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Application of Derivatives Class 12th

32. For what values of a the function f given by f(x) = x² + ax + 1 is increasing on [1, 2]?

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Vishal Baghel

Contributor-Level 10

We have, f (x) = x² + x + 1

So, f (x) = 2x + a

If f (x) is strictly increasing onx $(1, 2), f^{'} (x) > 0 .$

So, 1

$\to 2 * 1 < 2 * x < 2 * 2$

$\Rightarrow 2 < 2 x < 4$

$\Rightarrow 2 + a < 2 x + a < 4 + a .$

$\Rightarrow 2 + a < f^{'} (x) < 4 + a .$

The minimum value of f (x) is 2 + a and that of men value is 4 + a.

∴ 2 +a> 0and 4 + a> 0

$\Rightarrow$ a> -2 and a> -4.

$\Rightarrow$ a> -2.

∴The best value of a is -2.

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Application of Derivatives Class 12th

31. On which of the following intervals is the function f given by f(x) = x¹⁰⁰ + sin x –1 decreasing ?

(A) (0,1) (B) $(\frac{π}{2}, π)$ (C) $(0, \frac{π}{2})$ (D) None of these

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Vishal Baghel

Contributor-Level 10

We have, f (x) = x 100 + sin x - 1.

So, f (x) = 100x⁹⁹ + cosx.

(A) When x (0,1). We get.

x>0

$\Rightarrow$ x⁹⁹> 0

$\Rightarrow$ 100 x⁹⁹> 0.

Now, 0 radian = 0 degree

and 1 radian = $\frac{18 0^{°} * 1}{π} = \frac{18 0^{°}}{(\frac{22}{7})} = 5 7^{°} .$

So, cosx> 0 for x∈ (0,1) radian = (0,57)

∴f (x) > 0 for x∈ (0,1).

(B) When x ∈ $(\frac{π}{2}, π)$ we get,

So, x> 1

$\Rightarrow$ x⁹⁹> 1

$\Rightarrow$ 100x⁹⁹> 100. ${\begin{matrix} ? (\frac{π}{2}, π) = (1.57, 3.14) & which is great than 1 \end{matrix}}$

And cosx is negative between -1 and 0.

So, f (x) = 100x⁹⁹ + cosx> 100 - 1 = 99 > 0.

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

(c) When x ∈ $(0, \frac{π}{2}),$ we get,

$\Rightarrow$ x> 0

$\Rightarrow$ x⁹⁹> 0

$\Rightarrow$ 100x⁹⁹> 0

and cosx> 0. (firstquadrant).

I e, f(x) > 0.

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

Hence, option (D) is correct.

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Application of Derivatives Class 12th

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

(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<

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

...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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Application of Derivatives Class 12th

29. Prove that the function f given by f(x) = x² – x + 1 is neither strictly increasing nor decreasing on (– 1, 1).

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Vishal Baghel

Contributor-Level 10

We have, f (x) = x²-x + 1.

So, f (x) = $\frac{d}{d x} (x^{2} - x + 1) = 2 x - 1$

Atf (x) = 0.

$\Rightarrow$ 2x - 1 = 0

I e, x = $\frac{1}{2}$ divides the real line into two disjoint interval the interval ( -1, 1) into

Two disjoint interval

$(- 1, \frac{1}{2}) (\frac{1}{2}, 1)$

And f (x) is strictly increasing in $(\frac{1}{2}, 1)$ and strictly decreasing in $(- 1, \frac{1}{2})$

Hence, f (x) is with a increasing or decreasing on ( -1,1).

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Application of Derivatives Class 12th

28. Prove that the logarithmic function is increasing on (0, ∞).

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Vishal Baghel

Contributor-Level 10

We have, f (x) = log x.

