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Question : 107 of 150
Marks:
+1,
-0
Solution:
Given,
α and
β are the roots of
ax2+bx+c=0∴‌aα2+bα+c=0 . . . (i)
‌ and ‌‌aβ2+bβ+c=0 . . . (ii)
Also,
‌‌α+β=−‌ . . . (iii)
and
‌‌αβ=‌ . . . (iv)
Now,
‌| 1−cos(ax2+bx+c) |
| (x−α)2 |
(‌. form
)=‌| 0+sin‌(ax2+bx+c)⋅(2ax+b) |
| 2(x−α) |
(by L-Hospital rule)
=‌| (2ax+b)sin‌(ax2+bx+c) |
| 2(x−α) |
(‌. form
)‌(2ax+b)‌cos(ax2+bx+c)=‌‌| (2ax+b)+2asin‌(ax2+bx+c) |
| 2(1−0) |
=‌‌(2aα+b)‌cos(aα2+bα+c)(2aα+b) ‌+2asin‌(aα2+bα+c)=‌‌(2aα+b)2‌cos(0)+2asin‌(0)=‌‌=‌(2α+‌)2=‌‌[2α−(α+β)]2=‌‌(α−β)2
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