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Question : 65 of 160
Marks:
+1,
-0
Solution:
f(x)=|| 2‌cos‌x | 1 | 0 |
| x−‌ | 2‌cos‌x | 1 |
| 0 | 1 | 2‌cos‌x |
| f′(x)=|| −2sin‌x | 1 | 0 |
| 1 | 2‌cos‌x | 1 |
| 0 | 1 | 2‌cos‌x |
| +|| 2‌cos‌x | 0 | 0 |
| x−‌ | −2sin‌x | 1 |
| 0 | 0 | 2‌cos‌x |
| +|| 2‌cos‌x | 1 | 0 |
| x−‌ | 2‌cos‌x | 0 |
| 0 | 1 | −2sin‌x |
| f′(π)=|| −2sin‌π | 1 | 0 |
| 1 | 2‌cos‌π | 1 |
| 0 | 1 | 2‌cos‌π |
| +|| 2‌cos‌π | 0 | 0 |
| π−‌ | −2sin‌π | 1 |
| 0 | 0 | 2‌cos‌π |
| +|| 2‌cos‌π | 1 | 0 |
| π−‌ | 2‌cos‌π | 0 |
| 0 | 1 | −2sin‌π |
| f′(π)=|| +||+|| [| ∵sin‌π=0 |
| cos‌π=−1 |
] f′(π)=−(−2+0)=2
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