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

85. Find the maximum and minimum values of x + sin 2x on [0, 2π].

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Vishal Baghel | Contributor-Level 10

4 months ago

We have, f(x) = x + sin 2x ,x ∈ [0, 2π].

f(x) = 1 + 2cos 2x

At f(x) = 0

1 + 2 cos2x = 0

$\Rightarrow c o s 2 x = - \frac{1}{2} = - c o s \frac{π}{3} = c o s π - \frac{π}{3} = c o s \frac{2 π}{3}$

$∴ 2 x = 2 n π \pm \frac{2 π}{3}, n = 1, 2, 3 .$

$\Rightarrow x = n π \pm \frac{2 π}{3}$

$n = 0, x = \pm \frac{π}{3} \to x = \frac{π}{3} \in [0, 2 π] .$

$n = 1, x = π \pm \frac{π}{3} \to x = π + \frac{π}{3} a n d π - \frac{π}{3}$

$= \frac{4 π}{3} a d \frac{2 π}{3} \in [0, 2 π]$

$n = 2, x = 2 π \pm \frac{π}{3} \to x = 2 π + \frac{π}{3} a d 2 π - \frac{π}{3} .$

$\Rightarrow x = \frac{51}{3} \in [0, 2 π] .$

Hence, $x = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3} a n d \frac{5 π}{3}$

Missing

At $x = \frac{π}{3}, f (\frac{π}{3}) = \frac{π}{3} + s i n \frac{2 π}{3} = 1.05 +$ $s i n (π - \frac{π}{3}) = 1.05 + s i n \frac{π}{3}$

$= 1.05 +\sqrt3/2$

= 1.05 + 0.87

= 1.92

At $x = \frac{2 π}{3}, f (\frac{2 π}{3}) = \frac{2 π}{3} + s i n 2 \times \frac{2 π}{3}$ $= 2.10 + s i n (π + \frac{π}{3})$

$= 2.10 - s i n \frac{π}{3} = 2, 10 - 0.87$

= 1.23

At $x = \frac{4 π}{3}, f (\frac{4 π}{3}) = \frac{4 π}{3} + s i n 2 \times \frac{4 π}{3} = 4 \cdot 2 + s i n (3 x - \frac{π}{3})$

$= 4.2 + s i n \frac{π}{3} = 4.2 + 0.87 .$

=5.07.

At $x = \frac{5 π}{3}, f (\frac{5 π}{3}) = \frac{5 π}{3} + s i n 2 \times \frac{5 π}{3} = 5.25 + s i n (3 x + \frac{π}{3})$

$= 5 \cdot 25 - s i n \frac{1}{3}$

= 5.25 - 0.87 = 4.38

At and points,

f(0) = 0 + sin2 × 0 = 0

f(2π) = 2π + sin 2 × 2π = 6.2 + 0 = 6.28

∴Maximum value of f(x) = 6.28 at x = 2π and

minimum value of f(x) = 0 at x= 0

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By truth table

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