Let f(x) be a quadratic expression which is positive for all the real values of x. If g(x) = f(x) +

Q: Why can't we just assume $f(x) = x^2 + 1$ and solve it? It's much faster!

While testing a specific case like $f(x) = x^2 + 1$ is a brilliant way to build intuition and verify your answer during a time-pressured exam, it is not a formal proof. In JEE Advanced, we must demonstrate that the property holds for *any* valid quadratic $f(x)$. The general algebraic proof ensures that our conclusion is universal, not just a coincidence of a simple example.

Q: Why does the leading coefficient $a$ stay the same in $g(x)$?

That is a fantastic observation! Because $f'(x)$ and $f''(x)$ are linear and constant respectively, they do not contain any $x^2$ terms. When you sum $f(x) + f'(x) + f''(x)$, the only $x^2$ term comes from $f(x)$. Thus, the 'curvature' or the 'opening' of the parabola remains unchanged, which is the key to the entire problem.

Q: Is it possible for $g(x)$ to be zero?

No, it is not. Since we proved that the discriminant $D'$ is strictly negative and the leading coefficient $a$ is positive, the parabola $g(x)$ is strictly above the x-axis. It never touches or crosses it, meaning $g(x)$ can never be zero for any real $x$.

Question

Let $f(x)$ be a quadratic expression which is positive for all the real values of $x$. If $g(x) = f(x) + f'(x) + f''(x)$, then for any real $x$,

Options:

(A) $g(x) < 0$

(B) $g(x) > 0$

(C) $g(x) = 0$

(D) $g(x) \geq 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine you are standing on a vast, flat plane representing the $x$-axis. You are looking at a parabola, $f(x) = ax^2 + bx + c$, which is strictly positive for all real values of $x$. Geometrically, this means your parabola is a floating arch, never touching the ground. It is suspended in the air, defying gravity. For this to happen, two conditions must be met. First, the parabola must open upwards, meaning the leading coefficient $a$ must be greater than zero ($a > 0$). Second, it must never touch the $x$-axis, which implies that the discriminant $D = b^2 - 4ac$ must be strictly less than zero ($D < 0$). This is our starting point—the bedrock of our understanding. ### The Calculus of Change Now, we introduce a new function, $g(x)$, defined as the sum of the original function and its derivatives: $g(x) = f(x) + f'(x) + f''(x)$. We know $f(x) = ax^2 + bx + c$. Differentiating with respect to $x$, we find the first derivative: $$f'(x) = 2ax + b$$ Differentiating once more, we find the second derivative: $$f''(x) = 2a$$ Notice how the degree of the polynomial gracefully descends from quadratic to linear to constant. This is the elegance of calculus in action. ### The Synthesis Now, let us construct $g(x)$ by summing these components: $$g(x) = (ax^2 + bx + c) + (2ax + b) + (2a)$$ To make sense of this, we group the terms by their powers of $x$. We have the $x^2$ term: $ax^2$. We have the $x$ terms: $(b + 2a)x$. And finally, the constant terms: $(c + b + 2a)$. So, our new function is: $$g(x) = ax^2 + (b + 2a)x + (c + b + 2a)$$ It is still a quadratic! And crucially, its leading coefficient is still $a$. ### The Discriminant Reveal To determine if $g(x)$ is also always positive, we must examine its discriminant, $D'$. The formula for the discriminant of a quadratic $Ax^2 + Bx + C$ is $B^2 - 4AC$. Here, our $A = a$, our $B = (b + 2a)$, and our $C = (c + b + 2a)$. Substituting these into the formula, we get: $$D' = (b + 2a)^2 - 4a(c + b + 2a)$$ Let us expand this carefully. The first part, $(b + 2a)^2$, becomes $b^2 + 4ab + 4a^2$. The second part, $-4a(c + b + 2a)$, becomes $-4ac - 4ab - 8a^2$. Combining these, we see a beautiful cancellation: the $4ab$ and $-4ab$ terms vanish! We are left with: $$D' = (b^2 - 4ac) + 4a^2 - 8a^2$$ $$D' = (b^2 - 4ac) - 4a^2$$ ### The Final Conclusion Look closely at this result. We know that $(b^2 - 4ac)$ is the discriminant of the original function $f(x)$, which we established is negative. We also know that $4a^2$ is always positive, so $-4a^2$ is negative. A negative number minus a positive number is always negative! Therefore, $D' < 0$. Since $g(x)$ has a positive leading coefficient $a$ and a negative discriminant $D'$, it follows the exact same geometric rules as $f(x)$. It, too, is a floating parabola that never touches the $x$-axis. Thus, **$g(x) > 0$ for all real $x$**. We have successfully navigated the algebra to reveal a profound truth about the behavior of functions and their derivatives.

Animated Solution for Mathematics - Quadratic Equations: Let $f (x)$ be a quadratic expression which is positive for all the real values of $x$ . If $g (x) = f (x) + f^{'} (x) + f^{''} (x)$ , then for any real $x$ ,

Select Answer:

Visualized Solution

Visualizing $f (x)$

Conditions for $f (x) > 0$

Finding $f^{'} (x)$

Finding $f^{''} (x)$

Constructing $g (x)$

Simplifying $g (x)$

Discriminant of $g (x)$

Expanding $D^{'}$

Simplifying $D^{'}$

Analyzing the Sign of $D^{'}$

Final Conclusion

The Sigma Insight: Maximum and Minimum Values of Quadratic Expressions

Analyzing the Setup

The Calculus of Change

The Synthesis

The Discriminant Reveal

The Final Conclusion

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Visualized Solution

Visualizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Conditions for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)>0

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′′(x)

Constructing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Discriminant of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D′

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D′

Analyzing the Sign of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D′