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

78. Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i) f(x) =x³ , x ∈ [– 2, 2]

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Vishal Baghel | Contributor-Level 10

4 months ago

(i) We have,

f(x) = x³ , x ∈ [– 2, 2].

f(x) = 3x².

At, f(x) = 0

3x² = 0

x = 0 <--[-2, 2].

We shall absolute the value of f at x = 0 and points of interval [ -2, 2]. So,

f(0) = 0

f(- 2) = (- 2)³ = 8

f(2) = 2³ = 8.

∴ Absolute maximum value of f(x) = 8 at x = 2

and absolute minimum value of f(x) = -8 at x = -2.

(ii) f (x) = sin x + cos x , x ∈ [0, π]

A.(ii)

We have, f(x) = sin x + cos x , x ∈ [0, π]

f(x) = cos x - sin x.

atf(x) = 0

$\Rightarrow$ cosx - sin x = 0

$\Rightarrow$ sinx = cos x

$\Rightarrow \frac{s i n^{°} x}{c o r x} = 1 t a n x = 1 \Rightarrow t a n x = t a n \frac{π}{4}$

$\Rightarrow x = \frac{π}{4} \in [0, π]$

(iii) f(x) = 4x $- \frac{1}{2} x^{2}, x \in [- 2, \frac{9}{2}]$

A.(iii)

We have, f(x) = 4x $- \frac{1}{2} x^{2}, x \in [- 2, \frac{9}{2}]$

f(x) = 4 - x

atf(x) = 0

$\Rightarrow$ 4- x = 0

$\Rightarrow$ x = 4 $\in$ $[-2, \frac{9}{2}]$

$∴ f (4) = 4 (4) - \frac{1}{2} {(4)}^{2} = 16 - 8 = 8 .$

$f (- 2) = 4 (- 2) - \frac{1}{2} {(- 2)}^{2} = - 8 - 2 = - 10 .$

$f (\frac{9}{2}) = 4 (\frac{9}{2}) - \frac{1}{2} {(\frac{9}{2})}^{2} = 18 - \frac{81}{8} = \frac{94 - 81}{8} = \frac{63}{8} .$

= 7.87.5

Hence, absolute maximum value of f(x) = 8 at x = 4

and absolu

...more

(i) We have,

f(x) = x³ , x ∈ [– 2, 2].

f(x) = 3x².

At, f(x) = 0

3x² = 0

x = 0 <--[-2, 2].

We shall absolute the value of f at x = 0 and points of interval [ -2, 2]. So,

f(0) = 0

f(- 2) = (- 2)³ = 8

f(2) = 2³ = 8.

∴ Absolute maximum value of f(x) = 8 at x = 2

and absolute minimum value of f(x) = -8 at x = -2.

(ii) f (x) = sin x + cos x , x ∈ [0, π]

A.(ii)

We have, f(x) = sin x + cos x , x ∈ [0, π]

f(x) = cos x - sin x.

atf(x) = 0

$\Rightarrow$ cosx - sin x = 0

$\Rightarrow$ sinx = cos x

$\Rightarrow \frac{s i n^{°} x}{c o r x} = 1 t a n x = 1 \Rightarrow t a n x = t a n \frac{π}{4}$

$\Rightarrow x = \frac{π}{4} \in [0, π]$

(iii) f(x) = 4x $- \frac{1}{2} x^{2}, x \in [- 2, \frac{9}{2}]$

A.(iii)

We have, f(x) = 4x $- \frac{1}{2} x^{2}, x \in [- 2, \frac{9}{2}]$

f(x) = 4 - x

atf(x) = 0

$\Rightarrow$ 4- x = 0

$\Rightarrow$ x = 4 $\in$ $[-2, \frac{9}{2}]$

$∴ f (4) = 4 (4) - \frac{1}{2} {(4)}^{2} = 16 - 8 = 8 .$

$f (- 2) = 4 (- 2) - \frac{1}{2} {(- 2)}^{2} = - 8 - 2 = - 10 .$

$f (\frac{9}{2}) = 4 (\frac{9}{2}) - \frac{1}{2} {(\frac{9}{2})}^{2} = 18 - \frac{81}{8} = \frac{94 - 81}{8} = \frac{63}{8} .$

= 7.87.5

Hence, absolute maximum value of f(x) = 8 at x = 4

and absolute minimum value of f(x) = -10 at x = -2.

