NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3

Part (i)

Given:
$$f(x) = (2x-1)^2 + 3$$

Formula Used:
For any real number $x$, a squared term is always non-negative: $(2x-1)^2 \ge 0$.

Step 1: Analyze the bounds of the function.
Since $(2x-1)^2 \ge 0$ for all $x \in \mathbb{R}$:
$$(2x-1)^2 + 3 \ge 0 + 3$$
$$f(x) \ge 3$$

Step 2: Find the minimum point.
The minimum value occurs when $(2x-1)^2 = 0$
$$2x - 1 = 0 \implies x = \frac{1}{2}$$
Substituting $x = \frac{1}{2}$ into $f(x)$ yields $3$.
Since $f(x)$ can increase indefinitely as $x \to \pm\infty$, there is no maximum value.

Final Answer: Minimum Value $= 3$ at $x = \frac{1}{2}$; No Maximum Value.

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Part (ii)

Given:
$$f(x) = 9x^2 + 12x + 2$$

Formula Used:
Completing the square method: $a^2 + 2ab + b^2 = (a+b)^2$.

Step 1: Rewrite the function by completing the square.
$$f(x) = (3x)^2 + 2(3x)(2) + 4 - 2$$
$$f(x) = (3x+2)^2 - 2$$

Step 2: Analyze the minimum and maximum values.
Since $(3x+2)^2 \ge 0$ for all $x \in \mathbb{R}$:
$$(3x+2)^2 - 2 \ge -2$$
$$f(x) \ge -2$$

The minimum value occurs when $3x + 2 = 0 \implies x = -\frac{2}{3}$.
The function can grow infinitely large, so it has no maximum value.

Final Answer: Minimum Value $= -2$ at $x = -\frac{2}{3}$; No Maximum Value.

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Part (iii)

Given:
$$f(x) = -(x-1)^2 + 10$$

Formula Used:
Since $(x-1)^2 \ge 0$, multiplying by $-1$ reverses the inequality: $-(x-1)^2 \le 0$.

Step 1: Analyze the bounds.
$$-(x-1)^2 \le 0$$
Add 10 to both sides:
$$-(x-1)^2 + 10 \le 10$$
$$f(x) \le 10$$

Step 2: Find the maximum point.
The maximum value occurs when $-(x-1)^2 = 0 \implies x = 1$
$$f(1) = 10$$
As $x$ moves away from 1, $f(x)$ decreases without bound, so there is no minimum value.

Final Answer: Maximum Value $= 10$ at $x = 1$; No Minimum Value.

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Part (iv)

Given:
$$g(x) = x^3 + 1$$

Formula Used:
First Derivative Test: Critical points occur where $g'(x) = 0$.

Step 1: Differentiate the function with respect to $x$.
$$g'(x) = 3x^2$$

Step 2: Find critical points.
$$g'(x) = 0 \implies 3x^2 = 0 \implies x = 0$$

Step 3: Test the nature of the critical point.
For $x < 0$, $g'(x) = 3x^2 > 0$ (positive).
For $x > 0$, $g'(x) = 3x^2 > 0$ (positive).
Since the sign of $g'(x)$ does not change as $x$ passes through $0$, $x = 0$ is a point of inflection. The function increases continuously from $-\infty$ to $+\infty$.

Final Answer: The function has neither a maximum value nor a minimum value.

Part (i)

Given:
$$f(x) = |x+2| - 1$$

Step 1: Set up the inequality.
Absolute value property: $|x+2| \ge 0$ for all $x \in \mathbb{R}$.
Subtract 1 from both sides:
$$|x+2| - 1 \ge -1$$
$$f(x) \ge -1$$

Step 2: Find the point of minimum value.
$|x+2| = 0 \implies x = -2$
$$f(-2) = -1$$
The value can increase infinitely, so there is no maximum value.

Final Answer: Minimum Value $= -1$ at $x = -2$; No Maximum Value.

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Part (ii)

Given:
$$g(x) = -|x+1| + 3$$

Step 1: Set up the inequality.
$-|x+1| \le 0$ for all $x \in \mathbb{R}$.
Add 3 to both sides:
$$-|x+1| + 3 \le 3$$
$$g(x) \le 3$$

Step 2: Find the point of maximum value.
$-|x+1| = 0 \implies x = -1$
$$g(-1) = 3$$

Final Answer: Maximum Value $= 3$ at $x = -1$; No Minimum Value.

