If the function g(x) = { sqrt(kx+1), 0

Q: Why do we need continuity? Can't we just check if the derivatives are equal?

Great question! Differentiability is a stronger condition than continuity. A function can have equal derivatives on both sides but still have a 'jump' or 'break' at the point, which would make it non-differentiable. Think of it like a road: you cannot have a smooth slope if the road itself is broken!

Q: Why do we use the chain rule for the square root?

Excellent observation. The function is $\sqrt{kx+1}$. When we differentiate with respect to $x$, we treat $kx+1$ as an inner function. The derivative of $\sqrt{u}$ is $\frac{1}{2\sqrt{u}}$, and by the chain rule, we multiply by the derivative of the inside, which is $k$. So, we get $\frac{k}{2\sqrt{kx+1}}$.

Q: What happens if the function wasn't differentiable at $x=3$?

If it weren't differentiable, it would mean either the graph has a 'jump' (discontinuity) or a 'sharp corner' (cusp). At a sharp corner, the slope changes abruptly, so the left-hand derivative and right-hand derivative would not be equal. In this problem, we are *given* that it is differentiable, so we force those conditions to be true!

Question

If the function $g(x) = \begin{cases} \sqrt{kx+1}, & 0 \leq x \leq 3 \ mx + 2, & 3 < x \leq 5 \end{cases}$ is differentiable, then the value of $k + m$ is

Options:

(A) 10/3

(B) 4

(C) 2

(D) 16/5

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery! Today, we explore a problem that sits at the very heart of calculus: the concept of differentiability.

Imagine you are an engineer tasked with connecting two different roads. One road is a curved path, and the other is a straight highway. If you simply join them, you might end up with a sharp, dangerous corner.

In calculus, we demand more. We want a 'smooth' transition, which is exactly what it means for a function to be differentiable.

### Phase 1: The Bridge (Continuity)

Our function $g(x)$ is defined in two pieces:
$$g(x) = 
\begin{cases} 
k\sqrt{x+1} & 0 \leq x \leq 3 \ 
mx + 2 & 3 < x \leq 5 
\end{cases}$$

The problem states this function is differentiable everywhere. This implies two things: the function must be continuous (no jumps) and it must be smooth (no sharp corners).

Let's start with continuity. At the junction $x = 3$, the left-hand piece must meet the right-hand piece perfectly. Mathematically, we require:
$$\lim_{x 	o 3^-} g(x) = \lim_{x 	o 3^+} g(x)$$

Substituting $x = 3$ into both expressions, we get:
$$k\sqrt{3+1} = m(3) + 2$$

Simplifying this, we find $\sqrt{4} = 2$, leading to:
$$2k = 3m + 2$$
This is our first vital clue, which we will call **Equation 1**.

### Phase 2: The Smoothness (Differentiability)

Now that we have ensured the road is connected, we must ensure it is smooth. This means the slope of the curve on the left must match the slope of the line on the right at the exact moment they meet.

We need the Left Hand Derivative (LHD) to equal the Right Hand Derivative (RHD) at $x = 3$. For the left piece, $g(x) = k\sqrt{x+1}$, we use the chain rule to find the derivative:
$$g'(x) = \frac{k}{2\sqrt{x+1}}$$

For the right piece, $g(x) = mx + 2$, the derivative is simply the constant $m$. Equating these at $x = 3$, we get:
$$\frac{k}{2\sqrt{3+1}} = m$$

This simplifies to $\frac{k}{4} = m$, or:
$$k = 4m$$
This is our second vital clue, **Equation 2**.

### Phase 3: The Final Synthesis

We now have a system of two linear equations: $2k = 3m + 2$ and $k = 4m$. We substitute $k = 4m$ into our first equation:
$$2(4m) = 3m + 2$$

This gives us $8m = 3m + 2$. Subtracting $3m$ from both sides, we get $5m = 2$, which means:
$$m = \frac{2}{5}$$

With $m$ in hand, finding $k$ is a breeze:
$$k = 4\left(\frac{2}{5}ight) = \frac{8}{5}$$

The question asks for the sum $k + m$. Adding them together, we get:
$$\frac{8}{5} + \frac{2}{5} = \frac{10}{5} = 2$$

And there you have it! By enforcing the laws of continuity and smoothness, we have bridged the gap between these two functions. The final result is **2**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If the function $g (x) = {k x + 1, m x + 2, 0 \leq x \leq 3 3 < x \leq 5$ is differentiable, then the value of $k + m$ is

Select Answer:

Visualized Solution

Understanding the Piecewise Function

Condition 1: Continuity at $x = 3$

Applying Continuity at $x = 3$

Simplifying the Continuity Equation

Condition 2: Differentiability at $x = 3$

Calculating the Derivatives

Equating LHD and RHD at $x = 3$

Solving for $m$

Finding the Value of $m$

Finding the Value of $k$

Final Calculation: $k + m$

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

If $f (x) = {\frac{1}{∣ x ∣}, a x^{2} + b, ∣ x ∣ \geq 1 ∣ x ∣ < 1$ is differentiable at every point of the domain, then the values of $a$ and $b$ are respectively:

If the function $f (x) = {\frac{1}{∣ x ∣} a x^{2} + 2 b, ∣ x ∣ \geq 2, ∣ x ∣ < 2$ is differentiable on $R$ , then $48 (a + b)$ is equal to_____

Consider the function $f : (0, 2) \to R$ defined by $f (x) = \frac{x}{2} + \frac{2}{x}$ and the function $g (x)$ defined by $g (x) = {min {f (t)}, \frac{3}{2} + x, 0 < t \leq x and 0 < x \leq 1 1 < x < 2$ . Then

Suppose $f (x)$ is differentiable at $x = 1$ and $lim_{h \to 0} \frac{1}{h} f (1 + h) = 5$ , then $f^{'} (1)$ equals

Let $f (x) = ∣2 x^{2} + 5∣ x ∣ - 3∣, x \in R$ . If $m$ and $n$ denote the number of points where $f$ is not continuous and not differentiable respectively, then $m + n$ is equal to :

Let $f (x) = {0, x^{2}, x < 0 x \geq 0$ then for all $x$

Let $f : [a, b] \to [1, \infty)$ be a continuous function and let $g : R \to R$ be defined as $g (x) = ⎩ ⎨ ⎧ 0, \int_{a}^{x} f (t) d t, \int_{a}^{b} f (t) d t, if x < a, if a \leq x \leq b; if x > b .$ then

Let $f (x) = {- 1, - 2 \leq x < 0 x^{2} - 1, 0 \leq x \leq 2$ and $g (x) = ∣ f (x) ∣ + f (∣ x ∣)$ . Then, in the interval $(- 2, 2)$ , $g$ is :

Select Answer:

Visualized Solution

Understanding the Piecewise Function

Condition 1: Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=3

Applying Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=3

Simplifying the Continuity Equation

Condition 2: Differentiability at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=3

Calculating the Derivatives

Equating LHD and RHD at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=3

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

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

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

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k+m