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

75. Find the maximum and minimum values, if any, of the following functions given by

(i) f(x) = |x + 2| - 1

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Vishal Baghel | Contributor-Level 10

4 months ago

(i) we have, f(x) = |x + 2| - 1

We know that, for all $x \in ?, | x + 2 | \geq 0$

$\Rightarrow | x + 2 | - 1 \geq - 1 .$

$\Rightarrow$ f(x) $\geq$ - 1.

∴ Minimum value of f(x) = -1 when x + 2 = 0 $\Rightarrow$ x = - 2.

And maximum value of f(x) does not exist.

(ii) $g (x) = - | x + 1 | + 3$

A(ii)

We have, $g (x) = - | x + 1 | + 3$

For all $x \in, | x + 1 | \geq 0 .$

$\Rightarrow - | x + 1 | \leq 0$

$\Rightarrow - | x + 1 | + 3 \leq 0 + 3$

$\Rightarrow$ g(x) $\leq$ 3.

∴ Maximum value of g(x) = 3 when $| x + 1 | = 0 \Rightarrow x = - 1 .$

And minimum value does not exist.

(iii) h(x) = sin (2x) + 5.

A(iii)

we have, h(x) = sin (2x) + 5.

For all $x \in ?, - 1 \leq s i n 2 x \leq 1 .$ {range of sine function is [-1, 1]}

$\Rightarrow$ -1 + 5 $\leq$ sin 2x + 5 $\leq$ 1 + 5.

$\Rightarrow$ 4 $\leq$ h(x) $\leq$ 6.

∴ Maximum value of h(x) = 6.

Minimum value of h(x) = 4.

(iv) $f (x) = | s i n 4 x + 3 | .$

A(iv)

we have, $f (x) = | s i n 4 x + 3 | .$

As for all $x \in ?, - 1 \leq s i n 4 x \leq 1$

$\Rightarrow$ -1 + 3 $\leq$ sin 4x + 3 $\leq$ 1 + 3

$\Rightarrow | 2 | \leq | s i n 4 x + 3 | \leq | 4 | .$

$\Rightarrow$ 2 $\leq$ f(x) $\leq$ 4.

∴ Maximum value of f(x) = 4.

Minimum

...more

(i) we have, f(x) = |x + 2| - 1

We know that, for all $x \in ?, | x + 2 | \geq 0$

$\Rightarrow | x + 2 | - 1 \geq - 1 .$

$\Rightarrow$ f(x) $\geq$ - 1.

∴ Minimum value of f(x) = -1 when x + 2 = 0 $\Rightarrow$ x = - 2.

And maximum value of f(x) does not exist.

(ii) $g (x) = - | x + 1 | + 3$

A(ii)

We have, $g (x) = - | x + 1 | + 3$

For all $x \in, | x + 1 | \geq 0 .$

$\Rightarrow - | x + 1 | \leq 0$

$\Rightarrow - | x + 1 | + 3 \leq 0 + 3$

$\Rightarrow$ g(x) $\leq$ 3.

∴ Maximum value of g(x) = 3 when $| x + 1 | = 0 \Rightarrow x = - 1 .$

And minimum value does not exist.

(iii) h(x) = sin (2x) + 5.

A(iii)

we have, h(x) = sin (2x) + 5.

For all $x \in ?, - 1 \leq s i n 2 x \leq 1 .$ {range of sine function is [-1, 1]}

$\Rightarrow$ -1 + 5 $\leq$ sin 2x + 5 $\leq$ 1 + 5.

$\Rightarrow$ 4 $\leq$ h(x) $\leq$ 6.

∴ Maximum value of h(x) = 6.

Minimum value of h(x) = 4.

(iv) $f (x) = | s i n 4 x + 3 | .$

A(iv)

we have, $f (x) = | s i n 4 x + 3 | .$

As for all $x \in ?, - 1 \leq s i n 4 x \leq 1$

$\Rightarrow$ -1 + 3 $\leq$ sin 4x + 3 $\leq$ 1 + 3

$\Rightarrow | 2 | \leq | s i n 4 x + 3 | \leq | 4 | .$

$\Rightarrow$ 2 $\leq$ f(x) $\leq$ 4.

∴ Maximum value of f(x) = 4.

Minimum value of f(x) = 2.

(v) h(x) = x + 1.,x∈ ( -1, 1).

A(v)

we have, h(x) = x + 1.,x∈ ( -1, 1).

Given, -1

- 1 + 1 <x + 1 < 1 + 1

0 <h (x) < 2.

∴ Maximum value and minimum value of h(x) does not exist.

