It is given that at $x = 1$, the function $x^4 - 62x^2 + ax + 9$ attains its maximum value, on the interval $[0, 2]$. Find the value of $a$.

> [!TLDR] > For a function to attain its maximum at an interior point of a closed interval, the derivative must be zero there (Fermat’s theorem). Setting $f'(1)=0$ gives $a = 120$, and checking the second derivative confirms it’s a local maximum. We have a quartic polynomial $$f(x) = x^4 - 62x^2 + ax + 9$$ and we are told that on the interval $[0, 2]$, the maximum value occurs at $x = 1$. Since $1$ lies strictly inside $(0,2)$, this is an interior maximum. ## Why the derivative must be zero If a differentiable function has a local maximum at an interior point of an interval, the tangent line there must be horizontal — that is, the first derivative is zero. This is Fermat’s theorem (the interior critical point condition). It does **not** guarantee a maximum (it could be a minimum or a saddle), but it is a necessary condition. So the first step is always: **set $f'(1) = 0$**. > [!WARNING] > A common mistake is to forget that the maximum is *given* to be at $x=1$, so you don’t need to compare endpoints yet. The condition $f'(1)=0$ is forced by the problem statement — you are not *finding* the maximum, you are *using* the fact that it occurs at $1$. --- ## Step-by-step 1. **Differentiate** $$f'(x) = 4x^3 - 124x + a$$ 2. **Apply the condition** Since $x=1$ is a point of maximum, $f'(1) = 0$: $$4(1)^3 - 124(1) + a = 0 \quad\Rightarrow\quad 4 - 124 + a = 0$$ $$a = 120$$ So $a = 120$ is forced. 3. **Verify it’s actually a maximum (second derivative test)** $$f''(x) = 12x^2 - 124$$ At $x=1$: $$f''(1) = 12 - 124 = -112 < 0$$ A negative second derivative means the curve is concave down at $x=1$, confirming a local maximum. 4. **Check that this local maximum is indeed the global maximum on $[0,2]$** Since the interval is small and the polynomial is continuous, the global maximum on a closed interval occurs either at a critical point or at an endpoint. …

It is given that at x = 1, the function x⁴ - 62x² – Class 12 Mathematics

Exercise 6.3 · Q11

x = 1

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Concept understanding — Local Maximum Condition

Local Maximum Condition

Imagine hiking in a mountain range. You reach a point where, for a few steps in any direction, the ground drops away. You may not be the highest peak in the whole range, but right here every step goes downhill. That's a local maximum: a point higher than all nearby points.

The Intuition First

On a smooth, wavy curve, a local maximum is a "hilltop" — a point where the function peaks, then falls on both sides. Zoom in and the values just to the left and right are both lower.

"Local" means "in a small neighbourhood." The function might have higher values elsewhere (a global maximum), but that doesn't matter — a local maximum is king of its own tiny kingdom.

The Precise Mathematical Statement

Let $f$ be a real-valued function on a domain $D \subseteq \mathbb{R}$ . A point $c \in D$ is a local maximum if there exists some $\delta > 0$ such that for every $x$ in the domain within distance $\delta$ of $c$ :

f(x) \leq f(c)

That is, on some open interval $(c - \delta, c + \delta)$ , $f(c)$ is the largest value.

Note

The inequality is $f(x) \leq f(c)$ , not $f(x) < f(c)$ . If equality holds for some $x \neq c$ (a flat plateau), it's still a local maximum — just not a strict one.

The First Derivative Test

If $f$ is differentiable at $c$ and $c$ is a local maximum, then:

f'(c) = 0

This is the critical point condition — the tangent is horizontal. But $f'(c) = 0$ is necessary, not sufficient: a horizontal tangent could also be a local minimum or a saddle point (like $f(x) = x^3$ at $x = 0$ ).

Watch out

A common mistake: assuming $f'(c) = 0$ guarantees a local extremum. It does not. Check the sign change of the derivative around $c$ , or use the second derivative test.

The Second Derivative Test

If $f'(c) = 0$ and $f''(c) < 0$ , then $c$ is a local maximum: a negative second derivative means $f$ is concave down at $c$ — curving down like an upside-down bowl. If $f''(c) > 0$ , it's a local minimum. If $f''(c) = 0$ , the test is inconclusive.

A Concrete Example

Consider $f(x) = -x^2 + 4x - 3$ . …

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For a function to attain its maximum at an interior point of a closed interval, the derivative must be zero there (Fermat’s theorem). Setting $f'(1)=0$ gives $a = 120$ , and checking the second derivative confirms it’s a local maximum.

We have a quartic polynomial

f(x) = x^4 - 62x^2 + ax + 9

and we are told that on the interval $[0, 2]$ , the maximum value occurs at $x = 1$ . Since $1$ lies strictly inside $(0,2)$ , this is an interior maximum.

Why the derivative must be zero

If a differentiable function has a local maximum at an interior point of an interval, the tangent line there must be horizontal — that is, the first derivative is zero. This is Fermat’s theorem (the interior critical point condition). It does not guarantee a maximum (it could be a minimum or a saddle), but it is a necessary condition.

So the first step is always: set $f'(1) = 0$ .

Watch out

A common mistake is to forget that the maximum is given to be at $x=1$ , so you don’t need to compare endpoints yet. The condition $f'(1)=0$ is forced by the problem statement — you are not finding the maximum, you are using the fact that it occurs at $1$ .

Step-by-step

Differentiate

f'(x) = 4x^3 - 124x + a

Apply the condition Since $x=1$ is a point of maximum, $f'(1) = 0$ :

4(1)^3 - 124(1) + a = 0 \quad\Rightarrow\quad 4 - 124 + a = 0

a = 120

So $a = 120$ is forced.

Verify it’s actually a maximum (second derivative test)

f''(x) = 12x^2 - 124

At $x=1$ :

f''(1) = 12 - 124 = -112 < 0

A negative second derivative means the curve is concave down at $x=1$ , confirming a local maximum.

Check that this local maximum is indeed the global maximum on $[0,2]$ Since the interval is small and the polynomial is continuous, the global maximum on a closed interval occurs either at a critical point or at an endpoint. …

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Q.It is given that at $x = 1$ , the function $x^4 - 62x^2 + ax + 9$ attains its maximum value, on the interval $[0, 2]$ . Find the value of $a$ .

Concept understanding — Local Maximum Condition

Local Maximum Condition

The Intuition First

The Precise Mathematical Statement

The First Derivative Test

The Second Derivative Test

A Concrete Example

Why the derivative must be zero

Step-by-step

Method: Interior Extremum ⇒ First Derivative Vanishes (Fermat's Condition)

Steps

Common Mistakes

Q.It is given that at x=1x = 1x=1, the function x4−62x2+ax+9x^4 - 62x^2 + ax + 9x4−62x2+ax+9 attains its maximum value, on the interval [0,2][0, 2][0,2]. Find the value of aaa.

Why the derivative must be zero

Step-by-step

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Q.It is given that at $x = 1$ , the function $x^4 - 62x^2 + ax + 9$ attains its maximum value, on the interval $[0, 2]$ . Find the value of $a$ .

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