lim_{x \to 1^-} [x] = 0 $$
Calculate RHL at $x = 1$ ($x \to 1^+$). For values slightly greater than 1 (e.g., 1.01), the greatest integer is 1.
$$ \lim_{x \to 1^+} [x] = 1 $$
Since LHL $\neq$ RHL, the limit does not exist. The function is discontinuous at $x = 1$.
Problem 5: For what value of $k$ is the following function continuous at $x = 5$?
$$ f(x) = \begin{cases} \frac{x^2 - 25}{x - 5}, & x \neq 5 \\ k, & x = 5 \end{cases} $$
- A) $k = 5$
- B) $k = 10$ ✓ Correct
- C) $k = 0$
- D) $k = 25$
Solution:
For $f(x)$ to be continuous at $x = 5$, $\lim_{x \to 5} f(x) = f(5)$.
Evaluate the limit:
$$ \lim_{x \to 5} \frac{x^2 - 25}{x - 5} = \lim_{x \to 5} \frac{(x - 5)(x + 5)}{x - 5} $$
Cancel out the common factor $(x - 5)$ since $x \to 5$ implies $x \neq 5$:
$$ = \lim_{x \to 5} (x + 5) = 5 + 5 = 10 $$
Equating the limit to $f(5)$: $k = 10$.
Problem 6: Find all points of discontinuity for the function defined by:
$$ f(x) = \begin{cases} x + 5, & x \le 1 \\ x - 5, & x > 1 \end{cases} $$
- A) No points of discontinuity
- B) Discontinuous at $x = 5$
- C) Discontinuous at $x = 1$ ✓ Correct
- D) Discontinuous everywhere
Solution:
The function is defined by linear polynomials for $x < 1$ and $x > 1$, which are continuous. The only possible point of discontinuity is at $x = 1$.
Value at $x = 1$: $f(1) = 1 + 5 = 6$.
LHL ($x \to 1^-$): $\lim_{x \to 1^-} (x + 5) = 1 + 5 = 6$.
RHL ($x \to 1^+$): $\lim_{x \to 1^+} (x - 5) = 1 - 5 = -4$.
Since LHL $\neq$ RHL ($6 \neq -4$), the function is discontinuous exactly at $x = 1$.
Problem 7: Determine the continuity of $f(x)$ at $x = 0$:
$$ f(x) = \begin{cases} \frac{|x|}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases} $$
- A) Continuous at $x = 0$
- B) Discontinuous at $x = 0$ ✓ Correct
- C) LHL equals RHL but not $f(0)$
- D) Function is not defined properly
Solution:
We evaluate the left-hand and right-hand limits at $x = 0$.
For LHL ($x < 0$), $|x| = -x$. Thus, $f(x) = \frac{-x}{x} = -1$.
$$ \lim_{x \to 0^-} f(x) = -1 $$
For RHL ($x > 0$), $|x| = x$. Thus, $f(x) = \frac{x}{x} = 1$.
$$ \lim_{x \to 0^+} f(x) = 1 $$
Since LHL $\neq$ RHL, the limit does not exist, and $f(x)$ is discontinuous at $x = 0$.
Problem 8: Find the value of $k$ so that $f(x)$ is continuous at $x = 2$.
$$ f(x) = \begin{cases} kx^2, & x \le 2 \\ 3, & x > 2 \end{cases} $$
- A) $k = \frac{4}{3}$
- B) $k = \frac{3}{4}$ ✓ Correct
- C) $k = 3$
- D) $k = 12$
Solution:
For the function to be continuous at $x = 2$, LHL = RHL = $f(2)$.
LHL ($x \to 2^-$): $\lim_{x \to 2^-} kx^2 = k(2)^2 = 4k$.
RHL ($x \to 2^+$): $\lim_{x \to 2^+} 3 = 3$.
Value at $x = 2$: $f(2) = k(2)^2 = 4k$.
