Let f, g : R → R be two real valued functions defined as f(x) = { -|x + 3| , x

Q: Wait, why do we need to check continuity? Isn't differentiability enough?

That is a classic trap! Differentiability is a stronger condition than continuity. If a function is differentiable at a point, it *must* be continuous there. If you skip the continuity check, you might miss the constraint that links $k_1$ and $k_2$. Always check continuity first to ensure the 'bridge' between the two branches is solid.

Q: How do I know which branch of $g(x)$ to use when $x < 0$?

You must look at the output of $f(x)$. For $x < 0$, $f(x) = -|x+3|$. Since $x$ is near $0$, $x+3$ is positive, so $f(x) = -(x+3)$, which is negative. Because the input to $g$ is negative, you must use the $x < 0$ branch of $g(x)$. Always trace the range of the inner function to decide the domain of the outer function.

Q: Why did the constant $k_2$ vanish in the derivative?

The derivative of any constant is zero. When we differentiate $4e^x + k_2$, the $k_2$ term disappears because it represents a vertical shift, which doesn't affect the slope of the curve. However, $k_2$ is still vital for the continuity condition, as it determines the vertical position of the graph!

Question

Let $f, g : \mathbb{R} 	o \mathbb{R}$ be two real valued functions defined as $f(x) = \begin{cases} -|x + 3| & , x < 0 \ e^x & , x \ge 0 \end{cases}$ and $g(x) = \begin{cases} x^2 + k_1 x & , x < 0 \ 4x + k_2 & , x \ge 0 \end{cases}$, where $k_1$ and $k_2$ are real constants. If $(gof)$ is differentiable at $x = 0$, then $(gof)(-4) + (gof)(4)$ is equal to :

Options:

(A) $4(e^4 + 1)$

(B) $2(2e^4 + 1)$

(C) $4e^4$

(D) $2(2e^4 - 1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to the heart of calculus. Today, we are going to dissect a problem that is a favorite in the JEE Advanced arena. It is not just about crunching numbers; it is about understanding the 'smoothness' of a function.

We are dealing with a composite function, $h(x) = g(f(x))$, and we are tasked with ensuring it is differentiable at the critical point $x = 0$. Let us embark on this journey together.

### The Relay Race of Functions

Imagine $f(x)$ and $g(x)$ as two machines in a factory. $f(x)$ takes an input $x$, processes it, and spits out a value, which is then immediately fed into $g(x)$.

We are given:
$$f(x) = \begin{cases} -|x + 3| & , x < 0 \ e^x & , x \ge 0 \end{cases}$$

$$g(x) = \begin{cases} x^2 + k_1 x & , x < 0 \ 4x + k_2 & , x \ge 0 \end{cases}$$

For $x \ge 0$, $f(x) = e^x$. Since $e^x \ge 1$ for all $x \ge 0$, the output of $f(x)$ is strictly positive, requiring the branch $g(x) = 4x + k_2$. Thus, for $x \ge 0$, our composite function becomes:
$$h(x) = 4e^x + k_2$$

For $x < 0$, we look at $f(x) = -|x + 3|$. In the neighborhood of zero, $x + 3$ is positive, so $f(x) = -(x + 3)$. Since this value is negative, we use the branch $g(x) = x^2 + k_1 x$:
$$h(x) = (-(x + 3))^2 + k_1(-(x + 3)) = (x + 3)^2 - k_1(x + 3)$$

### The Bridge of Continuity

For a function to be differentiable at a point, it must first be continuous. It cannot have a 'jump' or a 'break' at $x = 0$.

We equate the limits of our two branches at $x = 0$:
$$\lim_{x 	o 0^-} h(x) = \lim_{x 	o 0^+} h(x)$$

Substituting our expressions:
$$LHL: (0 + 3)^2 - k_1(0 + 3) = 9 - 3k_1$$
$$RHL: 4e^0 + k_2 = 4 + k_2$$

Equating them gives us our first vital constraint:
$$3k_1 + k_2 = 5$$

### The Calculus of Smoothness

Differentiability means the slope of the curve coming from the left must perfectly match the slope of the curve coming from the right.

Let us find the derivatives of our branches:
For the right branch ($x \ge 0$):
$$h'(x) = \frac{d}{dx}(4e^x + k_2) = 4e^x \Rightarrow h'(0^+) = 4$$

For the left branch ($x < 0$):
$$h'(x) = \frac{d}{dx}((x + 3)^2 - k_1(x + 3)) = 2(x + 3) - k_1 \Rightarrow h'(0^-) = 6 - k_1$$

For the function to be differentiable, we must have $h'(0^-) = h'(0^+)$:
$$6 - k_1 = 4 \Rightarrow \mathbf{k_1 = 2}$$

Substituting $k_1 = 2$ into our continuity equation $3k_1 + k_2 = 5$:
$$3(2) + k_2 = 5 \Rightarrow 6 + k_2 = 5 \Rightarrow \mathbf{k_2 = -1}$$

### The Final Victory

We have defined our function $h(x)$ completely. Now, we evaluate the final expression: $(gof)(-4) + (gof)(4)$.

For $x = -4$, $f(-4) = -|-4 + 3| = -1$. Then $g(-1) = (-1)^2 + 2(-1) = 1 - 2 = -1$. So, $h(-4) = -1$.

For $x = 4$, $f(4) = e^4$. Since $e^4 > 0$, we use the second branch of $g$:
$$g(e^4) = 4e^4 + k_2 = 4e^4 - 1$$

Adding them together:
$$Sum = -1 + (4e^4 - 1) = \mathbf{4e^4 - 2}$$

There it is! You have navigated the piecewise definitions, enforced the continuity, matched the slopes, and arrived at the truth. This is the power of calculus—taking complex, fragmented pieces and weaving them into a single, coherent, and beautiful mathematical reality.

Select Answer:

Visualized Solution

Introduction to $h (x) = g (f (x))$

Analyze $f (x)$ for $x \geq 0$

Formulate $h (x)$ for $x \geq 0$

Analyze $f (x)$ for $x < 0$

Formulate $h (x)$ for $x < 0$

Continuity Condition at $x = 0$

Evaluate Continuity Equation

Right Hand Derivative (RHD)

Left Hand Derivative (LHD)

Equate Derivatives to find $k_{1}$

Solve for $k_{2}$

Calculate $h (- 4)$

Calculate $h (4)$

Final Summation

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Relay Race of Functions

The Bridge of Continuity

The Calculus of Smoothness

The Final Victory

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