Let f : R - R be a twice differentiable function such that f''(x) 0 for all x ∈ R and f'(a - 1) =

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

That is a fundamental property of calculus! The second derivative $f''(x)$ represents the rate of change of the slope $f'(x)$. If the rate of change of the slope is positive, the slope itself must be increasing. Think of it like acceleration: if your acceleration is positive, your velocity must be increasing.

Q: Why can we conclude $f'(u(x)) > 0$ so easily?

Great catch. We found that $u(x) = (\tan x - 1)^2 + a - 1$. This means $u(x)$ is always at least $a-1$. Since we know $f'(x)$ is strictly increasing and $f'(a-1) = 0$, any input greater than $a-1$ will result in a positive value for $f'$. Since $u(x) \ge a-1$, $f'(u(x))$ is always non-negative, and strictly positive whenever $u(x) \neq a-1$.

Q: Why do we complete the square for $u(x)$?

Completing the square is a powerful tool to find the range of a quadratic expression. By writing $u(x)$ as $(\tan x - 1)^2 + a - 1$, we immediately see the minimum value is $a-1$ when $\tan x = 1$. This allows us to compare the input of $f$ directly to the critical point $a-1$.

Question

Let $f : \mathbb{R} 	o \mathbb{R}$ be a twice differentiable function such that $f''(x) > 0$ for all $x \in \mathbb{R}$ and $f'(a - 1) = 0$, where $a$ is a real number. Let $g(x) = f(	an^2 x - 2 	an x + a), 0 < x < \frac{\pi}{2}$. Consider the following two statements: (I) $g$ is increasing in $(0, \frac{\pi}{4})$ (II) $g$ is decreasing in $(\frac{\pi}{4}, \frac{\pi}{2})$. Then,

Options:

(A) Neither (I) nor (II) is True

(B) Only (I) is True

(C) Both (I) and (II) are True

(D) Only (II) is True

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of Curvature Imagine you are standing on a path defined by a function $f(x)$. You are told that $f''(x) > 0$ for all $x$. In the language of calculus, this means the function is strictly convex, or "concave up." Geometrically, this tells us that the slope of the tangent line, $f'(x)$, is constantly increasing as you move from left to right. It is a function that is always "bending" upwards. We are also given a crucial piece of information: $f'(a-1) = 0$. Because $f'(x)$ is strictly increasing, it can cross the horizontal axis exactly once. That unique point is $x = a-1$. This gives us a clear sign scheme: for any $x < a-1$, the slope $f'(x)$ must be negative, and for any $x > a-1$, the slope $f'(x)$ must be positive. This is the heartbeat of our problem. ### The Inner Life of $g(x)$ Now, let us look at the composite function $g(x) = f(u(x))$, where $u(x) = an^2 x - 2 an x + a$. At first glance, this looks like a daunting composition. But let us simplify the inner function $u(x)$ by completing the square. We can rewrite it as: $$u(x) = ( an x - 1)^2 + a - 1$$ This is a beautiful transformation! Since a squared term $( an x - 1)^2$ can never be negative, the minimum value of $u(x)$ is $a-1$, which occurs precisely when $ an x = 1$, or $x = \frac{\pi}{4}$. For all other values of $x$ in the interval $(0, \frac{\pi}{2})$, $u(x)$ will be strictly greater than $a-1$. This is the key to unlocking the behavior of $g(x)$. ### The Chain Rule Symphony To determine if $g(x)$ is increasing or decreasing, we need its derivative. By the Chain Rule, we have: $$g'(x) = f'(u(x)) \cdot u'(x)$$ We already know that $f'(u(x))$ is positive for all $x eq \frac{\pi}{4}$ because $u(x) > a-1$. Now, let us differentiate $u(x) = an^2 x - 2 an x + a$. The derivative is $u'(x) = 2 an x \sec^2 x - 2 \sec^2 x$, which factors elegantly into: $$u'(x) = 2( an x - 1)\sec^2 x$$ Since $2$ and $\sec^2 x$ are always positive, the sign of $g'(x)$ depends entirely on the term $( an x - 1)$. ### The Final Verdict Let us test our intervals. In the interval $(0, \frac{\pi}{4})$, $ an x < 1$, so $( an x - 1) < 0$. This makes $g'(x) < 0$, meaning $g(x)$ is decreasing. Statement (I) claims $g$ is increasing, so it is **false**. In the interval $(\frac{\pi}{4}, \frac{\pi}{2})$, $ an x > 1$, so $( an x - 1) > 0$. This makes $g'(x) > 0$, meaning $g(x)$ is increasing. Statement (II) claims $g$ is decreasing, so it is also **false**. We have systematically dismantled the problem, and the conclusion is clear: **neither statement is true**. It is a wonderful reminder that in mathematics, a little bit of algebraic manipulation and a clear geometric intuition can turn a complex problem into a simple, elegant truth.

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Visualized Solution

Analyzing $f^{''} (x) > 0$

The Root at $x = a - 1$

Sign Scheme of $f^{'} (x)$

Defining $u (x)$

Minimum Value of $u (x)$

Evaluating $f^{'} (u (x))$

Applying the Chain Rule

Derivative of $u (x)$

The Expression for $g^{'} (x)$

Case 1: $x \in (0, \frac{π}{4})$

$g (x)$ in $(0, \frac{π}{4})$

Case 2: $x \in (\frac{π}{4}, \frac{π}{2})$

$g (x)$ in $(\frac{π}{4}, \frac{π}{2})$

Final Answer

The Sigma Insight: Monotonicity

Analyzing the Geometry of Curvature

The Inner Life of $g (x)$

The Chain Rule Symphony

The Final Verdict

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Which of the following is true?

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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 :

Let $f : (0, 1) \to R$ be a function defined by $f (x) = \frac{1}{1 - e ^{- x}}$ , and $g (x) = (f (- x) - f (x))$ . Consider two statements (I) $g$ is an increasing function in $(0, 1)$ (II) $g$ is one-one in $(0, 1)$ Then,

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Visualized Solution

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

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

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

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

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

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

Applying the Chain Rule

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

The Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g′(x)

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈(0,4π​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x) in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (0,4π​)

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈(4π​,2π​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x) in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (4π​,2π​)