Select all critical numbers of F in the interval (0, 27) from the list below. If none of the entries on the list are critical numbers of F, select "None of the above". TT/2 Зп /2 None of the above The absolute maximum value of F on the interval 0, 27 is equal to O F(r) O 27 O None of the above

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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For 0 < x < 27, define the function F(x)= |
sint
dt.
1+t
sint
The graph of
on the interval 0, 27 is shown below.
1+t
Sin t
1+ t
0.4
0.2
3 T
2 л
-0.2
Select all critical numbers of F in the interval (0, 27) from the list below. If none of the entries on the list are critical numbers of F, select "None of the
above".
TT/2
Зп /2
None of the above
The absolute maximum value of F on the interval 0, 27 is equal to
O F(T)
O None of the above
Transcribed Image Text:For 0 < x < 27, define the function F(x)= | sint dt. 1+t sint The graph of on the interval 0, 27 is shown below. 1+t Sin t 1+ t 0.4 0.2 3 T 2 л -0.2 Select all critical numbers of F in the interval (0, 27) from the list below. If none of the entries on the list are critical numbers of F, select "None of the above". TT/2 Зп /2 None of the above The absolute maximum value of F on the interval 0, 27 is equal to O F(T) O None of the above
Expert Solution
Step 1

Given function:

Fx=0xsint1+tdt ;     0x2π

Let y=sint1+t

Differentiating with respect to t,

y'=1+tcost-sint11+t2

For critical number, y'=0

1+tcost-sint1=0                 1+tcost=sint                           1+t=tant

Step 2

Now, given options are π2,π and 3π2

Critical numbers of tant are t=π2+nπ;   for n=0,±1,±2,..

Thus, the critical numbers of the given function are  π2and 3π2.

Therefore, options (1) and (3) are correct.

Now, checking on points 0,π2,3π2 and 2π,

t=0:     y=sin01+0=0t=π2:   y=11+1.57=0.39t=3π2: y=-11+4.7=-0.17

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