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Question

Number of ordered pair (a,b) from the set A={1,2,3,4,5} so that the function f(x)=x33+a2x2+bx+10 is an injective mapping xR:

A
13
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B
14
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C
15
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D
16
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Solution

The correct option is C 15
Given
f(x)=x³/3+ax²/2+bx+10
f'(x)=3x²/3+2ax/2+b
f'(x)=x²+ax+b
Since, D
We know that,
D=b²-4ac
There fore,
D=a²-4b
a=1,. b=(1,2,3,4,5) = 5 Pairs
a=2, b=(1,2,3,4,5) = 5 Pairs
a=3, b=(3,4,5). =3 pairs. (remaining doesn't satisfied the equ)
a=4, b=(4,5). =2 pairs
a=5, b=(5). =0 pairs
Hence,
No. Of ordered pair is 15. For injective function.

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