Equating LHL and RHL:
$$ 4k = 3 \implies k = \frac{3}{4} $$
Problem 9: Find the value of $k$ if $f(x)$ is continuous at $x = \pi$.
$$ f(x) = \begin{cases} kx + 1, & x \le \pi \\ \cos x, & x > \pi \end{cases} $$
- A) $k = \frac{2}{\pi}$
- B) $k = -\frac{2}{\pi}$ ✓ Correct
- C) $k = \pi$
- D) $k = -2$
Solution:
For $f(x)$ to be continuous at $x = \pi$, LHL = RHL = $f(\pi)$.
LHL ($x \to \pi^-$): $\lim_{x \to \pi^-} (kx + 1) = k\pi + 1$.
RHL ($x \to \pi^+$): $\lim_{x \to \pi^+} (\cos x) = \cos \pi = -1$.
Equating the two limits:
$$ k\pi + 1 = -1 \implies k\pi = -2 \implies k = -\frac{2}{\pi} $$
Problem 10: Check the continuity of the function $f(x)$ at $x = 0$.
$$ f(x) = \begin{cases} \frac{1 - \cos 4x}{x^2}, & x \neq 0 \\ 8, & x = 0 \end{cases} $$
- A) Discontinuous, limit does not exist
- B) Discontinuous, limit is 4
- C) Continuous at $x = 0$ ✓ Correct
- D) None of the above
Solution:
Evaluate the limit of $f(x)$ as $x \to 0$:
$$ \lim_{x \to 0} \frac{1 - \cos 4x}{x^2} $$
Using $1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right)$:
$$ \lim_{x \to 0} \frac{2 \sin^2 2x}{x^2} = 2 \lim_{x \to 0} \left( \frac{\sin 2x}{x} \right)^2 $$
Multiply and divide inside the bracket by 2:
$$ 2 \lim_{x \to 0} \left( \frac{\sin 2x}{2x} \cdot 2 \right)^2 = 2 (1 \cdot 2)^2 = 2(4) = 8 $$
Since $\lim_{x \to 0} f(x) = 8 = f(0)$, the function is continuous at $x = 0$.
Problem 11: Find the value of $k$ so that the function $f(x)$ is continuous at $x = \frac{\pi}{2}$.
$$ f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ 3, & x = \frac{\pi}{2} \end{cases} $$
- A) $k = 3$
- B) $k = 6$ ✓ Correct
- C) $k = \frac{3}{2}$
- D) $k = 0$
Solution:
Let $x = \frac{\pi}{2} - h$, where $h \to 0$ as $x \to \frac{\pi}{2}$.
$$ \lim_{h \to 0} \frac{k \cos\left(\frac{\pi}{2} - h\right)}{\pi - 2\left(\frac{\pi}{2} - h\right)} $$
Using $\cos(\frac{\pi}{2} - h) = \sin h$:
$$ = \lim_{h \to 0} \frac{k \sin h}{\pi - \pi + 2h} = \lim_{h \to 0} \frac{k \sin h}{2h} $$
$$ = \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h} = \frac{k}{2} (1) = \frac{k}{2} $$
Given $f\left(\frac{\pi}{2}\right) = 3$. Equating the limit:
$$ \frac{k}{2} = 3 \implies k = 6 $$
Problem 12: Determine the relationship between $a$ and $b$ so that the function $f(x)$ is continuous at $x = 3$.
$$ f(x) = \begin{cases} ax + 1, & x \le 3 \\ bx + 3, & x > 3 \end{cases} $$
- A) $a = b + 1$
- B) $a = 3b + 2$
- C) $a = b + \frac{2}{3}$ ✓ Correct
- D) $a = b - \frac{2}{3}$
Solution:
For the function to be continuous at $x = 3$, LHL = RHL.
LHL ($x \to 3^-$): $\lim_{x \to 3^-} (ax + 1) = 3a + 1$
RHL ($x \to 3^+$): $\lim_{x \to 3^+} (bx + 3) = 3b + 3$
Equating the LHL and RHL:
$$ 3a + 1 = 3b + 3 $$
$$ 3a - 3b = 2 \implies a - b = \frac{2}{3} \implies a = b + \frac{2}{3} $$
Problem 13: Examine the continuity of the function $f(x)$ at $x = 0$.
$$ f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases} $$
- A) Discontinuous because $\sin(1/x)$ oscillates
- B) Continuous at $x = 0$ ✓ Correct
- C) Limit evaluates to $\infty$
- D) Continuous everywhere except $x = 0$
Solution:
We must evaluate $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$.
