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

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

4 months ago

(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

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

less

(a) f (x) = x2 + 2x - 5.f(x) = 2x + 2 = 2 (x + 1).At, f(x) = 02 (x + 1) = 0<math><mrow><mo>⇒</mo></mrow></math> x = -1.At, x <math><mrow><mo stretchy="false">(</mo><mo>−</mo><mo>∞</mo><mo>,</mo><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo><mo>,</mo></mrow></math>f(x) = (- ) ve< 0.So, f (x)is strictly decreasing or <math><mrow><mo stretchy="false">(</mo><mo>−</mo><mo>∞</mo><mo>,</mo><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo><mo>.</mo></mrow></math>At x ∈<math><mrow><mo stretchy="false">(</mo><mo>−</mo><mn>1</mn><mo>,</mo><mo>∞</mo><mo stretchy="false">)</mo></mrow></math>f(x) = ( + ve) > 1f(x) is strictly increasing on <math><mrow><mo stretchy="false">(</mo><mo>−</mo><mn>1</mn><mo>,</mo><mo>∞</mo><mo stretchy="false">)</mo><mo>.</mo></mrow></math>(b) f(x) = 10 - 6x- 2x2So, f(x) = - 6 - 4x = - 2 (3 + 2x).Atf(x) = 0<math><mrow><mo>⇒</mo></mrow></math> 2 (3 + 2x) = 0.<math><mrow><mo>⇒</mo></mrow></math> x = <math><mrow><mfrac bevelled="true"><mrow><mo>−</mo><mn>3</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow></math>At x <math><mrow><mo stretchy="false">(</mo><mo>−</mo><mo>∞</mo><mo>,</mo><mfrac bevelled="true"><mrow><mo>−</mo><mn>3</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mo stretchy="false">)</mo><mo>,</mo></mrow></math>∴f(x) is strictly increasing on <math><mrow><mrow><mo>(</mo><mrow><mo>−</mo><mo>∞</mo><mo>,</mo><mfrac bevelled="true"><mrow><mo>−</mo><mn>3</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow><mo>)</mo></mrow></mrow></math>At x <math><mrow><mrow><mo>(</mo><mrow><mfrac bevelled="true"><mrow><mo>−</mo><mn>3</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mo>,</mo><mo>∞</mo></mrow><mo>)</mo></mrow></mrow></math>f(x) = ( -ve) ( + ve) = ( - ) ve< 0.∴f (x) is strictly decreasing on <math><mrow><mrow><mo>(</mo><mrow><mfrac bevelled="true"><mrow><mo>−</mo><mn>3</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mo>,</mo><mo>∞</mo></mrow><mo>)</mo></mrow></mrow></math>(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.<math><mrow><mo>⇒</mo></mrow></math> 6 (x + 1) (x + 2) = 0<math><mrow><mo>⇒</mo></mrow></math> x = -1 and x = -2.At, x <math><mrow><mo stretchy="false">(</mo><mo>−</mo><mo>∞</mo><mo>,</mo><mo>−</mo><mn>2</mn><mo stretchy="false">)</mo></mrow></math> , f(x) = ( -ve) ( -ve) ( -ve) = ( -ve) < 0So, 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-x2So, f(x) = - 9 - 2xAt, f(x) = 0<math><mrow><mo>⇒</mo></mrow></math> 9 - 2x = 0<math><mrow><mo>⇒</mo></mrow></math> x =<math><mrow><mrow><mrow><mo></mo><mfrac bevelled="true"><mrow><mo>−</mo><mn>9</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow></mrow></mrow></math>At, x ∈<math><mrow><mrow><mo>(</mo><mrow><mo>−</mo><mo>∞</mo><mo>,</mo><mfrac bevelled="true"><mrow><mo>−</mo><mn>9</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow><mo>)</mo></mrow><mo>,</mo></mrow></math>f(x) > 0So, f(x) is strictly increasing.At x ∈<math><mrow><mrow><mo>(</mo><mrow><mfrac bevelled="true"><mrow><mo>−</mo><mn>9</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mo>,</mo><mo>∞</mo></mrow><mo>)</mo></mrow><mo>.</mo></mrow></math>f(x) < 0.So, f(x) is strictly decreasing.(e) f(x) = (x + 1)3 (x- 3)3.So, f(x) = (x + 1)3 <math><mrow><mfrac><mrow><mi>d</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><msup><mrow><mo stretchy="false">(</mo><mi>x</mi><mo>−</mo><mn>3</mn><mo stretchy="false">)</mo></mrow><mrow><mn>3</mn></mrow></msup><mo>+</mo><msup><mrow><mo stretchy="false">(</mo><mi>x</mi><mo>−</mo><mn>3</mn><mo stretchy="false">)</mo></mrow><mrow><mn>3</mn></mrow></msup><mfrac><mrow><mi>d</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><msup><mrow><mo stretchy="false">(</mo><mi>x</mi><mo>+</mo><mn>1</mn><mo stretchy="false">)</mo></mrow><mrow><mn>3</mn></mrow></msup></mrow></math>= (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.<math><mrow><mo>⇒</mo></mrow></math> x> 1.∴f (x) is strictly increasing on <math><mrow><mo stretchy="false">(</mo><mn>1</mn><mo>,</mo><mo>∞</mo><mo stretchy="false">)</mo><mo>.</mo></mrow></math>And strictly decreasing on <math><mrow><mo stretchy="false">(</mo><mo>−</mo><mo>∞</mo><mo>,</mo><mn>1</mn><mo stretchy="false">)</mo><mo>.</mo></mrow></math>

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If the tangent to the curve, y = f(x) = xlog? x, (x > 0) at a point (c, f(c)) is parallel to the line - segment joining the points (1,0) and (e, e) then c is equal to:

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Let F₁(A,B,C) = $(A \land \sim B) \lor$ $[\sim C \land (A \lor B)] \lor \sim A a n d F_{2} (A, B) = (A \lor B) \lor (B \to \sim A)$ be two logical expressions. Then:

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By truth table

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Vishal Baghel

From option let it be isosceles where AB = AC then

$x = \sqrt[]{r^{2} - {(h - r)}^{2}}$

= $\sqrt[]{r^{2} - h^{2} - r^{2} + 2 r h}$

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then $x = \sqrt[]{2 \times \frac{3 r}{2} \times r - \frac{9 r^{2}}{4}} = \frac{\sqrt[]{3}}{2} r f r o m (i)$

$\Rightarrow B C = \sqrt[]{3} r$

So $A B = \sqrt[]{h^{2} + x^{2}} = \sqrt[]{\frac{9 r^{2}}{4} + \frac{3 r^{2}}{4}} = \sqrt[]{3} r$ .

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

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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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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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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