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Question : 78 of 160
Marks:
+1,
-0
Solution:
78. (d)
=| y3‌cos‌√x |
| √xe1∕y2 |
⇒∫dy=∫dx On putting
=t⇒−dy=dt and on putting
u=√x⇒du=,
we get
∫=∫2‌cos‌u‌d‌u ⇒−(et)=2‌sin‌u+k ⇒−(e1∕y2)=2‌sin‌√x+k ∵y(0)=1 means at
x=0,y=1 ⇒−(e1∕y1)=2‌sin‌√0+k ⇒k=− ∴−(e1∕y2)=2‌sin‌√x− Taking log both sides with base
e,
log(−e1∕y2)=log[(2‌sin‌√x)−] ⇒log(−)+log(e1∕y2)=log(2‌sin‌√x−) ⇒log‌1−log(−2)+=log(2‌sin‌√x−) ⇒=log[(2‌sin‌√x−)×(−2)] ⇒=log(e−4‌sin‌√x) ∴f(x)=e−4‌sin‌√x
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