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Ncert Solutions Maths class 12th Maths Class 12th Application of Derivatives

30. Which of the following functions are decreasing on $(0, \frac{π}{2})$ ?

(A) cos x (B) cos 2x (C) cos 3x (D) tan x

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Vishal Baghel | Contributor-Level 10

4 months ago

(A) We have,

f(x) = cosx

So, f(x) = -sin x

When x $\in [0, \frac{π}{2})$ we know that sin x> 0.

$\Rightarrow$ -sinx< 0. f(x) < 0 $\forall x \in (0, \frac{π}{2})$

∴f(x) is strictly decreasing on $(0, \frac{π}{2})$

(B) We have, f(x) = cos 2x

So, f(x) = -2sin 2x

When $x \in (0, \frac{π}{2})$ we know that sin x> 0.

i e, 0 $\frac{π}{2}$
=> 0<2x<π

So, sin 2x> 0 (sinØ is ( +ve) in 1st and 2ndquadrant).

$\Rightarrow$ -2sin 2x< 0.

$\Rightarrow$ f (x) < 0.

∴f (x) is strictly decreasing on $(0, \frac{π}{2}) .$

(c) We have, f(x) = cos 3x

$\Rightarrow$ f(x) = -3 sin 3x.

As $0 < x < \frac{π}{2}$

$\to 0 < 3 x < \frac{3 π}{2}$

We can divide the interval into two

Case I At, 0 < 3x

sin 3x> 0.

$\Rightarrow$ -3 sin 3x< 0 ${\begin{array}{l} ? 0 < B x < π \end{array}$

$\begin{array}{l} \to 0 < x < \frac{π}{3} . \end{array}$

$\Rightarrow$ f(x) < 0.

∴f(x) is strictly decreasing on $(0' \frac{π}{3})$

case II. At $π < 3 x < \frac{3 π}{2} w e g e t (i i i^{r d} q u a d r o u t) .$

sin 3x< 0.

$\Rightarrow$ -3 sin 3x> 0.

$\Rightarrow$ f(x) > 0. ${? π < 3 x < \frac{3 π}{2} \Rightarrow \frac{π}{3} < x < \frac{π}{2}}$

∴f(x) is strictly increasing on $(\frac{π}{3}, \frac{π}{2})$

Hence, f(x

...more

(A) We have,

f(x) = cosx

So, f(x) = -sin x

When x $\in [0, \frac{π}{2})$ we know that sin x> 0.

$\Rightarrow$ -sinx< 0. f(x) < 0 $\forall x \in (0, \frac{π}{2})$

∴f(x) is strictly decreasing on $(0, \frac{π}{2})$

(B) We have, f(x) = cos 2x

So, f(x) = -2sin 2x

When $x \in (0, \frac{π}{2})$ we know that sin x> 0.

i e, 0 $\frac{π}{2}$
=> 0<2x<π

So, sin 2x> 0 (sinØ is ( +ve) in 1st and 2ndquadrant).

$\Rightarrow$ -2sin 2x< 0.

$\Rightarrow$ f (x) < 0.

∴f (x) is strictly decreasing on $(0, \frac{π}{2}) .$

(c) We have, f(x) = cos 3x

$\Rightarrow$ f(x) = -3 sin 3x.

As $0 < x < \frac{π}{2}$

$\to 0 < 3 x < \frac{3 π}{2}$

We can divide the interval into two

Case I At, 0 < 3x

sin 3x> 0.

$\Rightarrow$ -3 sin 3x< 0 ${\begin{array}{l} ? 0 < B x < π \end{array}$

$\begin{array}{l} \to 0 < x < \frac{π}{3} . \end{array}$

$\Rightarrow$ f(x) < 0.

∴f(x) is strictly decreasing on $(0' \frac{π}{3})$

case II. At $π < 3 x < \frac{3 π}{2} w e g e t (i i i^{r d} q u a d r o u t) .$

sin 3x< 0.

$\Rightarrow$ -3 sin 3x> 0.

$\Rightarrow$ f(x) > 0. ${? π < 3 x < \frac{3 π}{2} \Rightarrow \frac{π}{3} < x < \frac{π}{2}}$

∴f(x) is strictly increasing on $(\frac{π}{3}, \frac{π}{2})$

Hence, f(x) is with a increasing or decreasing on $(0, \frac{π}{2}) .$

(d) We have, f(x) = tan x.

So, f(x) = sec²x.

$\Rightarrow$ f(x) = sec²x> 0 $\forall x \in (0, \frac{π}{2}) .$

∴f(x) is strictly increasing on $(0, \frac{π}{2}) .$

Hence, the fxⁿcosx and cos 2x are strictly decreasing on $(0, \frac{π}{2}) .$

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Hence $Δ$ be equilateral having each side of length $\sqrt[]{3} r .$

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30. Which of the following functions are decreasing on $(0, \frac{π}{2})$ ?

(A) cos x (B) cos 2x (C) cos 3x (D) tan x