So, f (x) = $\frac{d}{d x} (l o g x) = \frac{1}{x} .$

i e, f (x) $= \frac{1}{x}$ > 0. When $x \neq 0 x \in (0, \infty)$

Hence, the logarithmic fx is strictly increasing on $(0, \infty) .$

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Application of Derivatives Class 12th

27. Prove that $y = \frac{4 s i n θ}{(2 + c o s θ)} - θ$ is an increasing function of θ $(0, \frac{π}{2})$ .

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Vishal Baghel

Contributor-Level 10

We have, y = $\frac{4 s i n θ}{(2 + c o s θ)} - θ$

Differentiating w rt.Ø we get,

$\frac{d y}{d θ} = \frac{d}{d θ} [\frac{4 s i n θ}{2 + c o s θ} - θ]$

$= \frac{(2 + c o r θ) \frac{d}{d θ} (4 s i n θ) - 4 s i n θ \frac{d}{d θ} (2 + c o r θ)}{{(2 + c o s θ)}^{2}} - \frac{d θ}{d θ} .$

$= \frac{(2 + c o s θ) (4 c o s θ) - 4 s i n θ (- s i n θ)}{{(2 + c o s θ)}^{2}} - 1$

$= \frac{8 c o s θ + 4 c o s^{2} θ + 4 s i n^{2} θ \cdot}{{(2 + c o s θ)}^{2} .} - 1$

$= \frac{8 c o s θ + 4 (c o s^{2} θ + s i n^{2} θ) - {(2 + c o s θ)}^{2}}{{(2 + c o s θ)}^{2}}$

$= \frac{8 c o r θ + 4 - [4 + c o s^{2} θ + 4 c o s θ]}{{(2 + c o r θ)}^{2}} .$

$= \frac{8 c o s θ + 4 - 4 - c o s^{2} θ - 4 c o s θ}{{(2 + c o s θ)}^{2}}$

$= \frac{4 c o s θ - c o s^{2} θ}{{(2 + c o s θ)}^{2}} = \frac{c o s θ (4 - c o s θ)}{{(2 + c o s θ)}^{2}}$

When $θ \in [0, \frac{π}{2}]$ we know that, $0 \leq c o s θ \leq 1 .$

So, $4 - c o s θ > 0$

And also, (2 + cos $θ$ )²> 0.

$∴ \frac{d y}{d θ} \geq 0 \forall θ \in [0, \frac{π}{2}]$

Hence, y is an increasing fxⁿ of $θ$ in $[0, \frac{π}{2}]$

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4 months ago

Application of Derivatives Class 12th

26. Find the values of x for which y = [x(x – 2)]2 is an increasing function.

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Vishal Baghel

Contributor-Level 10

We have, y = [x (x- 2)]².

Differentiating the above w rt. x we get,

$\frac{d y}{d x} = \frac{d}{d x} [x] {(x - 2)}^{2}$

$= 2 [x (x - 2)] \frac{d}{d x} [x (x - 2)]$

$= 2 [x (x - 2)] [x \frac{d}{d x} (x - 2) + (x - 2) \frac{d x}{d x}]$

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

= 2x (x - 2) (2x - 2)

$\frac{d y}{d x}$ = 4x (x - 2) (x - 1).

Now, $\frac{d y}{d x} = 0$

$\Rightarrow$ 4x (x - 2) (x - 1) = 0.

i e, x = 0, x = 2, x = 1 divides the real line into

four disjoint interval. $(- \infty, 0],$ [0,1],[1,2] and $[2, \infty] .$

when x $[- \infty, 0] .$

$\frac{d y}{d x} = (- v e) (- v e) (- v e) =$ $(- v e) \leq 0, 0 x = 0$

∴f (x) is decreasing in $[- \infty, 0] .$

When $x \in [0, 1]$

$\frac{d y}{d x} = (+ v e) (- v e) (- v e)$ $= (+ v e) \geq 0, x = 0 x = 1$

∴f (x) is increasing in [0,1].

