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If f(x) is a quadratic expression such that f(x)>0 ∀ x ∈ R, and if g(x)=f(x)+f'(x)+f''(x), then prove that g(x) > 0 ∀ x ∈ R.

If f(x) is a quadratic expression such that f(x)>0 ∀ x ∈ R, and if g(x)=f(x)+f'(x)+f''(x), then prove that g(x) > 0 ∀ x ∈ R.

Grade:12

2 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question

Let
f(x) = ax^2 + bx + c
where a, b & c belong to real
Since f(x) > 0 for all x
a > 0 &
Discriminent < 0
b2< 4ac..............(1)
g(x) = f(x) + f'(x) + f''(x)
g(x) = ax^2 + bx + c + 2ax + b + 2a
g(x) = ax^2 + (2a+b)x + 2a + b + c
Let find discriminent
\Delta = (2a+b)^2 - 4a(2a+b+c)
\Delta = 4a^2 + b^2 + 4ab - 8a^2 - 4ab - 4ac
\Delta = b^2 - 4ac - 4a^2
So discriminent is -ve.
So g(x) does not have any real root.
g(x) > 0 for all x belong to real.

mycroft holmes
272 Points
9 years ago
We see that g(x) is a quadratic with the same leading coefficient as f(x) and hence g(x) has a minimum. Let that occur at x = x0.
 
Then at xo, we have 
 
 
Then we have 
 
 
Since the minimum value of g(x) is positive, we have that g(x) has no real roots.

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