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Question : 114 of 150
Marks:
+1,
-0
Solution:
‌| sin‌x4−x4‌cos‌x4+x20 |
| x4(e2x4−1−2x4) |
Using L'Hospital Rule
=| 4x7‌sin‌x4+20x19 |
| 8e2x4x7−16x7+4e2x4−4x3 |
=| 4x3(x4‌sin‌x4+5x16) |
| 4x3(2e2x4x4−4x4+e2x4−1) |
Using L'Hospital Rule
=| 4x3sin4+4x7‌cos‌x4+80x15 |
| 16e2x4x7+16e2x4x3−16x3 |
=| 4x3(sin‌x4+x4‌cos‌x4+20x12) |
| 4x3(4e2x4+4e2x4−4) |
Using L'Hospital Rule
=| 4x3‌cos‌x4+4x3‌cos‌x4−4x7‌sin‌x4+240x11 |
| 32e2x4.x7+48e2x4.x3 |
=| 4x3(2‌cos‌x4−x4‌sin‌x4+60x11) |
| 4x3(8e2x4.x4+12e2x4) |
=| 2‌cos‌x4−x4‌sin‌x4+60x11 |
| e2x4(8x4+12) |
=| 2‌cos(0)4−(0)4‌sin(0)4+60(0)11 |
| e2(0)2[8(0)4+12] |
==
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