We know that for any $x \neq 0$, the sine function is bounded:
$$ -1 \le \sin\left(\frac{1}{x}\right) \le 1 $$
Multiplying the entire inequality by $x^2$ (which is strictly positive for $x \neq 0$):
$$ -x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2 $$
Taking the limit as $x \to 0$ for the bounds:
$$ \lim_{x \to 0} (-x^2) = 0 \quad \text{and} \quad \lim_{x \to 0} (x^2) = 0 $$
By the Squeeze (Sandwich) Theorem, $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$. Since this equals $f(0)$, the function is continuous.
Problem 14: Find constants $a$ and $b$ if $f(x)$ is continuous at $x = 1$, given $f(1) = 3$.
$$ f(x) = \begin{cases} x^2 + a, & x \le 1 \\ bx + 2, & x > 1 \end{cases} $$
- A) $a = 2, b = 1$ ✓ Correct
- B) $a = 1, b = 2$
- C) $a = 3, b = 1$
- D) $a = 0, b = 3$
Solution:
For $f(x)$ to be continuous at $x = 1$, LHL = RHL = $f(1) = 3$.
LHL ($x \to 1^-$): $\lim_{x \to 1^-} (x^2 + a) = (1)^2 + a = 1 + a$.
Set LHL equal to $f(1)$: $1 + a = 3 \implies a = 2$.
RHL ($x \to 1^+$): $\lim_{x \to 1^+} (bx + 2) = b(1) + 2 = b + 2$.
Set RHL equal to $f(1)$: $b + 2 = 3 \implies b = 1$.
Thus, $a = 2$ and $b = 1$.
Problem 15: Examine the continuity of the function at $x = 0$.
$$ f(x) = \begin{cases} \frac{\sin 3x}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases} $$
- A) Continuous at $x = 0$
- B) Discontinuous at $x = 0$ (removable) ✓ Correct
- C) Limit evaluates to 0
- D) None of the above
Solution:
Evaluate the limit as $x \to 0$:
$$ \lim_{x \to 0} \frac{\sin 3x}{x} = \lim_{x \to 0} \left( \frac{\sin 3x}{3x} \times 3 \right) $$
We know standard limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$.
$$ = 1 \times 3 = 3 $$
The limit is 3, but the function value $f(0) = 1$. Since $3 \neq 1$, the function is discontinuous at $x = 0$.
Problem 16: Find the point of discontinuity for the function:
$$ f(x) = \begin{cases} x^3 - 3, & x \le 2 \\ x^2 + 1, & x > 2 \end{cases} $$
- A) No point of discontinuity (Continuous everywhere) ✓ Correct
- B) Discontinuous at $x = 2$
- C) Discontinuous at $x = 0$
- D) Discontinuous at $x = 3$
Solution:
The only potential point of discontinuity is the split point, $x = 2$.
LHL ($x \to 2^-$): $\lim_{x \to 2^-} (x^3 - 3) = (2)^3 - 3 = 8 - 3 = 5$.
RHL ($x \to 2^+$): $\lim_{x \to 2^+} (x^2 + 1) = (2)^2 + 1 = 4 + 1 = 5$.
Value at $x=2$: $f(2) = 2^3 - 3 = 5$.
Since LHL = RHL = $f(2)$, the function is continuous at $x = 2$ and thus continuous everywhere.
Problem 17: Identify the points of discontinuity for $f(x) = \frac{x + 1}{x^2 - 5x + 6}$.
- A) $x = -1$
- B) $x = 1, x = -6$
- C) $x = 2, x = 3$ ✓ Correct
- D) Continuous everywhere
Solution:
A rational function is discontinuous where its denominator is zero.
