Q.2
Find the range of values of a, such that f(x) = ax² + 2(a +1)x+ 9a + 4
is always negative.
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Answered by
3
Answer:
Step-by-step explanation:
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Answered by
1
Step-by-step explanation:
Considering denominator x
2
−8x+32
D<0 and a>0
So denominator is always positive
For f(x) to be negative, numerator ⇒ax
2
+2(a+1)x+9a+4<0 i.e a<0 and b
2
−4ac<0
⇒ax
2
+2(a+1)x+9a+4<0
⇒a<0 & 4(a+1)
2
−4a(9a+4)<0
⇒4(a
2
+2a+1−9a
2
−4a)<0
⇒4(−8
2
−2a+1)<0
8a
2
+2a−1>0
(4a−1)(2a+1)>0
a>
4
1
or a<−
2
1
but a<0
∴a∈(−∞,−
2
1
)
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