If 'R' is the least value of 'a' such that the function f(x) = x² + ax + 1 is increasing on [1, 2] a

Q: Wait, why do we look at the maximum of $-2x$ for the increasing case? Shouldn't we just plug in $x=1$ or $x=2$?

That is a brilliant observation! When we say $a \ge -2x$ for all $x$ in $[1, 2]$, we are demanding that $a$ stays above the entire graph of $-2x$ on that interval. If $a$ is greater than the highest point (the maximum) of $-2x$, it is guaranteed to be greater than every other point as well. That is why we hunt for the maximum!

Q: Why is $S = -4$ the greatest value for the decreasing case? Shouldn't it be larger?

Remember, for the function to be decreasing, we need $f'(x) \le 0$, which leads to $a \le -2x$. If $a$ must be less than or equal to $-2x$ for all $x$, it must be smaller than or equal to the smallest value $-2x$ ever reaches. If $a$ were any larger than the minimum of $-2x$, it would violate the inequality at that minimum point. Thus, $S$ must be the minimum of $-2x$.

Q: Can I just use the vertex of the parabola $f(x) = x^2 + ax + 1$ to solve this?

The vertex is at $x = -a/2$. While you could analyze where the vertex lies relative to the interval $[1, 2]$, it is much more elegant and less prone to error to use the derivative $f'(x) = 2x + a$. The derivative approach works universally, whether the vertex is inside or outside the interval.

Q: Is it possible for $f(x)$ to be both increasing and decreasing on the same interval?

Only if the derivative is identically zero, which would mean $f(x)$ is a constant function. In our case, $f'(x) = 2x + a$ depends on $x$, so it cannot be zero for all $x$ in $[1, 2]$. Therefore, it cannot be both increasing and decreasing.

Q: Why is the final answer $|R - S|$ and not just $R - S$?

The absolute value is a standard way to ask for the 'distance' between two values on the number line, ensuring the result is positive regardless of which value is larger. It is a safeguard to ensure your final answer is a magnitude!

Question

If 'R' is the least value of 'a' such that the function $f(x) = x^2 + ax + 1$ is increasing on $[1, 2]$ and 'S' is the greatest value of 'a' such that the function $f(x) = x^2 + ax + 1$ is decreasing on $[1, 2]$, then the value of $|R - S|$ is .

Sigma Percentile · Accepted Answer

### The Dance of the Derivative: Understanding Monotonicity

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to peel back the layers of a seemingly simple quadratic function: $f(x) = x^2 + ax + 1$.

At first glance, it looks like a standard parabola, but when we restrict its domain to the interval $[1, 2]$, it becomes a stage for a beautiful mathematical performance. We are tasked with finding the parameters $R$ and $S$ that dictate whether this function climbs (increases) or falls (decreases) across this specific window of $x$.

### The First Step: Unlocking the Slope

To understand how a function behaves, we must look at its rate of change. The derivative is our compass here.

By applying the power rule to $f(x) = x^2 + ax + 1$, we find the slope function:

$$f'(x) = 2x + a$$

This linear expression, $2x + a$, tells us everything we need to know. If $f'(x) > 0$, the function is climbing; if $f'(x) < 0$, it is sliding down.

But here is the catch: we need this to happen for *every* point in the interval $[1, 2]$.

### The Increasing Case: Finding the Floor

Imagine you are standing on the interval $[1, 2]$. For the function to be increasing, we need $f'(x) \ge 0$ for all $x$ in that range.

Substituting our derivative, we get $2x + a \ge 0$, or simply $a \ge -2x$.

Now, think about this logically. If $a$ must be greater than or equal to $-2x$ for every single $x$ between 1 and 2, $a$ must be at least as large as the 'toughest' requirement.

The function $g(x) = -2x$ is a downward-sloping line. Its maximum value on the interval $[1, 2]$ occurs at the smallest value of $x$, which is $x = 1$.

Thus, the maximum value is $-2(1) = -2$. To satisfy the condition for all $x$, $a$ must be at least $-2$. Therefore, our least value is **$R = -2$**.

### The Decreasing Case: Finding the Ceiling

Now, let us flip the script. For the function to be decreasing, we need $f'(x) \le 0$, which implies $2x + a \le 0$, or $a \le -2x$.

This time, $a$ must be smaller than or equal to every value that $-2x$ takes on the interval. To ensure $a$ is always below the curve, $a$ must be less than or equal to the *minimum* value of $-2x$.

Looking at our line $g(x) = -2x$, the minimum occurs at the largest $x$, which is $x = 2$. Thus, the minimum value is $-2(2) = -4$.

Consequently, the greatest value $a$ can take while still being 'less than or equal to' the function is **$S = -4$**.

### The Grand Finale: The Absolute Difference

We have navigated the slopes and found our boundaries. We have $R = -2$ and $S = -4$. The problem asks for the absolute difference $|R - S|$.

$$|R - S| = |-2 - (-4)| = |-2 + 4| = |2| = 2$$

And there it is! The elegance of the result lies in how the constraints on the derivative forced us to look at the extreme values of the interval.

You have successfully mastered the relationship between a function's monotonicity and its derivative's bounds. Keep this intuition close—it is the key to solving much more complex problems in calculus. You are doing fantastic work!

Enter Numerical Value:

Visualized Solution

The Function and Interval

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

Condition for Increasing Function

Isolating the Parameter $a$

Finding the Least Value $R$

Condition for Decreasing Function

Isolating $a$ for Decreasing Case

Finding the Greatest Value $S$

Calculating $∣ R - S ∣$

Final Result

The Sigma Insight: Monotonicity

The Dance of the Derivative

The First Step

The Increasing Case

The Decreasing Case

The Grand Finale

Similar Questions

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Comprehension Passage

Consider the statements: $P$ : There exists some $x \in R$ such that $f (x) + 2 x = 2 (1 + x^{2})$ , $Q$ : There exists some $x \in R$ such that $2 f (x) + 1 = 2 x (1 + x)$

Which of the following is true?

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Enter Numerical Value:

Visualized Solution

The Function and Interval

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

Condition for Increasing Function

Isolating the Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Finding the Least Value .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R