(iv) f(x) = (x - 1)² + 3, x∈[ -3, 1]

A.(iv)

We have, f(x) = (x - 1)² + 3, x∈[ -3, 1]

f(x) = 2(x -1)

atf(x) = 0

2(x - 1) = 0

x = 1∈ [ -3, 1]

∴f(1) = (1 - 1)² + 3 = 3

f(- 3) = (- 3 - 1)² + 3 = 16 + 3 = 1

∴ Absolute maximum value of f(x) = 19 at x = -3.

And absolute minimum value of f(x) = 3 at x = 1.

less

(i) We have,f(x) = x3 , x ∈ [– 2, 2].f(x) = 3x2.At, f(x) = 03x2 = 0x = 0 <--[-2, 2].We shall absolute the value of f at x = 0 and points of interval [ -2, 2]. So,f(0) = 0f(- 2) = (- 2)3 = 8f(2) = 23 = 8.∴ Absolute maximum value of f(x) = 8 at x = 2and absolute minimum value of f(x) = -8 at x = -2.(ii) f (x) = sin x + cos x , x ∈ [0, π]A.(ii)We have, f(x) = sin x + cos x , x ∈ [0, π]f(x) = cos x - sin x.atf(x) = 0<math><mrow><mo>⇒</mo></mrow></math> cosx - sin x = 0<math><mrow><mo>⇒</mo></mrow></math> sinx = cos x<math><mrow><mo>⇒</mo><mfrac><mrow><mi>s</mi><mi>i</mi><msup><mrow><mi>n</mi></mrow><mrow><mo>°</mo></mrow></msup><mi>x</mi></mrow><mrow><mi>c</mi><mi>o</mi><mi>r</mi><mi>x</mi></mrow></mfrac><mo>=</mo><mn>1</mn><mtext> </mtext><mi>t</mi><mi>a</mi><mi>n</mi><mi>x</mi><mo>=</mo><mn>1</mn><mo>⇒</mo><mi>t</mi><mi>a</mi><mi>n</mi><mi>x</mi><mo>=</mo><mi>t</mi><mi>a</mi><mi>n</mi><mfrac><mrow><mi>π</mi></mrow><mrow><mn>4</mn></mrow></mfrac></mrow></math><math><mrow><mo>⇒</mo><mi>x</mi><mo>=</mo><mfrac><mrow><mi>π</mi></mrow><mrow><mn>4</mn></mrow></mfrac><mo>∈</mo><mrow><mo>[</mo><mrow><mn>0</mn><mo>,</mo><mi>π</mi></mrow><mo>]</mo></mrow></mrow></math><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1734072978phpTrh5em_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1734072978phpTrh5em.jpeg" alt="" width="238" height="109"></picture></div></div>(iii) f(x) = 4x <math><mrow><mo>−</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>,</mo><mi>x</mi><mo>∈</mo><mrow><mo>[</mo><mrow><mo>−</mo><mn>2</mn><mo>,</mo><mfrac><mrow><mn>9</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow><mo>]</mo></mrow></mrow></math>A.(iii)We have, f(x) = 4x <math><mrow><mo>−</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>,</mo><mi>x</mi><mo>∈</mo><mrow><mo>[</mo><mrow><mo>−</mo><mn>2</mn><mo>,</mo><mfrac><mrow><mn>9</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow><mo>]</mo></mrow></mrow></math>f(x) = 4 - xatf(x) = 0<math><mrow><mo>⇒</mo></mrow></math> 4-  x = 0<math><mrow><mo>⇒</mo></mrow></math> x = 4 <math><mrow><mo>∈</mo></mrow></math><math><mrow><mrow><mo>[</mo><mrow><mo>−2,</mo><mfrac bevelled="true"><mrow><mn>9</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow><mo>]</mo></mrow></mrow></math><math><mrow><mo>∴</mo><mi>f</mi><mo stretchy="false">(</mo><mn>4</mn><mo stretchy="false">)</mo><mo>=</mo><mn>4</mn><mo stretchy="false">(</mo><mn>4</mn><mo stretchy="false">)</mo><mo>−</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><msup><mrow><mo stretchy="false">(</mo><mn>4</mn><mo stretchy="false">)</mo></mrow><mrow><mn>2</mn></mrow></msup><mo>=</mo><mn>1</mn><mn>6</mn><mo>−</mo><mn>8</mn><mo>=</mo><mn>8</mn><mo>.</mo></mrow></math><math><mrow><mi>f</mi><mo stretchy="false">(</mo><mo>−</mo><mn>2</mn><mo stretchy="false">)</mo><mo>=</mo><mn>4</mn><mo stretchy="false">(</mo><mo>−</mo><mn>2</mn><mo stretchy="false">)</mo><mo>−</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><msup><mrow><mo stretchy="false">(</mo><mo>−</mo><mn>2</mn><mo stretchy="false">)</mo></mrow><mrow><mn>2</mn></mrow></msup><mo>=</mo><mo>−</mo><mn>8</mn><mo>−</mo><mn>2</mn><mo>=</mo><mo>−</mo><mn>1</mn><mn>0</mn><mo>.</mo></mrow></math><math><mrow><mi>f</mi><mrow><mo>(</mo><mrow><mfrac><mrow><mn>9</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow><mo>)</mo></mrow><mo>=</mo><mn>4</mn><mrow><mo>(</mo><mrow><mfrac><mrow><mn>9</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow><mo>)</mo></mrow><mo>−</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><msup><mrow><mrow><mo>(</mo><mrow><mfrac><mrow><mn>9</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow><mo>)</mo></mrow></mrow><mrow><mn>2</mn></mrow></msup><mo>=</mo><mn>1</mn><mn>8</mn><mo>−</mo><mfrac><mrow><mn>8</mn><mn>1</mn></mrow><mrow><mn>8</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>9</mn><mn>4</mn><mo>−</mo><mn>8</mn><mn>1</mn></mrow><mrow><mn>8</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>6</mn><mn>3</mn></mrow><mrow><mn>8</mn></mrow></mfrac><mo>.</mo></mrow></math>= 7.87.5Hence, absolute maximum value of f(x) = 8 at x = 4and absolute minimum value of f(x) = -10 at x = -2.(iv) f(x) = (x - 1)2 + 3, x∈[ -3, 1]A.(iv)We have, f(x) = (x - 1)2 + 3, x∈[ -3, 1]f(x) = 2(x -1)atf(x) = 02(x - 1) = 0x = 1∈ [ -3, 1]∴f(1) = (1 - 1)2 + 3 = 3f(- 3) = (- 3 - 1)2 + 3 = 16 + 3 = 1∴ Absolute maximum value of f(x) = 19 at x = -3.And absolute minimum value of f(x) = 3 at x = 1.

