Continuity and Differentiability — NCERT Solutions

Given: $f(x) = 5x - 3$

Concept: A function $f$ is continuous at $x = c$ if $\lim_{x \to c} f(x) = f(c)$.

At $x = 0$:
$$f(0) = 5(0) - 3 = -3$$
$$\lim_{x \to 0} f(x) = \lim_{x \to 0}(5x - 3) = 5(0) - 3 = -3$$
Since $\lim_{x \to 0} f(x) = f(0) = -3$, $f$ is continuous at $x = 0$.

At $x = -3$:
$$f(-3) = 5(-3) - 3 = -15 - 3 = -18$$
$$\lim_{x \to -3} f(x) = 5(-3) - 3 = -18$$
Since $\lim_{x \to -3} f(x) = f(-3) = -18$, $f$ is continuous at $x = -3$.

At $x = 5$:
$$f(5) = 5(5) - 3 = 25 - 3 = 22$$
$$\lim_{x \to 5} f(x) = 5(5) - 3 = 22$$
Since $\lim_{x \to 5} f(x) = f(5) = 22$, $f$ is continuous at $x = 5$.

Hence, $f(x) = 5x - 3$ is continuous at $x = 0$, $x = -3$, and $x = 5$. $\blacksquare$

Given: $f(x) = 2x^2 - 1$

At $x = 3$:
$$f(3) = 2(3)^2 - 1 = 18 - 1 = 17$$
$$\lim_{x \to 3} f(x) = 2(3)^2 - 1 = 17$$

Since $\lim_{x \to 3} f(x) = f(3) = 17$, the function $f$ is continuous at $x = 3$.

(a) $f(x) = x - 5$

$f$ is a polynomial function. For any $c \in \mathbb{R}$:
$$\lim_{x \to c}(x-5) = c - 5 = f(c)$$
Hence $f$ is continuous at every real number.

(b) $f(x) = \dfrac{1}{x-5},\ x \neq 5$

The domain of $f$ is $\mathbb{R} - \{5\}$. For any $c \neq 5$:
$$\lim_{x \to c} \frac{1}{x-5} = \frac{1}{c-5} = f(c)$$
Hence $f$ is continuous at every point of its domain (i.e., at all $x \neq 5$). At $x = 5$, $f$ is not defined.

(c) $f(x) = \dfrac{x^2-25}{x+5},\ x \neq -5$

For $x \neq -5$:
$$f(x) = \frac{(x-5)(x+5)}{x+5} = x - 5$$
For any $c \neq -5$:
$$\lim_{x \to c} f(x) = c - 5 = f(c)$$
Hence $f$ is continuous at every point of its domain (i.e., at all $x \neq -5$). At $x = -5$, $f$ is not defined.

(d) $f(x) = |x - 5|$

We can write:
$$f(x) = \begin{cases} x - 5, & x \geq 5 \\ -(x-5), & x < 5 \end{cases}$$

For $c > 5$: $\lim_{x \to c}|x-5| = c - 5 = f(c)$. ✓

For $c < 5$: $\lim_{x \to c}|x-5| = -(c-5) = f(c)$. ✓

At $c = 5$:
$$\lim_{x \to 5^-}|x-5| = 0,\quad \lim_{x \to 5^+}|x-5| = 0,\quad f(5) = 0$$
All equal. ✓

Hence $f$ is continuous at every real number.

Given: $f(x) = x^n$, $n$ is a positive integer.

At $x = n$:
$$f(n) = n^n$$
$$\lim_{x \to n} f(x) = \lim_{x \to n} x^n = n^n$$

Since $\lim_{x \to n} f(x) = f(n) = n^n$, the function $f(x) = x^n$ is continuous at $x = n$. $\blacksquare$

At $x = 0$:
$$f(0) = 0$$
$$\lim_{x \to 0} f(x) = 0$$
Since $\lim_{x \to 0} f(x) = f(0)$, $f$ is continuous at $x = 0$.

