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Grade:11

1 Answers

Arun
25763 Points
2 years ago
Dear student
 
for positive values
it means Coefficient of x^2 should be negative
hence equation will be
ax^2 – 2 ax + 3 x – 6 = 0
D should be grewter than 0
9 + 4a^2 – 12 a + 24 a > 0
4a^2 + 12 a + 9 >0
(2a + 3)^2
always positive
f(0) = -6
f(1) = a + 3 – 2 a – 6 = – a – 3
f(2) = 0
f (3) = 3a + 3
f (4) = 8a + 6
hence f(3) and f(4) should be positive and f(5) should be neagtive or zero.
a> – 1
a > – 3/4
f(5) = 15 a + 9
15 a + 9
a
hence no integral valurpe of alphais obtained.

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