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Question

Let ax2+bx+6=0 does not have distinct real roots. If the least value of 3a+b is k, then the value of |k| is

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Solution

Let f(x)=ax2+bx+6
As ax2+bx+6=0 does not have distinct real roots, so
D0
Putting x=0
f(0)=6>0
So,
f(x)0, xR

Now,
f(3)09a+3b+603a+b2

Therefore, the least value of 3a+b is 2.
Hence, |k|=2

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