NCERT Solutions Class 12 Maths Continuity and Differentiability Exercise 5.1

Understanding Mathematical Continuity at a Point

Graphically, a function is said to be continuous if you can draw its curve without lifting your pen from the paper. Mathematically, we define continuity at a specific point $x = c$ using limits. A real-valued function $f(x)$ is continuous at a point $c$ in its domain if the limit of the function as $x$ approaches $c$ is equal to the value of the function at $c$.

Formally, $f(x)$ is continuous at $x = c$ if:
$\lim_{x \to c} f(x) = f(c)$

This single equation implies three critical conditions must be met simultaneously:

1. The function value $f(c)$ is defined.
2. The limit of $f(x)$ as $x \to c$ exists.
3. The limit of $f(x)$ as $x \to c$ matches $f(c)$.

For piecewise-defined functions, the limit exists if and only if the Left-Hand Limit (LHL) equals the Right-Hand Limit (RHL) at that point. Thus, the working rule for proving continuity is establishing that:
$\text{LHL} = \text{RHL} = f(c)$
If any of these three values differ, the function is discontinuous at $x = c$.

Step-by-Step Method to Solve Continuity Problems

1. Identify the Point of Interest — Locate the point $x = c$ where you need to check the continuity. This is usually the transition point in a piecewise function.
2. Evaluate the Left-Hand Limit (LHL) — Find $\lim_{x \to c^-} f(x)$. Substitute $x = c - h$, and evaluate the limit as $h \to 0$.
3. Evaluate the Right-Hand Limit (RHL) — Find $\lim_{x \to c^+} f(x)$. Substitute $x = c + h$, and evaluate the limit as $h \to 0$.
4. Find the Function Value — Calculate the value of the function exactly at $x = c$, which is $f(c)$.
5. Check for Equality — Compare your values. If LHL = RHL = $f(c)$, then the function is continuous. If not, state the point of discontinuity.

Worked Solutions from NCERT Exercise 5.1 Principles

• Example 1: Prove that the function $f(x) = 5x - 3$ is continuous at $x = 0$. Step 1: Find the value of the function at $x = 0$. $f(0) = 5(0) - 3 = -3$. Step 2: Evaluate the limit as $x \to 0$. $\lim_{x \to 0} f(x) = \lim_{x \to 0} (5x - 3) = 5(0) - 3 = -3$. Step 3: Check equality. Since $\lim_{x \to 0} f(x) = f(0) = -3$, the function is continuous at $x = 0$.
• Example 2: Examine the continuity of the function $f(x) = |x|$ at $x = 0$. Step 1: Write $f(x)$ as a piecewise function. $f(x) = -x$ if $x < 0$, and $f(x) = x$ if $x \ge 0$. Step 2: Find the Left-Hand Limit (LHL) at $x = 0$. $\text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h) = \lim_{h \to 0} (-(-h)) = 0$. Step 3: Find the Right-Hand Limit (RHL) at $x = 0$. $\text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) = \lim_{h \to 0} (h) = 0$. Step 4: Find $f(0)$. $f(0) = 0$. Step 5: Check equality. Since LHL = RHL = $f(0) = 0$, the modulus function is continuous at $x = 0$.

Board Exam Tips & Common Pitfalls

1. Don't Forget the Formal Notation: In the CBSE board exam, writing $\lim_{h \to 0}$ is highly critical. Skipping limit notation during calculations can cost you precious step-marking points.
2. Modulus Functions Trap: Always rewrite modulus terms like $|x-a|$ as a piecewise function before attempting to calculate LHL and RHL.
3. Indeterminate Forms: If substituting $x = c$ yields a $0/0$ form, algebraic simplification, factorization, or rationalization is required before calculating limits. Do not plug the value directly to declare it continuous or discontinuous.

Practice Questions with Solutions

• Q: Find the value of $k$ so that the function $f(x)$ is continuous at $x = 5$, where $f(x) = kx + 1$ if $x \le 5$ and $f(x) = 3x - 5$ if $x > 5$. A: Step 1: Find the value of the function at $x = 5$. $f(5) = k(5) + 1 = 5k + 1$. Step 2: Find the Left-Hand Limit (LHL) as $x \to 5^-$. $\text{LHL} = \lim_{x \to 5^-} f(x) = 5k + 1$. Step 3: Find the Right-Hand Limit (RHL) as $x \to 5^+$. $\text{RHL} = \lim_{x \to 5^+} (3x - 5) = 3(5) - 5 = 10$. Step 4: Since the function is continuous, LHL = RHL = $f(5)$. $5k + 1 = 10 \implies 5k = 9 \implies k = 9/5$. Final answer: $k = 9/5$.
• Q: Show that the function $f(x) = x^2$ is continuous at any real number $c$. A: Step 1: Let $c$ be any arbitrary real number. Step 2: Find the value of the function at $x = c$. $f(c) = c^2$. Step 3: Evaluate the limit as $x \to c$. $\lim_{x \to c} f(x) = \lim_{x \to c} x^2 = c^2$. Step 4: Since $\lim_{x \to c} f(x) = f(c) = c^2$, the polynomial function $f(x) = x^2$ is continuous at every real number. Final answer: Continuous for all $x \in \mathbb{R}$.
• Q: Find all points of discontinuity of the function $f(x) = 2x + 3$ if $x \le 2$ and $f(x) = 2x - 3$ if $x > 2$. A: Step 1: The function is a polynomial for $x < 2$ and $x > 2$, hence continuous in those intervals. We only need to check at $x = 2$. Step 2: Find the Left-Hand Limit (LHL) at $x = 2$. $\text{LHL} = \lim_{x \to 2^-} (2x + 3) = 2(2) + 3 = 7$. Step 3: Find the Right-Hand Limit (RHL) at $x = 2$. $\text{RHL} = \lim_{x \to 2^+} (2x - 3) = 2(2) - 3 = 1$. Step 4: Compare LHL and RHL. Since LHL (7) $\neq$ RHL (1), the limit does not exist at $x = 2$. Final answer: $x = 2$ is the point of discontinuity.
• Q: Examine the continuity of the function $f(x) = \sin(x) \cdot \cos(x)$ for all real numbers. A: Step 1: Let $g(x) = \sin(x)$ and $h(x) = \cos(x)$. Step 2: We know that sine and cosine are continuous functions for all real values of $x$. Step 3: Apply the algebra of continuous functions. If two functions $g(x)$ and $h(x)$ are continuous, their product $g(x) \cdot h(x)$ is also continuous. Step 4: Therefore, $f(x) = \sin(x) \cdot \cos(x)$ is continuous everywhere. Final answer: Continuous for all real numbers.

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