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Question : 123 of 150
Marks:
+1,
-0
Solution:
Given,
f(x)=|| sin‌3x | 1 | 2(cos‌+sin‌‌)2 |
| cos‌3‌x | −1 | 2(cos2‌−sin‌2‌) |
| tan‌3‌x | 4 | 1+2‌tan‌3‌x |
|On differentiating w.r.t.
x, we get
f′(x)=|| ‌(sin‌3x) | 1 | 2(cos‌+sin‌‌)2 |
| ‌(cos‌3‌x) | −1 | 2(cos2‌−sin‌2‌) |
| ‌(tan‌3‌x) | 4 | 1+2‌tan‌3‌x |
| +|| sin‌3x | ‌(1) | 2(cos‌+sin‌‌)2 |
| cos‌3‌x | ‌(−1) | 2(cos2‌−sin‌2‌) |
| tan‌3‌x | ‌(4) | 1+2‌tan‌3‌x |
|+|| sin‌3x | 1 | 2‌(cos‌+sin‌‌)2 |
| cos‌3‌x | −1 | 2‌(cos2‌−sin‌2‌) |
| tan‌3‌x | 4 | ‌(1+2‌tan‌3‌x) |
|=|| 3‌cos‌3‌x | 1 | 2(cos‌+sin‌‌)2 |
| 3sin‌3x | −1 | 2(cos2‌−sin‌2‌) |
| 3 sec23x | 4 | 1+2‌tan‌3‌x |
|+|| sin‌3x | 0 | 2(cos‌+sin‌‌)2 |
| cos‌3‌x | 0 | 2(cos2‌−sin‌2‌) |
| tan‌3‌x | 0 | 1+2‌tan‌3‌x |
| sin‌3x‌‌1‌‌2×2(cos‌+sin‌‌)×(‌sin‌‌+‌‌cos‌)+cos‌3‌x−1‌‌2(−2‌cos‌×‌sin‌‌.−2sin‌‌×‌‌cos‌)tan‌3‌x‌‌4(0+2×3 sec23x)At
x=(2n+1)Ï€,
f′(x)=|| 3(−1) | 1 | 2(1) |
| 0 | −1 | 2(−1) |
| 3 | 4 | 1+0 |
|+0+|| 0 | 1 | 4(0−1)×[−‌(−1)+‌×0] |
| 0 |
| 0 | −1 | 0 |
|=||+0+||=[−3(−1+8)−1(0+6)+2(0+3)]+[0−1(0−0)−6(−4)]=−21+24=3
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