The value of limx→1α sincx2+bx+axα−1 (where α and β are the roots ax2+bx+c=0) is
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a
cβ−ααβ
b
β−αα
c
1α
d
cα1α−1β
answer is D.
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Detailed Solution
∵ Roots of ax2+bx+c=0 are α and βso roots of cx2+bx+a=0 are 1α and 1β∴ cx2+bx+a=cx−1αx−1β Now, limx→1α sincx−1αx−1βαcx−1αx−1β⋅cx−1β =cαlimx→1α sincx−1αx−1βcx−1αx−1β⋅limx→1α x−1β=cα1α−1β