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Hatshy [7]
3 years ago
13

Consider the function f(x) = 6sin(x^2) on the interval 0 ≤ x ≤ 3.

Mathematics
2 answers:
Vitek1552 [10]3 years ago
4 0
<span> a) F' = 6 sin(x^2) = 0
x^2 = pi
x = sqrt(pi)

b) Fmax = F(1) + integral [1, pi] f(x) dx = 9.7743 </span>
stira [4]3 years ago
4 0

Solution:

The given function is

f(x)= 6 sin x²→→→0 ≤ x ≤ 3.

Differentiating once

f'(x)= 6 cos x²× 2 x=12 x cos x²

For Maximum or Minimum

f'(x)=0

12 x cos x²=0

cos x²=0 ∧ x=0

cos x²=cos \frac{\pi }{2}

x²=  \frac{\pi}{2}

x= \pm \sqrt\frac{\pi}{2}

As, the interval given is,  0 ≤ x ≤ 3.

x=  \sqrt\frac{\pi}{2}= \frac{22}{14}

f"(x)=12 (-sin x²) × 2 x × x + 12 cos x²= 24 x². (-sin x²)+ 12 cos x²

At, x=  \sqrt\frac{\pi}{2}= \frac{22}{14}

f"(x)=-24 × \frac{\pi}{2} ×1 + 0= A negative number

Showing the function attains maxima at this point.

f(0)=6 sin 0=0

f(3)= 6 sin 3²= 6 sin 9(radian)=6 × 0.40(approx)= 2.40

f( \sqrt\frac{\pi}{2})=6 × sin(\frac{\pi}{2} )= 6 ×1=6→→Maximum Value

(b) f(1)=5

I.e at x=1, y=5

Maximum value attained by , f(x)=6 sin x² is 6 at , x=  \sqrt\frac{\pi}{2}.


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\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ X(\stackrel{x_1}{-6}~,~\stackrel{y_1}{2})\qquad Z(\stackrel{x_2}{5}~,~\stackrel{y_2}{8})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ XZ=\sqrt{[5-(-6)]^2+[8-2]^2}\implies XZ=\sqrt{(5+6)^2+(8-2)^2} \\\\\\ XZ=\sqrt{121+36}\implies \boxed{XZ=\sqrt{157}} \\\\[-0.35em] ~\dotfill

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Because 14x = 25y    Which means x must equal more than y. And 25/14 is the only answer like this.

For the second question, the answer is 30 x 18.

30/5 = 6     18/3 = 6      18 + 18 + 30 + 30 = 96

So, both the perimeter and the ratio is correct with this answer.

For the third question, the first answer is correct.

x < -4    and only a negative number can be less than another negative. Also, the negative must have a higher value to be less than another negative.

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sdas [7]

Given:

\tan \theta > 0 and \sec \theta.

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The quadrant.

Solution:

We know that,

In I quadrant, all trigonometric ratios are positive.

In II quadrant, only \sin \theta, \text{cosec}\theta are positive and others are negative.

In III quadrant, only \cot \theta, \tan\theta are positive and others are negative.

In IV quadrant, only \cos \theta, \sec\theta are positive and others are negative.

We have,

\tan \theta > 0 and \sec \theta.

Here, \tan \theta is positive and \sec \theta is negative. So, \theta lies in the III quadrant.

Therefore, the correct option is C, i.e., III.

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