At $x = 1$:
$$f(1) = 1$$
$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1$$
$$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 5 = 5$$

Since LHL $\neq$ RHL, $\lim_{x \to 1} f(x)$ does not exist. Hence $f$ is not continuous at $x = 1$.

At $x = 2$:
$$f(2) = 5$$
$$\lim_{x \to 2} f(x) = 5$$
Since $\lim_{x \to 2} f(x) = f(2) = 5$, $f$ is continuous at $x = 2$.

The only possible point of discontinuity is $x = 2$ (where the definition changes).

$$f(2) = 2(2)+3 = 7$$
$$\lim_{x \to 2^-} f(x) = 2(2)+3 = 7$$
$$\lim_{x \to 2^+} f(x) = 2(2)-3 = 1$$

Since LHL $\neq$ RHL, $f$ is discontinuous at $x = 2$.

For $x < 2$: $f(x) = 2x+3$ is a polynomial, hence continuous.
For $x > 2$: $f(x) = 2x-3$ is a polynomial, hence continuous.

Conclusion: $x = 2$ is the only point of discontinuity.

Possible points of discontinuity: $x = -3$ and $x = 3$.

At $x = -3$:
$$f(-3) = |-3|+3 = 3+3 = 6$$
$$\lim_{x \to -3^-} f(x) = |-3|+3 = 6$$
$$\lim_{x \to -3^+} f(x) = -2(-3) = 6$$
LHL = RHL = $f(-3) = 6$. Hence $f$ is continuous at $x = -3$.

At $x = 3$:
$$f(3) = 6(3)+2 = 20$$
$$\lim_{x \to 3^-} f(x) = -2(3) = -6$$
$$\lim_{x \to 3^+} f(x) = 6(3)+2 = 20$$
LHL $\neq$ RHL. Hence $f$ is discontinuous at $x = 3$.

Conclusion: $x = 3$ is the only point of discontinuity.

For $x > 0$: $\dfrac{|x|}{x} = \dfrac{x}{x} = 1$

For $x < 0$: $\dfrac{|x|}{x} = \dfrac{-x}{x} = -1$

At $x = 0$:
$$f(0) = 0$$
$$\lim_{x \to 0^-} f(x) = -1$$
$$\lim_{x \to 0^+} f(x) = 1$$
LHL $\neq$ RHL, so $f$ is discontinuous at $x = 0$.

For $x \neq 0$: $f$ equals a constant on each side, hence continuous.

Conclusion: $x = 0$ is the only point of discontinuity.

For $x < 0$: $\dfrac{x}{|x|} = \dfrac{x}{-x} = -1$

So $f(x) = -1$ for all $x < 0$ and $f(x) = -1$ for all $x \geq 0$.

At $x = 0$:
$$f(0) = -1$$
$$\lim_{x \to 0^-} f(x) = -1$$
$$\lim_{x \to 0^+} f(x) = -1$$
All equal. Hence $f$ is continuous at $x = 0$.

For all other points, $f$ is clearly continuous.

Conclusion: $f$ has no points of discontinuity.

The only possible point of discontinuity is $x = 1$.

$$f(1) = 1+1 = 2$$
$$\lim_{x \to 1^-} f(x) = (1)^2+1 = 2$$
$$\lim_{x \to 1^+} f(x) = 1+1 = 2$$

LHL = RHL = $f(1) = 2$. Hence $f$ is continuous at $x = 1$.

For $x > 1$: $f(x) = x+1$ is a polynomial — continuous.
For $x < 1$: $f(x) = x^2+1$ is a polynomial — continuous.

Conclusion: $f$ has no points of discontinuity.

The only possible point of discontinuity is $x = 2$.

$$f(2) = (2)^3 - 3 = 8 - 3 = 5$$
$$\lim_{x \to 2^-} f(x) = (2)^3 - 3 = 5$$
$$\lim_{x \to 2^+} f(x) = (2)^2 + 1 = 5$$

LHL = RHL = $f(2) = 5$. Hence $f$ is continuous at $x = 2$.