When x $\in [1, 2]$

$\frac{d y}{d x} = (+ v e) (- v e) (+ v e) = (- v e) \leq 0,$ $0for x = 1 + x = 2$

∴f (x) is decreasing.

When $x \in (2, \infty) .$

$\frac{d y}{d x} = (+ v e) (+ v e) (+ v e) = (+ v e) \geq 0, 0,for x = 2$

∴f (x) is in creasing

Hence, f (x) is increasing for x $\in [0, 1) x \in [2, \infty)$

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Application of Derivatives Class 12th

25. Show that 2 log(1 ) 2 x y x x = +− + , x > – 1, is an increasing function of x throughout its domain.

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Vishal Baghel

Contributor-Level 10

We have, y = log $(1 + x) - \frac{2 x}{2 + x}, x > - 1$

Differentiating the above wrt.x.we get,

$\frac{d y}{d x} = \frac{1}{1 + x} \frac{d}{d x} (1 + x) -$ $\frac{(2 + x) \frac{d}{d x} 2 x - 2 x \frac{d}{d x} (2 + x)}{{(2 + x)}^{2}}$

$\Rightarrow \frac{d y}{d x} = \frac{1}{1 + x} - \frac{(2 + x) \cdot 2 - 2 x}{{(2 + x)}^{2}} .$

$\Rightarrow \frac{d y}{d x} = \frac{1}{1 + x} - \frac{4 + 2 x - 2 x}{{(2 + x)}^{2}} = \frac{1}{1 + x} - \frac{4}{{(2 + x)}^{2}} .$

$= \frac{{(2 + x)}^{2} - 4 (1 + x)}{(1 + x) {(2 + x)}^{2}}$

$= \frac{4 + x^{2} + 4 x - 4 - 4 x}{(1 + x) {(2 + x)}^{2}}$

$\frac{d y}{d x} = \frac{x^{2}}{(1 + x) {(2 + x)}^{2}} .$

The given domain of the given function isx> -1.

$\Rightarrow$ (x + 1) > 0.

Also, $x^{2} \geq 0$

(2 + x)²> 0.

Hence, $\frac{d y}{d x} = \frac{(+ v e)}{(+ v) (+ v e)} =(+ve) > 0 .$

∴ y is an increasing function of x throughout its domain.

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Application of Derivatives Class 12th

24. Find the intervals in which the following functions are strictly increasing or decreasing:

(a) x² + 2x – 5 (b) 10 – 6x – 2x² (c) –2x³ – 9x² – 12x + 1 (d) 6 – 9x – x²

(e) (x + 1)³ (x – 3)³

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Vishal Baghel

Contributor-Level 10

(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 = $\frac{- 3}{2}$

At x $(- \infty, \frac{- 3}{2}),$

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

At x $(\frac{- 3}{2}, \infty)$

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

∴f (x) is strictly decreasing on $(\frac{- 3}{2}, \infty)$

(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 =

...more

(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 = $\frac{- 3}{2}$

At x $(- \infty, \frac{- 3}{2}),$

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

At x $(\frac{- 3}{2}, \infty)$

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

∴f (x) is strictly decreasing on $(\frac{- 3}{2}, \infty)$

(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 = $\frac{- 9}{2}$

At, x ∈ $(- \infty, \frac{- 9}{2}),$

f(x) > 0

So, f(x) is strictly increasing.

At x ∈ $(\frac{- 9}{2}, \infty) .$

f(x) < 0.

So, f(x) is strictly decreasing.

(e) f(x) = (x + 1)³ (x- 3)³.

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

= (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 $(- \infty, 1) .$

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Application of Derivatives Class 12th

23. Find the intervals in which the function f given by f(x) = 2x³ – 3x² – 36x + 7 is

(a) Increasing (b) Decreasing

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Vishal Baghel

Contributor-Level 10

We have, f (x) = 2x³- 3x²- 36x + 7.

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

= 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 decr

...more

We have, f (x) = 2x³- 3x²- 36x + 7.

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

= 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)

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