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Part (iii)

Given:
$$h(x) = \sin(2x) + 5$$

Step 1: Apply the range of sine.
Trigonometric range property: $-1 \le \sin(\theta) \le 1$.
$$-1 \le \sin(2x) \le 1$$
Add 5 across the inequality:
$$-1 + 5 \le \sin(2x) + 5 \le 1 + 5$$
$$4 \le h(x) \le 6$$

Final Answer: Maximum Value $= 6$, Minimum Value $= 4$.

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Part (iv)

Given:
$$f(x) = |\sin(4x) + 3|$$

Step 1: Apply the range of sine.
$$-1 \le \sin(4x) \le 1$$
Add 3 across the inequality:
$$-1 + 3 \le \sin(4x) + 3 \le 1 + 3$$
$$2 \le \sin(4x) + 3 \le 4$$

Step 2: Apply the absolute value.
Since all values inside the modulus are already positive (between 2 and 4), taking the modulus does not alter the boundaries:
$$2 \le |\sin(4x) + 3| \le 4$$

Final Answer: Maximum Value $= 4$, Minimum Value $= 2$.

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Part (v)

Given:
$$h(x) = x+1 \quad \text{for } x \in (-1, 1)$$

Step 1: Analyze the open interval boundaries.
Given $-1 < x < 1$. Add 1 to all parts of the inequality:
$$-1 + 1 < x + 1 < 1 + 1$$
$$0 < h(x) < 2$$

Step 2: Evaluate the presence of extrema.
Since the interval is open, $h(x)$ approaches 0 as $x \to -1^+$ and approaches 2 as $x \to 1^-$, but it never actually reaches these boundary values within the given domain.

Final Answer: The function has neither a maximum value nor a minimum value on the given open interval.

Part (i)

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

Step 1: Differentiate w.r.t $x$: $f'(x) = 2x$.

Step 2: Set $f'(x) = 0$ to find critical points: $2x = 0 \implies x = 0$.

Step 3: Apply the first derivative test:
For $x < 0$, $f'(x) < 0$ (decreasing).
For $x > 0$, $f'(x) > 0$ (increasing).
Since $f'(x)$ changes sign from negative to positive at $x = 0$, $x=0$ is a point of local minimum.

Step 4: Compute the value: $f(0) = 0^2 = 0$.

Final Answer: Local minimum point: $x = 0$; Local minimum value $= 0$; No local maximum.

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Part (ii)

Given: $g(x) = x^3 - 3x$

Step 1: Differentiate: $g'(x) = 3x^2 - 3$.

Step 2: Set $g'(x) = 0 \implies 3(x^2 - 1) = 0 \implies x = 1 \text{ or } x = -1$.

Step 3: Analyze using the First Derivative Test.
Around $x = -1$: For $x < -1$, $g'(x) > 0$. For $x > -1$ (close to -1), $g'(x) < 0$. Sign changes from $+$ to $-$. Thus, $x = -1$ is a point of local maximum.
Around $x = 1$: For $x < 1$ (close to 1), $g'(x) < 0$. For $x > 1$, $g'(x) > 0$. Sign changes from $-$ to $+$. Thus, $x = 1$ is a point of local minimum.

Step 4: Compute the values.
Local maximum value: $g(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2$
Local minimum value: $g(1) = (1)^3 - 3(1) = 1 - 3 = -2$

Final Answer: Local maximum at $x = -1$ with value 2; Local minimum at $x = 1$ with value -2.

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Part (iii)

Given: $h(x) = \sin x + \cos x, \quad 0 < x < \frac{\pi}{2}$

Step 1: Differentiate: $h'(x) = \cos x - \sin x$.

Step 2: Set $h'(x) = 0 \implies \cos x - \sin x = 0 \implies \tan x = 1$.
Since $0 < x < \frac{\pi}{2}$, we have $x = \frac{\pi}{4}$.

Step 3: Test the point $x = \frac{\pi}{4}$.
If $x < \frac{\pi}{4}$, $\cos x > \sin x \implies h'(x) > 0$.
If $x > \frac{\pi}{4}$, $\cos x < \sin x \implies h'(x) < 0$.
The sign of $h'(x)$ changes from positive to negative, making $x = \frac{\pi}{4}$ a point of local maximum.

Step 4: Find the local maximum value.
$h\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.

Final Answer: Local maximum at $x = \frac{\pi}{4}$ with value $\sqrt{2}$.

