Let f(x) = 2xⁿ + λ, λ ∈ R, n ∈ N, and f(4) = 133, f(5) = 255. Then the sum of all the positive integ

Q: Why can't we just solve for $\lambda$ first?

You absolutely could, but it would be algebraically tedious. By subtracting the two equations, the $\lambda$ terms cancel out immediately, leaving you with a much cleaner expression involving only $n$. In competitive exams like JEE, recognizing these shortcuts is key to saving time.

Q: How do I know $n$ is a small integer?

The expression $5^n - 4^n$ grows extremely rapidly as $n$ increases. For $n=1$, the value is $1$. For $n=2$, it is $9$. For $n=3$, it is $61$. If $n$ were any larger, the result would far exceed $61$. This rapid growth is a common characteristic of exponential functions, which helps narrow down the search space significantly.

Q: What if I forgot the formula for the sum of divisors?

Don't panic! For a small number like $38$, you don't need a complex formula. Simply list the pairs of factors: $1 \times 38 = 38$ and $2 \times 19 = 38$. The divisors are $1, 2, 19,$ and $38$. Just add them up: $1 + 2 + 19 + 38 = 60$. It is often faster to list them for small numbers than to use the prime factorization method.

Question

Let $f(x) = 2x^n + \lambda, \lambda \in \mathbb{R}, n \in \mathbb{N}$, and $f(4) = 133, f(5) = 255$. Then the sum of all the positive integer divisors of $(f(3) - f(2))$ is

Options:

(A) 61

(B) 60

(C) 58

(D) 59

Sigma Percentile · Accepted Answer

### The Beauty of Functional Detective Work

Welcome, future engineer! Today, we are going to unravel a problem that might look like a simple algebra exercise, but it is actually a beautiful dance of logic and number theory.

We are given a function $f(x) = 2x^n + \lambda$, where $\lambda$ is a real number and $n$ is a natural number. We are also given two clues: $f(4) = 133$ and $f(5) = 255$. Our mission is to find the sum of all positive integer divisors of $f(3) - f(2)$.

### Phase 1: The Setup

First, we must translate the given information into the language of mathematics. We have two unknowns, $n$ and $\lambda$, and two data points to solve for them.

Let us write down our two equations:

$$2(4^n) + \lambda = 133$$

$$2(5^n) + \lambda = 255$$

Imagine you are a detective. The most elegant way to solve this is to make one of the variables disappear. Notice that both equations contain a $+\lambda$; if we subtract the first equation from the second, the $\lambda$ will vanish.

### Phase 2: The Elimination

Let us perform the subtraction:

$$(2(5^n) + \lambda) - (2(4^n) + \lambda) = 255 - 133$$

The $\lambda$ terms cancel out, leaving us with:

$$2(5^n) - 2(4^n) = 122$$

We can factor out the $2$ on the left side and divide both sides by $2$:

$$5^n - 4^n = 61$$

This is where the magic happens. Because exponential functions grow so quickly, we do not need complex logarithms; we can simply test small integers for $n$.

If $n=1$, $5^1 - 4^1 = 1$. If $n=2$, $5^2 - 4^2 = 25 - 16 = 9$.

If $n=3$, $5^3 - 4^3 = 125 - 64 = 61$. Bingo! We have found our $n = 3$.

### Phase 3: Finding the Constants

Now that we know $n=3$, we can easily find $\lambda$. Let us plug $n=3$ back into our first equation:

$$2(4^3) + \lambda = 133$$

$$2(64) + \lambda = 133$$

$$128 + \lambda = 133 \Rightarrow \lambda = 5$$

Our function is now fully revealed: $f(x) = 2x^3 + 5$.

### Phase 4: The Final Calculation

The problem asks for the sum of the divisors of $f(3) - f(2)$. Let us calculate this difference:

$$f(3) - f(2) = (2(3^3) + 5) - (2(2^3) + 5)$$

Notice how the constant $\lambda = 5$ cancels out again. We are left with:

$$f(3) - f(2) = 2(3^3 - 2^3) = 2(27 - 8) = 2(19) = 38$$

Finally, we need the sum of the positive integer divisors of $38$. The divisors of $38$ are $1, 2, 19,$ and $38$.

Adding them together:

$$1 + 2 + 19 + 38 = 60$$

And there you have it! The final sum is **60**. You have successfully navigated the algebra, the exponential growth, and the number theory.

Animated Solution for Mathematics - Functions: Let $f (x) = 2 x^{n} + λ, λ \in R, n \in N$ , and $f (4) = 133, f (5) = 255$ . Then the sum of all the positive integer divisors of $(f (3) - f (2))$ is

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Visualized Solution

Defining the Function $f (x)$

Equation for $f (4)$

Equation for $f (5)$

Eliminating $λ$

Simplifying the Result

Solving for $n$

Finding $λ$

The Final Function

Calculating $f (3) - f (2)$

Finding Divisors of 38

Sum of Divisors

The Sigma Insight: Classification of Functions

The Beauty of Functional Detective Work

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=2xn+λ,λ∈R,n∈N, and f(4)=133,f(5)=255. Then the sum of all the positive integer divisors of (f(3)−f(2)) is

Select Answer:

Visualized Solution

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

Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(4)

Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(5)

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