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(i) we have, f(x) = |x + 2| - 1We know that, for all <math><mrow><mi>x</mi><mo>∈</mo><mi>?</mi><mo>,</mo><mo>|</mo><mi>x</mi><mo>+</mo><mn>2</mn><mo>|</mo><mo>≥</mo><mn>0</mn></mrow></math><math><mrow><mo>⇒</mo><mo>|</mo><mi>x</mi><mo>+</mo><mn>2</mn><mo>|</mo><mo>−</mo><mn>1</mn><mo>≥</mo><mo>−</mo><mn>1</mn><mo>.</mo></mrow></math><math><mrow><mo>⇒</mo></mrow></math> f(x)<math><mrow><mo>≥</mo></mrow></math>- 1.∴ Minimum value of f(x) = -1 when x + 2 = 0 <math><mrow><mo>⇒</mo></mrow></math> x = - 2.And maximum value of f(x) does not exist.(ii) <math><mrow><mi>g</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mo>−</mo><mo>|</mo><mi>x</mi><mo>+</mo><mn>1</mn><mo>|</mo><mo>+</mo><mn>3</mn></mrow></math>A(ii)We have, <math><mrow><mi>g</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mo>−</mo><mo>|</mo><mi>x</mi><mo>+</mo><mn>1</mn><mo>|</mo><mo>+</mo><mn>3</mn></mrow></math>For all <math><mrow><mi>x</mi><mo>∈</mo><mo>,</mo><mtext> </mtext><mo>|</mo><mi>x</mi><mo>+</mo><mn>1</mn><mo>|</mo><mo>≥</mo><mn>0</mn><mo>.</mo></mrow></math><math><mrow><mo>⇒</mo><mo>−</mo><mo>|</mo><mi>x</mi><mo>+</mo><mn>1</mn><mo>|</mo><mo>≤</mo><mn>0</mn></mrow></math><math><mrow><mo>⇒</mo><mo>−</mo><mo>|</mo><mi>x</mi><mo>+</mo><mn>1</mn><mo>|</mo><mo>+</mo><mn>3</mn><mo>≤</mo><mn>0</mn><mo>+</mo><mn>3</mn></mrow></math><math><mrow><mo>⇒</mo></mrow></math> g(x) <math><mrow><mo>≤</mo><mn></mn></mrow></math>3.∴ Maximum value of g(x) = 3 when <math><mrow><mo>|</mo><mi>x</mi><mo>+</mo><mn>1</mn><mo>|</mo><mo>=</mo><mn>0</mn><mo>⇒</mo><mi>x</mi><mo>=</mo><mo>−</mo><mn>1</mn><mo>.</mo></mrow></math>And minimum value does not exist.(iii) h(x) = sin (2x) + 5.A(iii)we have, h(x) = sin (2x) + 5.For all <math><mrow><mi>x</mi><mo>∈</mo><mi>?</mi><mo>,</mo><mtext> </mtext><mo>−</mo><mn>1</mn><mo>≤</mo><mi>s</mi><mi>i</mi><mi>n</mi><mn>2</mn><mi>x</mi><mo>≤</mo><mn>1</mn><mo>.</mo></mrow></math> {range of sine function is [-1, 1]}<math><mrow><mo>⇒</mo></mrow></math> -1 + 5 <math><mrow><mo>≤</mo></mrow></math>sin 2x + 5 <math><mrow><mo>≤</mo></mrow></math>1 + 5.<math><mrow><mo>⇒</mo></mrow></math> 4 <math><mrow><mo>≤</mo></mrow></math>h(x) <math><mrow><mo>≤</mo></mrow></math>6.∴ Maximum value of h(x) = 6.Minimum value of h(x) = 4.(iv) <math><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mo>|</mo><mi>s</mi><mi>i</mi><mi>n</mi><mn>4</mn><mi>x</mi><mo>+</mo><mn>3</mn><mo>|</mo><mo>.</mo></mrow></math>A(iv)we have, <math><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mo>|</mo><mi>s</mi><mi>i</mi><mi>n</mi><mn>4</mn><mi>x</mi><mo>+</mo><mn>3</mn><mo>|</mo><mo>.</mo></mrow></math>As for all <math><mrow><mi>x</mi><mo>∈</mo><mi>?</mi><mo>,</mo><mo>−</mo><mn>1</mn><mo>≤</mo><mi>s</mi><mi>i</mi><mi>n</mi><mn>4</mn><mi>x</mi><mo>≤</mo><mn>1</mn></mrow></math><math><mrow><mo>⇒</mo></mrow></math> -1 + 3 <math><mrow><mo>≤</mo></mrow></math>sin 4x + 3<math><mrow><mo>≤</mo></mrow></math> 1 + 3<math><mrow><mo>⇒</mo><mtext> </mtext><mo>|</mo><mn>2</mn><mo>|</mo><mo>≤</mo><mo>|</mo><mi>s</mi><mi>i</mi><mi>n</mi><mn>4</mn><mi>x</mi><mo>+</mo><mn>3</mn><mo>|</mo><mo>≤</mo><mo>|</mo><mn>4</mn><mo>|</mo><mo>.</mo></mrow></math><math><mrow><mo>⇒</mo></mrow></math> 2 <math><mrow><mo>≤</mo></mrow></math>f(x) <math><mrow><mo>≤</mo></mrow></math>4.∴ Maximum value of f(x) = 4.Minimum value of f(x) = 2.(v) h(x) = x + 1.,x∈ ( -1, 1).A(v)we have, h(x) = x + 1.,x∈ ( -1, 1).Given, -1<x<1< strong=""></x<1<>- 1 + 1 <x + 1 < 1 + 10 <h (x) < 2.∴ Maximum value and minimum value of h(x) does not exist.

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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.
For tangent at (a, b), slope is m = dx/dy = 1 / (dy/dx) = 1 / (2a² - 15a + 10).
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2a² - 15a + 10 = -3
2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
Following the image: 2a² - 15a + 7 = 0
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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:

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

Area

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

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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

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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75. Find the maximum and minimum values, if any, of the following functions given by

(i) f(x) = |x + 2| - 1

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75. Find the maximum and minimum values, if any, of the following functions given by

(i) f(x) = |x + 2| - 1