Consider the function f : (0, ∞) → R defined by f(x) = e^(-|log_6 x|). If m and n be respectively th

Q: Wait, why is the function continuous everywhere?

Great question! The function $f(x) = e^{-|\log_6 x|}$ is a composition of three continuous functions: the logarithm, the modulus, and the exponential function. Since the composition of continuous functions is always continuous within their domain, there are no jumps or breaks in the graph. Thus, $m=0$.

Q: How do I handle the modulus in the exponent?

Think of the modulus as a switch. When the argument $\log_6 x$ is positive (which happens when $x > 1$), the modulus is just $\log_6 x$. When the argument is negative (when $0 < x < 1$), the modulus becomes $-\log_6 x$. By splitting the function into these two cases, you remove the modulus and can work with standard algebraic expressions.

Q: Why is the derivative at $x=1$ not defined?

Differentiability at a point requires the slope to be the same from both sides. At $x=1$, the left-hand derivative is $1/\ln 6$ and the right-hand derivative is $-1/\ln 6$. Because these two slopes are not equal, the graph has a sharp, jagged corner at $x=1$, making it non-differentiable.

Question

Consider the function $f : (0, \infty) 	o \mathbb{R}$ defined by $f(x) = e^{-|\log_6 x|}$. If $m$ and $n$ be respectively the number of points at which $f$ is not continuous and $f$ is not differentiable, then $m + n$ is

Options:

(A) 0

(B) 3

(C) 1

(D) 2

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine you are standing at the edge of a mathematical landscape, looking at the function $f(x) = e^{-|\log_6 x|}$. It looks intimidating, but in the world of JEE Advanced, we dismantle such complexity systematically. Let us embark on a journey to understand why this function behaves the way it does. ### The Continuity Check: A Smooth Foundation First, we must ask: where does this function break? Continuity is the soul of a function. We are looking at a composition of three standard, well-behaved functions: the logarithmic function $g(x) = \log_6 x$, the modulus function $h(u) = |u|$, and the exponential function $k(v) = e^{-v}$. Each of these is continuous across its entire domain. A fundamental theorem of calculus states that the composition of continuous functions is, itself, continuous. Therefore, for all $x \in (0, \infty)$, our function $f(x)$ flows without any jumps, holes, or vertical asymptotes. This means the number of points of non-continuity, $m$, is exactly $0$. We have cleared the first hurdle with elegance. ### The Differentiability Crisis: The Sharp Corner Now, we turn our attention to differentiability. This is where the modulus function, $|\log_6 x|$, demands our respect. The modulus function is notorious for creating 'sharp corners'—points where the graph makes a sudden, jagged turn. These corners occur precisely where the expression inside the modulus equals zero. Setting $\log_6 x = 0$, we find $x = 6^0 = 1$. This is our critical point. To see if the function is differentiable here, we must look at it from both sides. ### The Calculus Showdown: Left vs. Right For $x > 1$, the term $\log_6 x$ is positive, so the modulus simply vanishes: $f(x) = e^{-\log_6 x}$. Using the change of base formula, $\log_6 x = \frac{\ln x}{\ln 6}$, we rewrite this as: $$f(x) = e^{-\frac{\ln x}{\ln 6}} = x^{-\frac{1}{\ln 6}}$$ For $0 < x < 1$, the term $\log_6 x$ is negative. The modulus flips the sign, making the exponent positive: $$f(x) = e^{-(-\log_6 x)} = e^{\log_6 x} = x^{\frac{1}{\ln 6}}$$ Now, we apply the power rule to find the derivatives. For $x > 1$, the derivative is: $$f'(x) = -\frac{1}{\ln 6} x^{-\frac{1}{\ln 6} - 1}$$ Evaluating this at $x=1$, we get $f'(1^+) = -\frac{1}{\ln 6}$. For $0 < x < 1$, the derivative is: $$f'(x) = \frac{1}{\ln 6} x^{\frac{1}{\ln 6} - 1}$$ Evaluating this at $x=1$, we get $f'(1^-) = \frac{1}{\ln 6}$. ### The Conclusion: A Jagged Peak Comparing our results, we see that $f'(1^+) = -\frac{1}{\ln 6}$ and $f'(1^-) = \frac{1}{\ln 6}$. They are not equal! Geometrically, this confirms that the graph has a sharp, jagged peak at $x=1$. The slope changes abruptly from positive to negative. Thus, the function is not differentiable at $x=1$. We have found our $n = 1$. With $m=0$ and $n=1$, the sum $m+n$ is simply **$1$**. You have successfully navigated the complexity of this function, proving that even the most intimidating problems yield to a systematic, step-by-step approach.

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Visualized Solution

Analyze the Function $f (x) = e^{- ∣ l o g_{6} x ∣}$

Continuity of Component Functions

Determining $m$ (Points of Non-Continuity)

Analyzing Differentiability

Function Behavior for $x > 1$

Function Behavior for $0 < x < 1$

Calculating Right Hand Derivative (RHD)

Calculating Left Hand Derivative (LHD)

Comparing LHD and RHD

Final Calculation of $m + n$

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Continuity Check

The Differentiability Crisis

The Calculus Showdown

The Conclusion

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