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Part (iv)

Given: $f(x) = \sin x - \cos x, \quad 0 < x < 2\pi$

Step 1: Find $f'(x) = \cos x + \sin x$.

Step 2: Set $f'(x) = 0 \implies \cos x + \sin x = 0 \implies \tan x = -1$.
In the interval $(0, 2\pi)$, $\tan x = -1$ at:
$x = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ and $x = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$.

Step 3: Test using the first derivative values around critical points.
At $x = \frac{3\pi}{4}$: $f'(x)$ changes from $+$ to $-$. Thus, it is a point of local maximum.
At $x = \frac{7\pi}{4}$: $f'(x)$ changes from $-$ to $+$. Thus, it is a point of local minimum.

Step 4: Compute the extreme values.
$f\left(\frac{3\pi}{4}\right) = \sin\left(\frac{3\pi}{4}\right) - \cos\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}} - \left(-\frac{1}{\sqrt{2}}\right) = \sqrt{2}$.
$f\left(\frac{7\pi}{4}\right) = \sin\left(\frac{7\pi}{4}\right) - \cos\left(\frac{7\pi}{4}\right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2}$.

Final Answer: Local maximum at $x = \frac{3\pi}{4}$ with value $\sqrt{2}$; Local minimum at $x = \frac{7\pi}{4}$ with value $-\sqrt{2}$.

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Part (v)

Given: $f(x) = x^3 - 6x^2 + 9x + 15$

Step 1: Find $f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3)$.

Step 2: Factor and set to 0: $3(x-1)(x-3) = 0 \implies x = 1, \: x = 3$.

Step 3: Check sign changes for $f'(x)$.
At $x = 1$: For $x < 1$, $f'(x) > 0$. For $1 < x < 3$, $f'(x) < 0$. Sign changes from $+$ to $-$. Local Maximum.
At $x = 3$: For $1 < x < 3$, $f'(x) < 0$. For $x > 3$, $f'(x) > 0$. Sign changes from $-$ to $+$. Local Minimum.

Step 4: Calculate values.
$f(1) = 1^3 - 6(1)^2 + 9(1) + 15 = 1 - 6 + 9 + 15 = 19$
$f(3) = 3^3 - 6(3)^2 + 9(3) + 15 = 27 - 54 + 27 + 15 = 15$

Final Answer: Local maximum at $x = 1$ with value 19; Local minimum at $x = 3$ with value 15.

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Part (vi)

Given: $g(x) = \frac{x}{2} + \frac{2}{x}, \quad x > 0$

Step 1: Compute $g'(x) = \frac{1}{2} - \frac{2}{x^2}$.

Step 2: Set $g'(x) = 0 \implies \frac{1}{2} = \frac{2}{x^2} \implies x^2 = 4 \implies x = 2$ (since $x > 0$).

Step 3: Apply the first derivative test around $x = 2$.
For $0 < x < 2$, $\frac{2}{x^2} > \frac{1}{2} \implies g'(x) < 0$.
For $x > 2$, $\frac{2}{x^2} < \frac{1}{2} \implies g'(x) > 0$.
Sign changes from negative to positive. Hence, $x = 2$ is a local minimum.

Step 4: Find the local minimum value: $g(2) = \frac{2}{2} + \frac{2}{2} = 1 + 1 = 2$.

Final Answer: Local minimum at $x = 2$ with value 2.

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Part (vii)

Given: $g(x) = \frac{1}{x^2+2}$

Step 1: Find $g'(x)$ using the quotient/chain rule: $g'(x) = \frac{-2x}{(x^2+2)^2}$.

Step 2: Find critical points by setting $g'(x) = 0 \implies -2x = 0 \implies x = 0$.

Step 3: Analyze the first derivative behavior.
The denominator $(x^2+2)^2$ is always positive.
For $x < 0$, $g'(x) = \frac{\text{positive}}{\text{positive}} > 0$.
For $x > 0$, $g'(x) = \frac{\text{negative}}{\text{positive}} < 0$.
The derivative changes from positive to negative at $x = 0$, confirming a local maximum.

Step 4: Compute the value: $g(0) = \frac{1}{0^2+2} = \frac{1}{2}$.

Final Answer: Local maximum at $x = 0$ with value $\frac{1}{2}$.