Conclusion: $f$ has no points of discontinuity.

The only possible point of discontinuity is $x = 1$.

$$f(1) = (1)^{10} - 1 = 0$$
$$\lim_{x \to 1^-} f(x) = (1)^{10} - 1 = 0$$
$$\lim_{x \to 1^+} f(x) = (1)^2 = 1$$

LHL $= 0 \neq 1 =$ RHL. Hence $f$ is discontinuous at $x = 1$.

Conclusion: $x = 1$ is the only point of discontinuity.

The only possible point of discontinuity is $x = 1$.

$$f(1) = 1+5 = 6$$
$$\lim_{x \to 1^-} f(x) = 1+5 = 6$$
$$\lim_{x \to 1^+} f(x) = 1-5 = -4$$

LHL $= 6 \neq -4 =$ RHL.

Hence $f$ is not continuous at $x = 1$.

For $x < 1$ and $x > 1$, $f$ is a polynomial, hence continuous.

Conclusion: $f$ is not a continuous function (it is discontinuous at $x = 1$).

Possible points of discontinuity: $x = 1$ and $x = 3$.

At $x = 1$:
$$f(1) = 3$$
$$\lim_{x \to 1^-} f(x) = 3, \quad \lim_{x \to 1^+} f(x) = 4$$
LHL $\neq$ RHL. Hence $f$ is discontinuous at $x = 1$.

At $x = 3$:
$$f(3) = 5$$
$$\lim_{x \to 3^-} f(x) = 4, \quad \lim_{x \to 3^+} f(x) = 5$$
LHL $\neq$ RHL. Hence $f$ is discontinuous at $x = 3$.

At all other points in $[0,10]$, $f$ is constant on open intervals, hence continuous.

Conclusion: $f$ is discontinuous at $x = 1$ and $x = 3$.

Possible points of discontinuity: $x = 0$ and $x = 1$.

At $x = 0$:
$$f(0) = 0$$
$$\lim_{x \to 0^-} f(x) = 2(0) = 0, \quad \lim_{x \to 0^+} f(x) = 0$$
LHL = RHL = $f(0) = 0$. Hence $f$ is continuous at $x = 0$.

At $x = 1$:
$$f(1) = 0$$
$$\lim_{x \to 1^-} f(x) = 0, \quad \lim_{x \to 1^+} f(x) = 4(1) = 4$$
LHL $\neq$ RHL. Hence $f$ is discontinuous at $x = 1$.

Conclusion: $f$ is discontinuous only at $x = 1$.

Possible points of discontinuity: $x = -1$ and $x = 1$.

At $x = -1$:
$$f(-1) = -2$$
$$\lim_{x \to -1^-} f(x) = -2, \quad \lim_{x \to -1^+} f(x) = 2(-1) = -2$$
LHL = RHL = $f(-1) = -2$. Hence $f$ is continuous at $x = -1$.

At $x = 1$:
$$f(1) = 2(1) = 2$$
$$\lim_{x \to 1^-} f(x) = 2(1) = 2, \quad \lim_{x \to 1^+} f(x) = 2$$
LHL = RHL = $f(1) = 2$. Hence $f$ is continuous at $x = 1$.

Conclusion: $f$ is continuous for all real $x$.

For $f$ to be continuous at $x = 3$:
$$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)$$

$$f(3) = 3a + 1$$
$$\lim_{x \to 3^-} f(x) = 3a + 1$$
$$\lim_{x \to 3^+} f(x) = 3b + 3$$

Setting LHL = RHL:
$$3a + 1 = 3b + 3$$
$$3a - 3b = 2$$
$$\boxed{3a - 3b = 2}$$

This is the required relationship between $a$ and $b$.

Continuity at $x = 0$:

$$f(0) = \lambda(0 - 0) = 0$$
$$\lim_{x \to 0^-} f(x) = \lambda(0 - 0) = 0$$
$$\lim_{x \to 0^+} f(x) = 4(0)+1 = 1$$

For continuity: LHL = RHL
$$0 = 1$$
This is a contradiction. Hence no value of $\lambda$ makes $f$ continuous at $x = 0$.