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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..........

Raj Pandey

y (x) = ∫? (2t² - 15t + 10)dt
dy/dx = 2x² - 15x + 10.
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2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
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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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f' (c) = 1 + lnc = e/ (e-1)
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The lengths of the sides of a triangle are 10 + x2, 10 + x2 and 20 – 2x2. If for x = k, the area of the triangle is maximum, then 3k2 is equal to:

Raj Pandey

$C D = \sqrt[]{{(10 + x^{2})}^{2} - {(10 - x^{2})}^{2}} = 2 \sqrt[]{10} | x |$

Area

$= \frac{1}{2} \times C D \times A B = \frac{1}{2} \times 2 \sqrt[]{10} | x | (20 - 2 x^{2})$

$\Rightarrow 10 - x^{2} = 2 x$

3x² = 10

x = k

3k² = 10

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

So F₁ (A, B, C) is not a tautology

Now again by truth table

So F₂ (A, B) be a tautology.

The triangle of maximum area that can be inscribed in a given circle of radius ‘r’ is:

Vishal Baghel

From option let it be isosceles where AB = AC then

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

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

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

Now ar $(Δ A B C) = Δ = \frac{1}{2} B C \times A L$

$Δ = \frac{1}{2} \times 2 \sqrt[]{2 h r - a^{2}} \times h$

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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78. Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i) f(x) =x³ , x ∈ [– 2, 2]

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78. Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (i) f(x) =x3 , x ∈ [– 2, 2]

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78. Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i) f(x) =x³ , x ∈ [– 2, 2]