Let a ∈ R and let f : R → R be given by f(x) = x⁵ - 5x + a. Then (a) f(x) has three real roots if a

Q: Wait, why do we only look at the local maximum and minimum? What about the rest of the curve?

That is a fantastic question! For a polynomial like $f(x) = x^5 - 5x + a$, the shape is essentially a 'snake' with one peak and one valley. Because the function is continuous and goes to positive infinity as $x \to \infty$ and negative infinity as $x \to -\infty$, the only way to have three roots is to 'trap' the $x$-axis between the peak and the valley. If the peak is above the axis and the valley is below, the curve must cross the axis once before the peak, once between the peak and valley, and once after the valley. The rest of the curve is just 'tails' that go off to infinity, so they can't create any more roots.

Q: Why does $f(x_{max}) \cdot f(x_{min}) < 0$ guarantee three roots? Could it have more?

Great intuition! For this specific quintic polynomial, the derivative $f'(x) = 5x^4 - 5$ has only two real roots ($x = 1$ and $x = -1$). This means there are exactly two turning points. A continuous function with only two turning points can cross the $x$-axis at most three times. Since we have established that the peak is above the axis and the valley is below, the Intermediate Value Theorem guarantees at least one root in each of the three intervals: $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$. Thus, exactly three roots.

Q: Why does the derivative $5x^4 - 5 = 0$ only give two real roots when it's a degree 4 polynomial?

Remember that the Fundamental Theorem of Algebra tells us a degree 4 polynomial has 4 roots in the complex plane. When we solve $5(x^4 - 1) = 0$, we get $x^4 = 1$. The roots are $x = 1, x = -1, x = i, x = -i$. Since we are only interested in real-valued functions and real roots, we discard the imaginary roots $i$ and $-i$. Only $x = 1$ and $x = -1$ are relevant for the turning points on the real coordinate plane.

Question

Let $a \in R$ and let $f : R o R$ be given by $f(x) = x^5 - 5x + a$. Then Options: (A) $f(x)$ has three real roots if $a > 4$ (B) $f(x)$ has only real root if $a > 4$ (C) $f(x)$ has three real roots if $a < -4$ (D) $f(x)$ has three real roots if $-4 < a < 4$

Sigma Percentile · Accepted Answer

### The Dance of the Quintic: Unlocking the Secrets of $f(x) = x^5 - 5x + a$ Have you ever looked at a complex polynomial and felt like it was hiding a secret? Today, we are going to peel back the layers of the quintic function $f(x) = x^5 - 5x + a$. This isn't just an algebraic exercise; it is a journey into the geometry of functions. Imagine you are standing in front of a graph. You have a fixed, snake-like curve defined by $y = x^5 - 5x$. The parameter $a$ is like an elevator—it shifts the entire curve up or down. Our mission is to find the exact range of $a$ that forces this curve to cross the $x$-axis three times. ### Phase 1: The Calculus Toolkit To understand where the curve turns, we need to find its peaks and valleys. We turn to the derivative, our most powerful tool for analyzing change. We differentiate $f(x)$ with respect to $x$: $$f'(x) = \frac{d}{dx}(x^5 - 5x + a) = 5x^4 - 5$$ Setting $f'(x) = 0$ gives us the critical points where the slope of the tangent is horizontal: $$5x^4 - 5 = 0 \implies x^4 = 1$$ This yields two real critical points: $x = 1$ and $x = -1$. These are the 'hinges' of our curve. ### Phase 2: The Geometric Dance Now, let's find the height of our peak and valley. At $x = -1$, we find the local maximum: $$f(-1) = (-1)^5 - 5(-1) + a = -1 + 5 + a = a + 4$$ At $x = 1$, we find the local minimum: $$f(1) = (1)^5 - 5(1) + a = 1 - 5 + a = a - 4$$ Notice the elegance here: the peak is always at $a + 4$ and the valley is always at $a - 4$. The distance between them is a constant 8 units, regardless of the value of $a$. ### Phase 3: The Condition for Three Roots For the curve to cross the $x$-axis three times, the peak must be above the $x$-axis ($f(-1) > 0$) and the valley must be below the $x$-axis ($f(1) < 0$). If both were above, the curve would only cross once, far to the left. If both were below, it would only cross once, far to the right. Mathematically, we need the product of these two values to be negative: $$f(-1) \cdot f(1) < 0 \implies (a + 4)(a - 4) < 0$$ Solving this inequality is straightforward. The roots are $a = -4$ and $a = 4$. The product is negative when $a$ lies between these two values: $$-4 < a < 4$$ ### The Final Revelation When $a$ is in this 'sweet spot' of $(-4, 4)$, the curve is perfectly positioned to pierce the $x$-axis three times. If $a > 4$, the entire valley is lifted above the $x$-axis, leaving us with only one root. If $a < -4$, the entire peak is pushed below the $x$-axis, again leaving us with only one root. You have just mastered the behavior of a quintic polynomial! Keep this geometric intuition close—it is the key to solving even more complex problems in your **JEE journey**.

Animated Solution for Mathematics - Differentiation: Let $a \in R$ and let $f : R \to R$ be given by $f (x) = x^{5} - 5 x + a$ . Then

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Visualized Solution

Visualizing the Function $f (x)$

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

Solving $f^{'} (x) = 0$

Local Maximum at $x = - 1$

Local Minimum at $x = 1$

Condition for Three Real Roots

Solving $(a + 4) (a - 4) < 0$

One Real Root when $a > 4$

One Real Root when $a < - 4$

Final Summary

The Sigma Insight: Maxima and Minima

The Dance of the Quintic

Phase 1

Phase 2

Phase 3

The Final Revelation

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a∈R and let f:R→R be given by f(x)=x5−5x+a. Then

Select Answer:

Visualized Solution

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

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

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

Local Maximum at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−1

Local Minimum at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1

Condition for Three Real Roots

Solving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a+4)(a−4)<0

One Real Root when .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a>4

One Real Root when .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a<−4