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Part (viii)

Given: $f(x) = x\sqrt{1-x}, \quad 0 < x < 1$

Step 1: Apply product rule to find $f'(x)$.
$$f'(x) = 1 \cdot \sqrt{1-x} + x \cdot \left(\frac{-1}{2\sqrt{1-x}}\right)$$
$$f'(x) = \sqrt{1-x} - \frac{x}{2\sqrt{1-x}} = \frac{2(1-x)-x}{2\sqrt{1-x}} = \frac{2-3x}{2\sqrt{1-x}}$$

Step 2: Find critical points ($f'(x) = 0$).
$$2 - 3x = 0 \implies x = \frac{2}{3}$$

Step 3: Evaluate sign changes around $x = \frac{2}{3}$.
For $x < \frac{2}{3}$, $2-3x > 0 \implies f'(x) > 0$.
For $x > \frac{2}{3}$, $2-3x < 0 \implies f'(x) < 0$.
Sign changes from $+$ to $-$. Thus, $x = \frac{2}{3}$ is a local maximum point.

Step 4: Calculate maximum value.
$$f\left(\frac{2}{3}\right) = \frac{2}{3}\sqrt{1-\frac{2}{3}} = \frac{2}{3}\sqrt{\frac{1}{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}$$

Final Answer: Local maximum at $x = \frac{2}{3}$ with value $\frac{2\sqrt{3}}{9}$.

Part (i)

Given: $f(x) = x^3 \quad \text{on } [-2, 2]$

Step 1: Find critical points.
$f'(x) = 3x^2 = 0 \implies x = 0 \in [-2, 2]$

Step 2: Evaluate the function at the critical point and interval endpoints: $x = -2, 0, 2$.
$f(-2) = (-2)^3 = -8$
$f(0) = 0^3 = 0$
$f(2) = 2^3 = 8$

Step 3: Compare values to find absolute extremes.
Max value $= 8$, Min value $= -8$.

Final Answer: Absolute Maximum Value $= 8$ at $x = 2$; Absolute Minimum Value $= -8$ at $x = -2$.

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Part (ii)

Given: $f(x) = \sin x + \cos x \quad \text{on } [0, \pi]$

Step 1: Identify critical points in $(0, \pi)$.
$f'(x) = \cos x - \sin x = 0 \implies \tan x = 1 \implies x = \frac{\pi}{4}$

Step 2: Evaluate $f(x)$ at $x = 0, \frac{\pi}{4}, \pi$.
$f(0) = \sin 0 + \cos 0 = 0 + 1 = 1$
$f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \approx 1.414$
$f(\pi) = \sin \pi + \cos \pi = 0 - 1 = -1$

Step 3: Compare the values.
Highest value: $\sqrt{2}$
Lowest value: $-1$

Final Answer: Absolute Maximum Value $= \sqrt{2}$ at $x = \frac{\pi}{4}$; Absolute Minimum Value $= -1$ at $x = \pi$.

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Part (iii)

Given: $f(x) = 4x - \frac{1}{2}x^2 \quad \text{on } \left[-2, \frac{9}{2}\right]$

Step 1: Differentiate to find critical points.
$f'(x) = 4 - x = 0 \implies x = 4$.
Since $4 \in \left[-2, \frac{9}{2}\right]$, it is an internal critical point.

Step 2: Compute values at $x = -2, 4, \frac{9}{2}$.
$f(-2) = 4(-2) - \frac{1}{2}(-2)^2 = -8 - 2 = -10$
$f(4) = 4(4) - \frac{1}{2}(4)^2 = 16 - 8 = 8$
$f\left(\frac{9}{2}\right) = 4\left(\frac{9}{2}\right) - \frac{1}{2}\left(\frac{9}{2}\right)^2 = 18 - \frac{81}{8} = \frac{144-81}{8} = \frac{63}{8} = 7.875$

Final Answer: Absolute Maximum Value $= 8$ at $x = 4$; Absolute Minimum Value $= -10$ at $x = -2$.

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Part (iv)

Given: $f(x) = (x-1)^2 + 3 \quad \text{on } [-3, 1]$

Step 1: Find critical points.
$f'(x) = 2(x-1) = 0 \implies x = 1$
The critical point matches the right boundary of our interval.

Step 2: Evaluate at the endpoints $x = -3$ and $x = 1$.
$f(-3) = (-3-1)^2 + 3 = (-4)^2 + 3 = 16 + 3 = 19$
$f(1) = (1-1)^2 + 3 = 3$

Final Answer: Absolute Maximum Value $= 19$ at $x = -3$; Absolute Minimum Value $= 3$ at $x = 1$.