Let f : [0, 2] → R be a twice differentiable function such that f''(x) 0, for all x ∈ (0, 2). If φ

Q: Wait, why does $f''(x) > 0$ imply $f'(x)$ is increasing?

That is a fundamental question! By definition, the derivative of a function $f'(x)$ is $f''(x)$. If $f''(x) > 0$, the rate of change of the slope is positive. If the rate of change of a quantity is positive, that quantity must be increasing. Thus, $f'(x)$ is strictly increasing.

Q: Can I just assume $f(x) = x^2$ and solve it? It's much faster.

In a high-pressure exam like JEE Advanced, using a test function like $f(x) = x^2$ is a brilliant 'trap-catcher' strategy! Since $f''(x) = 2 > 0$, it satisfies the condition. $\phi(x) = x^2 + (2-x)^2 = 2x^2 - 4x + 4$. The vertex is at $x = 1$. It decreases on $(0, 1)$ and increases on $(1, 2)$. It works perfectly! However, always understand the general proof for conceptual mastery.

Q: Why does the sign of $\phi'(x)$ flip exactly at $x = 1$?

Because $x = 1$ is the axis of symmetry for the interval $[0, 2]$. At $x = 1$, $x = 2 - x$. So, $\phi'(1) = f'(1) - f'(1) = 0$. This is the critical point where the function stops decreasing and starts increasing.

Question

Let $f : [0, 2] 	o R$ be a twice differentiable function such that $f''(x) > 0$, for all $x \in (0, 2)$. If $\phi(x) = f(x) + f(2 - x)$, then $\phi$ is :

Options:

(A) decreasing on $(0, 2)$

(B) decreasing on $(0, 1)$ and increasing on $(1, 2)$

(C) increasing on $(0, 2)$

(D) increasing on $(0, 1)$ and decreasing on $(1, 2)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup We are given a twice-differentiable function $f: [0, 2] o \mathbb{R}$ with the condition $f''(x) > 0$ for all $x \in (0, 2)$. Geometrically, $f''(x) > 0$ implies that the function is **concave upwards**. Imagine a bowl where the slope is constantly increasing as you move from left to right. This condition serves as our "master key." Because $f''(x) > 0$, the derivative $f'(x)$ is a **strictly increasing function**. This means that for any two points $a$ and $b$ in the interval, if $a > b$, then $f'(a) > f'(b)$. This property is the engine that will drive our entire solution. ### The Derivative Dance We are asked to analyze the monotonicity of the function $\phi(x) = f(x) + f(2 - x)$. To determine whether $\phi(x)$ is increasing or decreasing, we must examine its rate of change, $\phi'(x)$. Differentiating $\phi(x)$ with respect to $x$ gives: $$\phi'(x) = \frac{d}{dx}[f(x)] + \frac{d}{dx}[f(2 - x)]$$ The first term is simply $f'(x)$. The second term requires the **Chain Rule**: we differentiate the outer function $f$ to get $f'(2 - x)$ and multiply by the derivative of the inner function $(2 - x)$, which is $-1$. Thus, we arrive at the crucial expression: $$\phi'(x) = f'(x) - f'(2 - x)$$ ### The Symmetry Pivot We must now analyze the sign of $\phi'(x)$ on the interval $(0, 2)$. The expression $f'(x) - f'(2 - x)$ depends entirely on the relationship between $x$ and $2 - x$. Note that $x = 1$ is the midpoint of our interval $[0, 2]$. We split our analysis into two regions: **Case 1: $x \in (0, 1)$** In this region, $x < 1$, which implies $2 - x > 1$. Consequently, $x < 2 - x$. Since $f'(x)$ is strictly increasing, it follows that $f'(x) < f'(2 - x)$. Therefore, $\phi'(x) < 0$, meaning $\phi(x)$ is **strictly decreasing** on $(0, 1)$. **Case 2: $x \in (1, 2)$** Here, $x > 1$, which implies $2 - x < 1$. Thus, $x > 2 - x$. Since $f'(x)$ is strictly increasing, a larger input $x$ yields a larger output $f'(x)$, so $f'(x) > f'(2 - x)$. Therefore, $\phi'(x) > 0$, meaning $\phi(x)$ is **strictly increasing** on $(1, 2)$. ### The Final Conclusion We have successfully dissected the behavior of $\phi(x)$. It decreases on the interval $(0, 1)$ and, after passing the symmetry point at $x = 1$, it increases on the interval $(1, 2)$. This result is a testament to the power of symmetry. By understanding the relationship between the function and its reflection, we have unlocked the behavior of the sum. Always look for the symmetry, trust the calculus, and never fear the derivative.

Animated Solution for Mathematics - Differentiation: Let $f : [0, 2] \to R$ be a twice differentiable function such that $f^{''} (x) > 0$ , for all $x \in (0, 2)$ . If $ϕ (x) = f (x) + f (2 - x)$ , then $ϕ$ is :

Select Answer:

Visualized Solution

Given Condition $f^{''} (x) > 0$

Defining $ϕ (x)$

Differentiating $ϕ (x)$

Applying Chain Rule

Monotonicity of $f^{'} (x)$

Analysis for $x \in (0, 1)$

Comparing Inputs

Sign of $ϕ^{'} (x)$ on $(0, 1)$

Conclusion for $(0, 1)$

Analysis for $x \in (1, 2)$

Comparing Inputs

Sign of $ϕ^{'} (x)$ on $(1, 2)$

Conclusion for $(1, 2)$

Final Conclusion

The Sigma Insight: Monotonicity

Analyzing the Setup

The Derivative Dance

The Symmetry Pivot

The Final Conclusion

Similar Questions

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If $P (x)$ is a polynomial of degree less than or equal to $2$ and $S$ is the set of all such polynomials so that $P (0) = 0, P (1) = 1$ and $P^{'} (x) > 0\forall x \in [0, 1]$ , then

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Let $f : R \to R$ be defined as $f (x) = {- \frac{4}{3} x^{3} + 2 x^{2} + 3 x 3 x e^{x}, x > 0, x \leq 0$ . Then $f$ is increasing function in the interval

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?

Select Answer:

Visualized Solution

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

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

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

Applying Chain Rule

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

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

Comparing Inputs

Sign of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ϕ′(x) on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (0,1)

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

Analysis for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈(1,2)

Comparing Inputs

Sign of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ϕ′(x) on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,2)

Conclusion for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,2)

Final Conclusion

The Sigma Insight: Monotonicity

Analyzing the Setup

The Derivative Dance

The Symmetry Pivot

The Final Conclusion

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Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider the statements: P: There exists some x∈R such that f(x)+2x=2(1+x2), Q: There exists some x∈R such that 2f(x)+1=2x(1+x)

Which of the following is true?

Given Condition $f^{''} (x) > 0$

Defining $ϕ (x)$

Differentiating $ϕ (x)$

Monotonicity of $f^{'} (x)$

Analysis for $x \in (0, 1)$

Sign of $ϕ^{'} (x)$ on $(0, 1)$

Conclusion for $(0, 1)$

Analysis for $x \in (1, 2)$

Sign of $ϕ^{'} (x)$ on $(1, 2)$

Conclusion for $(1, 2)$

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If $f (x) = \int_{x^{2}}^{x^{2} + 1} e^{- t^{2}} d t$ , then $f (x)$ increases in

The function $f (x) = \frac{l n ( π + x )}{l n ( e + x )}$ is

Let $f : R \to R$ be defined as $f (x) = {- \frac{4}{3} x^{3} + 2 x^{2} + 3 x 3 x e^{x}, x > 0, x \leq 0$ . Then $f$ is increasing function in the interval

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