JEE MainMathematics

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How many real roots does a strictly increasing odd-degree polynomial have?

The number of real solutions of x^7+5 x^3+3 x+1=0 is equal to.

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A0B1C3D5

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Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

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Students also ask

Why does

f^{'} (x) > 0

everywhere imply at most one real root?

A strictly increasing function is strictly monotonic (injective). It can take any specific value (including zero) at most once. By Rolle's Theorem, if there were 2 or more real roots, $f^{'} (x)$ would have to be zero somewhere between them, which is impossible since $f^{'} (x) \geq 3$ .

Answer

The polynomial has degree $7$ and its first derivative is strictly positive for all real $x$ , meaning it is strictly increasing and has exactly $1$ real root.

Why each option works or fails

A: 0 - Believing that an equation with all positive coefficients cannot have any real solutions. Every polynomial of odd degree has at least one real root by the Intermediate Value Theorem because its ends tend to $- \infty$ and $+ \infty$ .
B · correct: 1 - The polynomial is strictly increasing on $R$ ( $f^{'} (x) > 0$ ), so it crosses the $x$ -axis exactly once. This is correct.
C: 3 - Mistakenly concluding that the three non-constant terms imply three real roots or miscalculating the number of sign changes. Differentiate $f (x)$ to check for local extrema. Since $f^{'} (x) = 7 x^{6} + 15 x^{2} + 3 > 0$ for all real $x$ , the function has no turning points and can only intersect the axis once.
D: 5 - Confusing the number of complex roots with real roots, or misinterpreting Descartes' Rule of Signs on $f (- x)$ . Descartes' Rule of Signs gives an upper bound on positive or negative roots, not the exact count. Combine it with the derivative $f^{'} (x) > 0$ to rule out multiple real roots.

Step-by-step solution

Given: Let $f (x) = x^{7} + 5 x^{3} + 3 x + 1 = 0$ .
Approach: Examine the derivative $f^{'} (x)$ to check for monotonicity. If $f^{'} (x) > 0$ for all $x \in R$ , $f (x)$ is strictly increasing and can cross the x-axis at most once. Since it is an odd-degree polynomial, it must cross at least once by the Intermediate Value Theorem.
Execute: Differentiate $f (x)$ : $f^{'} (x) = 7 x^{6} + 15 x^{2} + 3$ Since $x^{6} \geq 0$ and $x^{2} \geq 0$ for all real $x$ , we have: $f^{'} (x) \geq 3 > 0, \forall x \in R$ Thus, $f (x)$ is strictly increasing on $(- \infty, \infty)$ .
Execute: As $x \to - \infty$ , $f (x) \to - \infty$ and as $x \to + \infty$ , $f (x) \to + \infty$ . Since $f (x)$ is continuous and strictly increasing from $- \infty$ to $+ \infty$ , it crosses the $x$ -axis exactly once. Hence, there is exactly $1$ real solution.
Verify: At $x = 0$ , $f (0) = 1 > 0$ . At $x = - 1$ , $f (- 1) = (- 1)^{7} + 5 (- 1)^{3} + 3 (- 1) + 1 = - 1 - 5 - 3 + 1 = - 8 < 0$ . By IVT, a real root lies in $(- 1, 0)$ . Monotonicity guarantees uniqueness.

Shortcut:

When to use it: Direct inspection using Descartes' Rule of Signs and odd degree property.

Concept: Descartes' Rule of Signs: $f (- 1) = (- 1)^{7} + 5 (- 1)^{3} + 3 (- 1) + 1 = - 1 - 5 - 3 + 1 = - 8 < 0$ has all positive coefficients, so 0 sign changes $⟹ 0$ positive real roots. For $f (- x) = - x^{7} - 5 x^{3} - 3 x + 1$ , the signs are $-, -, -, +$ , having exactly 1 sign change $⟹$ exactly 1 negative real root. Since $x = 0$ is not a root, total real roots $= 0 + 1 = 1$ .

Option verdict: An odd-degree polynomial with real coefficients must have at least one real root because complex roots appear in conjugate pairs. Thus 0 real roots is impossible.

Option verdict: Descartes' Rule gives exactly 1 negative real root and 0 positive real roots, yielding exactly 1 real root.

Option verdict: Mistakenly guessing that a 7th degree polynomial with few terms might have 3 real roots.

Option verdict: Assuming the non-zero degree powers (7, 3, 1) allow up to 5 real roots without checking the derivative's strictly positive nature.

Discriminator: All coefficients are positive, so no positive root; $f (- x)$ has 1 sign change, guaranteeing exactly 1 negative real root.

How many real roots does a strictly increasing odd-degree polynomial have?

What feels right?

The polynomial has degree $7$ and its first derivative is strictly positive for all real $x$ , meaning it is strictly increasing and has exactly $1$ real root.

Why each option works or fails

Solution

01Given

02Approach

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04Execute

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