Application of Derivatives — Class 12 Maths Solution

(i) We have, $f(x) = |x + 2| - 1$ for all $x \in R$.
Since, $|x + 2| \ge 0 \Rightarrow |x + 2| - 1 \ge - 1$
Minimum $f(x) = - 1$, which occurs when $x + 2 = 0$ i.e., when $x = - 2$.

$f(x)$ has no maximum value because $f(x\rangle \to \infty$ as $|x| \to \infty$.

(ii) We have, $g(x) = - |x + 1| + 3$ for all $x \in R$.

Since, $|x + 1| \ge 0 \Rightarrow - |x + 1| \le 0 \Rightarrow - |x + 1| + 3 \le 3$

Therefore, Maximum value of$g(x)$ is $3$, which occurs when $x + 1 = 0$, i.e., when $x = - 1$.

hence we can say that $g(x)$ has no minimum value because $g(x) \to - \infty$ as $|x| \to \infty$.

(iii) We have, $h(x) = \sin 2x + 5\,\,\forall \,\,x \in R$

We know that, $- 1 \le \sin 2x \le 1$ for all $x \in R$
$\Rightarrow 5 - 1 \le 5 + \sin 2x \le 5 + 1$ for all $x \in R$
$\Rightarrow 4 \le f(x) \le 6$ for all $x \in R$.

Therefore maximum value of $f(x) = 6$, which occurs when $\sin 2x = 1$ and minimum value of $f(x) = 4$, which occurs when $\sin 2x = - 1$.

(iv) We have, $f(x) = |\sin 4x + 3|\forall x \in R$.
We know that, $- 1 \le \sin 4x \le 1$ for all $x \in R$

$\Rightarrow 3 - 1 \le \sin 4x + 3 \le 1 + 3$ for all $x \in R$.
$\Rightarrow |2|\,\, \le \,|\,\sin 4x + 3|\,\, \le \,|4|\forall \,\,x \in R$

$\Rightarrow 2 \le f(x) \le 4\,\,\forall \,\,x \in R$

Therefore minimum value of $f(x) = 2$, which occurs when $\sin 4x = - 1$ and maximum value of $f(x) = 4$, which occurs when $\sin 4x = 1$.

(v) We have, $h(x) = x + 1,\forall - 1 < x < 1$.
$- 1 < x < 1 = - 1 + 1 < x + 1 < 1 + 1 \Rightarrow 0 < x + 1 < 2$
Here, range of $h = (0,\;2)$

There is no definite value for maximum or minimum of $h(x)$.
Therefore, $h$ has neither a maximum nor a minimum value.