Set the denominator to zero: $x^2 - 5x + 6 = 0$.
Factorize the quadratic equation:
$$ x^2 - 3x - 2x + 6 = 0 \implies x(x - 3) - 2(x - 3) = 0 $$
$$ (x - 2)(x - 3) = 0 $$
The denominator is zero at $x = 2$ and $x = 3$. Therefore, the points of discontinuity are $x = 2$ and $x = 3$.
Problem 18: Discuss the continuity of the composite function $f(x) = \sin(x^2)$.
- A) Discontinuous at $x = 0$
- B) Continuous only for $x > 0$
- C) Continuous everywhere ✓ Correct
- D) Discontinuous at multiples of $\pi$
Solution:
Let $g(x) = \sin x$ and $h(x) = x^2$.
The function $h(x) = x^2$ is a polynomial function, hence continuous everywhere.
The function $g(x) = \sin x$ is a trigonometric function, known to be continuous everywhere.
The composition of two continuous functions is also continuous. Therefore, $f(x) = g(h(x)) = \sin(x^2)$ is continuous for all real numbers.
Problem 19: Find $k$ if $f(x)$ is continuous at $x = 0$.
$$ f(x) = \begin{cases} \frac{\sqrt{1 + kx} - \sqrt{1 - kx}}{x}, & x < 0 \\ \frac{2x + 1}{x - 1}, & x \ge 0 \end{cases} $$
- A) $k = 1$
- B) $k = -1$ ✓ Correct
- C) $k = 0$
- D) $k = 2$
Solution:
For $f(x)$ to be continuous at $x = 0$, LHL = RHL = $f(0)$.
RHL ($x \to 0^+$) and $f(0)$: $\lim_{x \to 0^+} \frac{2x + 1}{x - 1} = \frac{0 + 1}{0 - 1} = -1$.
LHL ($x \to 0^-$): We evaluate the radical expression limit by rationalizing.
$$ \lim_{x \to 0^-} \frac{\sqrt{1 + kx} - \sqrt{1 - kx}}{x} \times \frac{\sqrt{1 + kx} + \sqrt{1 - kx}}{\sqrt{1 + kx} + \sqrt{1 - kx}} $$
$$ = \lim_{x \to 0^-} \frac{(1 + kx) - (1 - kx)}{x(\sqrt{1 + kx} + \sqrt{1 - kx})} = \lim_{x \to 0^-} \frac{2kx}{x(\sqrt{1 + kx} + \sqrt{1 - kx})} $$
Cancel $x$ and apply the limit $x \to 0$:
$$ \frac{2k}{\sqrt{1+0} + \sqrt{1-0}} = \frac{2k}{2} = k $$
Equating LHL to RHL: $k = -1$.
Problem 20: Examine the continuity of $f(x) = |x| + |x - 1|$ at $x = 0$ and $x = 1$.
- A) Discontinuous at $x = 0$ only
- B) Discontinuous at $x = 1$ only
- C) Discontinuous at both $x = 0$ and $x = 1$
- D) Continuous at both $x = 0$ and $x = 1$ ✓ Correct
Solution:
We redefine the function intervals removing the modulus sign:
$$ f(x) = \begin{cases} -x - (x - 1) = -2x + 1, & x < 0 \\ x - (x - 1) = 1, & 0 \le x < 1 \\ x + (x - 1) = 2x - 1, & x \ge 1 \end{cases} $$
Check continuity at $x = 0$:
LHL ($x \to 0^-$): $\lim_{x \to 0^-} (-2x + 1) = 1$.
RHL ($x \to 0^+$): $\lim_{x \to 0^+} (1) = 1$.
Since LHL = RHL = $f(0) = 1$, continuous at $x = 0$.
Check continuity at $x = 1$:
LHL ($x \to 1^-$): $\lim_{x \to 1^-} (1) = 1$.
RHL ($x \to 1^+$): $\lim_{x \to 1^+} (2x - 1) = 2(1) - 1 = 1$.
Since LHL = RHL = $f(1) = 1$, continuous at $x = 1$.