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

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

4 months ago

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

...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)

less

We have, f (x) = 2x3- 3x2- 36x + 7.So, f (x) = <math><mrow><mfrac><mrow><mi>d</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mrow><mo> (</mo><mrow><mn>2</mn><msup><mrow><mi>x</mi></mrow><mrow><mn>3</mn></mrow></msup><mo>−</mo><mn>3</mn><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><mn>3</mn><mn>6</mn><mi>x</mi><mo>+</mo><mn>7</mn></mrow><mo>)</mo></mrow><mo>=</mo><mn>6</mn><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><mn>6</mn><mi>x</mi><mo>−</mo><mn>3</mn><mn>6</mn><mo>.</mo></mrow></math>= 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<math><mrow><mo>⇒</mo></mrow></math> 6 (x- 3) (x + 2) = 0.So, when x- 3 = 0 or x + 2 = 0.<math><mrow><mo>⇒</mo></mrow></math> x = 3 or x = -2.Hence we an divide the real line into three disjoint internalI <math><mrow><mo stretchy="false"> (</mo><mo>−</mo><mo>∞</mo><mo>, </mo><mo>−</mo><mn>2</mn><mo stretchy="false">) II</mo><mo stretchy="false"> (</mo><mo>−</mo><mn>2</mn><mo>, </mo><mn>3</mn><mo stretchy="false">)andIII</mo><mo stretchy="false"> (</mo><mn>3</mn><mo>, </mo><mo>∞</mo><mo stretchy="false">)</mo></mrow></math>At x ∈<math><mrow><mo stretchy="false"> (</mo><mo>−</mo><mo>∞</mo><mo>, </mo><mo>−</mo><mn>2</mn><mo stretchy="false">)</mo><mo>, </mo></mrow></math>f (x) = ( + ve) ( -ve) ( -ve) = ( + ve) > 0.So, f (x) is strictly increasing in <math><mrow><mo stretchy="false"> (</mo><mo>−</mo><mo>∞</mo><mo>, </mo><mo>−</mo><mn>2</mn><mo stretchy="false">)</mo></mrow></math>At, x∈ ( -2,3), f (x) = ( + ve) ( + ve) ( -ve) = ( -ve) < 0.So, f (x) is strictly decreasing in ( -2,3).At, x ∈<math><mrow><mo stretchy="false"> (</mo><mn>3</mn><mo>, </mo><mo>∞</mo><mo stretchy="false">)</mo></mrow></math>f (x) = ( + ve) ( + ve) ( + ve) = ( + )ve> 0.So, f (x) is strictly increasing in <math><mrow><mo stretchy="false"> (</mo><mn>3</mn><mo>, </mo><mo>∞</mo><mo stretchy="false">)</mo></mrow></math>∴ (a) f (x) is strictly increasing in <math><mrow><mo stretchy="false"> (</mo><mo>−</mo><mo>∞</mo><mo>, </mo><mo>−</mo><mn>2</mn><mo stretchy="false">)and</mo><mo stretchy="false"> (</mo><mn>3</mn><mo>, </mo><mo>−</mo><mo>∞</mo><mo stretchy="false">)</mo></mrow></math> (b) f (x) is strictly decreasing in ( -2,3)

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If the normal to the curve y(x) = ∫(from 2 to x) (2t² - 15t + 10)dt at a point (a, b) is parallel to the line x + 3y = -5, a > 1, then the value of |a + 6b| is equal to..........

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y (x) = ∫? (2t² - 15t + 10)dt
dy/dx = 2x² - 15x + 10.
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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:

Vishal Baghel

By truth table

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From option let it be isosceles where AB = AC then

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

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

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

(a) Increasing (b) Decreasing