Maths

(A) We have,

f(x) = cosx

So, f(x) = -sin x

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

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

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

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

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

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

i e, 0 $\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
=> 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 $\left(0,\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right).$

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

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

As $0<x<\frac{\pi }{2}$

$\to 0<3x<\frac{3\pi }{2}$

We can divide the interval into two

Case I At, 0 < 3x

sin 3x> 0.

$\Rightarrow$ -3 sin 3x< 0 $\{\begin{array}{l}?0<Bx<\pi \hfill \end{array}$

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

$\Rightarrow$ f(x) < 0.

∴f(x) is strictly decreasing on $\left(0\text{'}\frac{\pi }{3}\right)$

case II. At $\pi <3x<\frac{3\pi }{2}weget(ii{i}^{rd}quadrout).$

sin 3x< 0.

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

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

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

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

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

So, f(x) = sec²x.

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

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

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

(a) f (x) = x² + 2x - 5.

f(x) = 2x + 2 = 2 (x + 1).

At, f(x) = 0

2 (x + 1) = 0

$\Rightarrow$ x = -1.

At, x $(-\infty ,-1),$

f(x) = (- ) ve< 0.

So, f (x)is strictly decreasing or $(-\infty ,-1).$

At x ∈$(-1,\infty )$

f(x) = ( + ve) > 1

f(x) is strictly increasing on $(-1,\infty ).$

(b) f(x) = 10 - 6x- 2x²

So, f(x) = - 6 - 4x = - 2 (3 + 2x).

Atf(x) = 0

$\Rightarrow$ 2 (3 + 2x) = 0.

$\Rightarrow$ x = $\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

At x $(-\infty ,\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.),$

∴f(x) is strictly increasing on $\left(-\infty ,\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

At x $\left(\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\infty \right)$

f(x) = ( -ve) ( + ve) = ( - ) ve< 0.

∴f (x) is strictly decreasing on $\left(\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\infty \right)$

(c) f (x) = 2x³- 9x²- 12x.

So, f (x) =- 6x²- 18x- 12 = - 6 (x² + 3x + 2).

= -6 [x² + x + 2x + 2]

= -6 [x (x + 1) + 2 (x + 1)]

= -6 (x + 1) (x + 2)

At, f (x) = 0.

$\Rightarrow$ 6 (x + 1) (x + 2) = 0

$\Rightarrow$ x = -1 and x = -2.

At, x $(-\infty ,-2)$ , f(x) = ( -ve) ( -ve) ( -ve) = ( -ve) < 0

So, f(x) is strictly decreasing.

At, x( -2, -1), f(x) = ( -ve) ( -ve) ( -ve) = ( -ve) < 0.

So, f(x) is strictly decreasing

(d) f(x) = 6 - 9x-x²

So, f(x) = - 9 - 2x

At, f(x) = 0

$\Rightarrow$ 9 - 2x = 0

$\Rightarrow$ x =$\raisebox{1ex}{$-9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

At, x ∈$\left(-\infty ,\raisebox{1ex}{$-9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right),$

f(x) > 0

So, f(x) is strictly increasing.

At x ∈$\left(\raisebox{1ex}{$-9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\infty \right).$

f(x) < 0.

So, f(x) is strictly decreasing.

(e) f(x) = (x + 1)³ (x- 3)³.

So, f(x) = (x + 1)³ $\frac{d}{dx}{(x-3)}^3+{(x-3)}^3\frac{d}{dx}{(x+1)}^3$

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

= 3 (x + 1)² (x- 3)² (x + 1 + x- 3).

= 3 (x + 1)² (x- 3)² (2x- 2).

= 6 (x + 1)² (x- 3)² (x- 1).

For strictly increasing,

f(x) > 0.

As 6 > 0,(x + 1)²> 0 and (x- 3)²> 0. We need

(x 1) > 0.

$\Rightarrow$ x> 1.

∴f (x) is strictly increasing on $(1,\infty ).$

And strictly decreasing on $(-\infty ,1).$

We have, f (x) = 2x³- 3x²- 36x + 7.

So, f (x) = $\frac{d}{dx}\left(2x^3-3x^2-36x+7\right)=6x^2-6x-36.$

= 6 (x²-x- 6).

= 6 (x²- 3x + 2x- 6)

= 6 [x (x- 3) + 2 (x- 3)]

= 6 (x- 3) (x + 2).

At, f (x) = 0

$\Rightarrow$ 6 (x- 3) (x + 2) = 0.

So, when x- 3 = 0 or x + 2 = 0.

$\Rightarrow$ x = 3 or x = -2.

Hence we an divide the real line into three disjoint internal

I $(-\infty ,-2)\; II(-2,3)andIII(3,\infty )$

At x ∈$(-\infty ,-2),$

f (x) = ( + ve) ( -ve) ( -ve) = ( + ve) > 0.

So, f (x) is strictly increasing in $(-\infty ,-2)$

At, x∈ ( -2,3),

f (x) = ( + ve) ( + ve) ( -ve) = ( -ve) < 0.

So, f (x) is strictly decreasing in ( -2,3).

At, x ∈$(3,\infty )$

f (x) = ( + ve) ( + ve) ( + ve) = ( + )ve> 0.

So, f (x) is strictly increasing in $(3,\infty )$

∴ (a) f (x) is strictly increasing in $(-\infty ,-2)and(3,-\infty )$

(b) f (x) is strictly decreasing in ( -2,3)