Continuity at $x = 1$:

Since $1 > 0$, $f(x) = 4x + 1$ near $x = 1$.
$$f(1) = 4(1)+1 = 5$$
$$\lim_{x \to 1} f(x) = 4(1)+1 = 5$$

Hence $f$ is continuous at $x = 1$ for any value of $\lambda$.

Given: $g(x) = x - [x]$

Let $n$ be any integer. We check continuity at $x = n$.

$$g(n) = n - [n] = n - n = 0$$

LHL:
$$\lim_{x \to n^-} g(x) = \lim_{x \to n^-}(x - [x])$$
For $x$ slightly less than $n$, $[x] = n-1$, so:
$$\lim_{x \to n^-} g(x) = n - (n-1) = 1$$

RHL:
$$\lim_{x \to n^+} g(x) = \lim_{x \to n^+}(x - [x])$$
For $x$ slightly greater than $n$, $[x] = n$, so:
$$\lim_{x \to n^+} g(x) = n - n = 0$$

Since LHL $= 1 \neq 0 =$ RHL, the limit does not exist at $x = n$.

Hence $g(x) = x - [x]$ is discontinuous at every integer $n$. $\blacksquare$

Given: $f(x) = x^2 - \sin x + 5$

$$f(\pi) = \pi^2 - \sin\pi + 5 = \pi^2 - 0 + 5 = \pi^2 + 5$$

$$\lim_{x \to \pi} f(x) = \pi^2 - \sin\pi + 5 = \pi^2 + 5$$

Since $\lim_{x \to \pi} f(x) = f(\pi) = \pi^2 + 5$, the function is continuous at $x = \pi$.

We know that $\sin x$ and $\cos x$ are continuous for all $x \in \mathbb{R}$.

Since the sum, difference, and product of continuous functions are continuous:

(a) $f(x) = \sin x + \cos x$ is continuous for all $x \in \mathbb{R}$.

(b) $f(x) = \sin x - \cos x$ is continuous for all $x \in \mathbb{R}$.

(c) $f(x) = \sin x \cdot \cos x$ is continuous for all $x \in \mathbb{R}$.

Cosine: $\cos x$ is continuous for all $x \in \mathbb{R}$.

Cosecant: $\csc x = \dfrac{1}{\sin x}$

$\sin x$ is continuous everywhere and $\sin x = 0$ at $x = n\pi$, $n \in \mathbb{Z}$.
Hence $\csc x$ is continuous for all $x \in \mathbb{R} - \{n\pi : n \in \mathbb{Z}\}$.

Secant: $\sec x = \dfrac{1}{\cos x}$

$\cos x = 0$ at $x = (2n+1)\dfrac{\pi}{2}$, $n \in \mathbb{Z}$.
Hence $\sec x$ is continuous for all $x \in \mathbb{R} - \left\{(2n+1)\dfrac{\pi}{2} : n \in \mathbb{Z}\right\}$.

Cotangent: $\cot x = \dfrac{\cos x}{\sin x}$

$\sin x = 0$ at $x = n\pi$, $n \in \mathbb{Z}$.
Hence $\cot x$ is continuous for all $x \in \mathbb{R} - \{n\pi : n \in \mathbb{Z}\}$.

The only possible point of discontinuity is $x = 0$.

$$f(0) = 0 + 1 = 1$$
$$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin x}{x} = 1 \quad \left(\text{standard limit: } \lim_{x \to 0}\frac{\sin x}{x} = 1\right)$$
$$\lim_{x \to 0^+} f(x) = 0 + 1 = 1$$

LHL = RHL = $f(0) = 1$. Hence $f$ is continuous at $x = 0$.

For $x < 0$: $\dfrac{\sin x}{x}$ is continuous (ratio of continuous functions, denominator $\neq 0$).
For $x > 0$: $x + 1$ is a polynomial, hence continuous.

