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Question

If (a2−1)x2+(a−1)x+a2−4a+3=0 be an identity in x, then the value of a is/are ?

A
1
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B
1
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C
3
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D
1,1,3
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Solution

The correct option is B 1
(a21)x2+(a1)x+a24a+3=0
(a1)[(a+1)x2+x+(a3)]=0
a=1.

1190633_1155206_ans_8c035a0a601d4e2fa122e07b5855b0cd.jpg

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