CBSE Class 12 Maths Exemplar Chapter 5: Continuity and Differentiability NCERT Solutions

To examine the continuity of the function $f(x)\; =\; x^3\; +\; 2x^2\; -\; 1$ at $x\; =\; 1$, we need to check if the limit of the function as x approaches 1 exists and is equal to the function's value at x = 1.

The condition for continuity at a point $x\; =\; a$ is $\backslash lim\_\{x\; o\; a^-\}\; f(x)\; =\; \backslash lim\_\{x\; o\; a^+\}\; f(x)\; =\; f(a)$.

Here, $a\; =\; 1$.

First, let's find the value of the function at $x\; =\; 1$:

$f(1)\; =\; (1)^3\; +\; 2(1)^2\; -\; 1\; =\; 1\; +\; 2\; -\; 1\; =\; 2$.

Next, let's find the Left-Hand Limit (LHL):

$$LHL\; =\; \backslash lim\_\{x\; o\; 1^\{-\}\}\; f(x)\; =\; \backslash lim\_\{x\; o\; 1^\{-\}\}\; (x^3\; +\; 2x^2\; -\; 1)$$

Substitute $x\; =\; 1\; -\; h$ where $h\; o\; 0$:

$$=\; \backslash lim\_\{h\; o\; 0\}\; ((1-h)^3\; +\; 2(1-h)^2\; -\; 1)$$

$$=\; (1)^3\; +\; 2(1)^2\; -\; 1\; =\; 1\; +\; 2\; -\; 1\; =\; 2$$.

Now, let's find the Right-Hand Limit (RHL):

$$RHL\; =\; \backslash lim\_\{x\; o\; 1^\{+\}\}\; f(x)\; =\; \backslash lim\_\{x\; o\; 1^\{+\}\}\; (x^3\; +\; 2x^2\; -\; 1)$$

Substitute $x\; =\; 1\; +\; h$ where $h\; o\; 0$:

$$=\; \backslash lim\_\{h\; o\; 0\}\; ((1+h)^3\; +\; 2(1+h)^2\; -\; 1)$$

$$=\; (1)^3\; +\; 2(1)^2\; -\; 1\; =\; 1\; +\; 2\; -\; 1\; =\; 2$$.

Since $LHL\; =\; RHL\; =\; f(1)\; =\; 2$, the function $f(x)$ is continuous at $x\; =\; 1$.

Note: As mentioned in the source, every polynomial function is continuous at any real point, which is confirmed by this calculation.

We need to check the continuity of the piecewise function f(x) at x = 2. The condition for continuity is \lim_{x o 2^-} f(x) = \lim_{x o 2^+} f(x) = f(2).

First, let's find the value of the function at x = 2. Since x \ge 2 for f(x) = 3x + 5:

f(2) = 3(2) + 5 = 6 + 5 = 11.

Now, let's calculate the Left-Hand Limit (LHL) as x approaches 2 from the left (i.e., x < 2):

LHL = \lim_{x o 2^{-}} f(x) = \lim_{x o 2^{-}} x^2

Substitute x = 2 - h where h o 0:

= \lim_{h o 0} (2 - h)^2 = \lim_{h o 0} (4 - 4h + h^2) = 4.

Next, let's calculate the Right-Hand Limit (RHL) as x approaches 2 from the right (i.e., x > 2):

RHL = \lim_{x o 2^{+}} f(x) = \lim_{x o 2^{+}} (3x + 5)

Substitute x = 2 + h where h o 0:

= \lim_{h o 0} [3(2 + h) + 5] = \lim_{h o 0} (6 + 3h + 5) = 11.

We found that LHL = 4, RHL = 11, and f(2) = 11.

Since LHL

eq RHL (4 ≠ 11), the limit of the function does not exist at x = 2.

Therefore, the function f(x) is discontinuous at x = 2.

We need to check the continuity of the given function f(x) at x = 0. The condition for continuity is \lim_{x o 0^-} f(x) = \lim_{x o 0^+} f(x) = f(0).

The value of the function at x = 0 is given as:

f(0) = 5.

Now, let's calculate the Left-Hand Limit (LHL) as x approaches 0:

LHL = \lim_{x o 0^{-}} f(x) = \lim_{x o 0^{-}} \frac{1 - \cos 2x}{x^2}

Using the trigonometric identity \cos 2\theta = 1 - 2\sin^2 \theta, we get 1 - \cos 2x = 2\sin^2 x.

= \lim_{x o 0^{-}} \frac{2\sin^2 x}{x^2}

We can rewrite this as:

= 2 \lim_{x o 0^{-}} \left(\frac{\sin x}{x}\right)^2

We know that \lim_{x o 0} \frac{\sin x}{x} = 1. Therefore,

= 2 \times (1)^2 = 2.

Now, let's calculate the Right-Hand Limit (RHL) as x approaches 0:

RHL = \lim_{x o 0^{+}} f(x) = \lim_{x o 0^{+}} \frac{1 - \cos 2x}{x^2}

Using the same trigonometric identity:

= \lim_{x o 0^{+}} \frac{2\sin^2 x}{x^2} = 2 \lim_{x o 0^{+}} \left(\frac{\sin x}{x}\right)^2

= 2 \times (1)^2 = 2.

We have LHL = 2, RHL = 2, and f(0) = 5.

Since LHL = RHL

eq f(0) (2 ≠ 5), the function f(x) is not continuous at x = 0.

We need to check the continuity of the function f(x) at x = 2. The condition for continuity is \lim_{x o 2^-} f(x) = \lim_{x o 2^+} f(x) = f(2).

The value of the function at x = 2 is given as:

f(2) = 5.

Now, let's calculate the Left-Hand Limit (LHL) as x approaches 2:

LHL = \lim_{x o 2^{-}} f(x) = \lim_{x o 2^{-}} \frac{2x^2 - 3x - 2}{x - 2}

We can factor the numerator. We look for two numbers that multiply to (2)(-2) = -4 and add to -3. These numbers are -4 and 1.

2x^2 - 3x - 2 = 2x^2 - 4x + x - 2 = 2x(x - 2) + 1(x - 2) = (2x + 1)(x - 2).

So, the expression becomes:

= \lim_{x o 2^{-}} \frac{(2x + 1)(x - 2)}{x - 2}

For x

eq 2, we can cancel out the (x - 2) term:

= \lim_{x o 2^{-}} (2x + 1) = 2(2) + 1 = 4 + 1 = 5.

Now, let's calculate the Right-Hand Limit (RHL) as x approaches 2:

RHL = \lim_{x o 2^{+}} f(x) = \lim_{x o 2^{+}} \frac{2x^2 - 3x - 2}{x - 2}

Using the same factorization:

= \lim_{x o 2^{+}} \frac{(2x + 1)(x - 2)}{x - 2}

For x

eq 2, we cancel out the (x - 2) term:

= \lim_{x o 2^{+}} (2x + 1) = 2(2) + 1 = 4 + 1 = 5.

We have LHL = 5, RHL = 5, and f(2) = 5.

Since LHL = RHL = f(2), the function f(x) is continuous at x = 2.