Conclusion: $f$ has no points of discontinuity.

For $x \neq 0$: $f(x) = x^2 \sin\dfrac{1}{x}$ is a product of continuous functions, hence continuous.

At $x = 0$:
$$f(0) = 0$$
We need $\lim_{x \to 0} x^2 \sin\dfrac{1}{x}$.

Since $\left|\sin\dfrac{1}{x}\right| \leq 1$ for all $x \neq 0$:
$$\left|x^2 \sin\frac{1}{x}\right| \leq x^2$$

As $x \to 0$, $x^2 \to 0$, so by Squeeze Theorem:
$$\lim_{x \to 0} x^2 \sin\frac{1}{x} = 0 = f(0)$$

Hence $f$ is continuous at $x = 0$ and therefore continuous everywhere.

At $x = 0$:
$$f(0) = -1$$
$$\lim_{x \to 0} f(x) = \lim_{x \to 0}(\sin x - \cos x) = \sin 0 - \cos 0 = 0 - 1 = -1$$

Since $\lim_{x \to 0} f(x) = f(0) = -1$, $f$ is continuous at $x = 0$.

For $x \neq 0$: $\sin x - \cos x$ is continuous everywhere.

Conclusion: $f$ is continuous for all $x \in \mathbb{R}$.

For continuity at $x = \dfrac{\pi}{2}$:
$$\lim_{x \to \pi/2} \frac{k\cos x}{\pi - 2x} = 3$$

Let $x = \dfrac{\pi}{2} + h$, so as $x \to \dfrac{\pi}{2}$, $h \to 0$.

$$\cos x = \cos\left(\frac{\pi}{2}+h\right) = -\sin h$$
$$\pi - 2x = \pi - 2\left(\frac{\pi}{2}+h\right) = -2h$$

$$\lim_{h \to 0} \frac{k(-\sin h)}{-2h} = \lim_{h \to 0} \frac{k\sin h}{2h} = \frac{k}{2} \cdot 1 = \frac{k}{2}$$

Setting $\dfrac{k}{2} = 3$:
$$\boxed{k = 6}$$

For continuity at $x = 2$:
$$\lim_{x \to 2^-} f(x) = f(2) = \lim_{x \to 2^+} f(x)$$

$$\lim_{x \to 2^-} kx^2 = k(4) = 4k$$
$$\lim_{x \to 2^+} f(x) = 3$$
$$f(2) = k(4) = 4k$$

Setting $4k = 3$:
$$\boxed{k = \frac{3}{4}}$$

For continuity at $x = \pi$:
$$\lim_{x \to \pi^-} f(x) = f(\pi) = \lim_{x \to \pi^+} f(x)$$

$$f(\pi) = k\pi + 1$$
$$\lim_{x \to \pi^-} f(x) = k\pi + 1$$
$$\lim_{x \to \pi^+} f(x) = \cos\pi = -1$$

Setting $k\pi + 1 = -1$:
$$k\pi = -2$$
$$\boxed{k = -\frac{2}{\pi}}$$

For continuity at $x = 5$:
$$\lim_{x \to 5^-} f(x) = f(5) = \lim_{x \to 5^+} f(x)$$

$$f(5) = 5k + 1$$
$$\lim_{x \to 5^-} f(x) = 5k + 1$$
$$\lim_{x \to 5^+} f(x) = 3(5) - 5 = 10$$

Setting $5k + 1 = 10$:
$$5k = 9$$
$$\boxed{k = \frac{9}{5}}$$

For $f$ to be continuous, it must be continuous at $x = 2$ and $x = 10$.

At $x = 2$:
$$f(2) = 5$$
$$\lim_{x \to 2^+} f(x) = 2a + b$$
Setting equal: $2a + b = 5$ ... (i)

At $x = 10$:
$$f(10) = 21$$
$$\lim_{x \to 10^-} f(x) = 10a + b$$
Setting equal: $10a + b = 21$ ... (ii)

Subtracting (i) from (ii):
$$8a = 16 \implies a = 2$$