JEE MainMathematics

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Sets, Relations and Functions: Mathematics | JEE Main

Q: Why do we not need to find the value of lambda?

Because lambda cancels out in differences of the form .

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))

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

60

59

61

58

Step-by-step solutionView

Correct answer

The value of $f(3) - f(2)$ is $38$ , and the sum of all its positive divisors is $60$ .

Option analysis

Why each option works or fails

A ·

60

None. The student solved for $n=3$ , found $f(3) - f(2) = 2(27 - 8) = 38$ , and correctly summed its divisors $1 + 2 + 19 + 38 = 60$ . Correctly determined $n=3$ , evaluated the difference, and summed all positive divisors.

B ·

59

The student computed the sum of proper divisors (excluding $1$ or forgetting to include $1$ ) instead of all positive divisors. Remember that the set of positive integer divisors of $N$ includes $1$ and $N$ itself. For $38$ , the divisors are $1, 2, 19, 38$ , which sum to $60$ , not $59$ .

C ·

61

The student made an arithmetic error when adding the divisors, adding $1$ extra or misidentifying a divisor. Check the divisor sum carefully: $\sigma(38) = (1 + 2)(1 + 19) = 3 \times 20 = 60$ .

D ·

58

The student made an arithmetic slip in evaluating $f(3) - f(2)$ or when summing the divisors. Ensure $f(3) - f(2) = 2(3^3 - 2^3) = 2(27 - 8) = 38$ , whose prime factorization is $2 \times 19$ .

Reviewed route

Solution

StepWorking

01given

$f(x) = 2x^n + \lambda$ where $\lambda \in \mathbb{R}, n \in \mathbb{N}$ , $f(4) = 133$ , and $f(5) = 255$ .

02goal

Find the sum of all positive integer divisors of $f(3) - f(2)$ .

03approach

Subtract $f(4)$ from $f(5)$ to eliminate $\lambda$ and solve for $n \in \mathbb{N}$ . Then calculate $f(3) - f(2) = 2(3^n - 2^n)$ and compute the divisor sum formula $\sigma_1(N)$ .

04execute

Subtracting the equations gives $f(5) - f(4) = 2(5^n - 4^n) = 255 - 133 = 122$ , hence $5^n - 4^n = 61$ . For $n=3$ , $5^3 - 4^3 = 125 - 64 = 61$ , so $n = 3$ .

05execute

Evaluate $f(3) - f(2) = 2(3^3 - 2^3) = 2(27 - 8) = 2 \times 19 = 38$ .

06execute

The prime factorization of 38 is $2^1 \times 19^1$ . The sum of divisors is $\sigma(38) = (1 + 2)(1 + 19) = 3 \times 20 = 60$ .

✓verify

Divisors of 38 are 1, 2, 19, 38. Their direct sum is $1 + 2 + 19 + 38 = 60$ , matching the result.

Hints that build this answer step by step

What equation results from subtracting $f(4)$ from $f(5)$ to eliminate $\lambda$ ?
$2(5^n - 4^n) = 122$
From $5^n - 4^n = 61$ with $n \in \mathbb{N}$ , what is the value of $n$ ?
$n = 3$
What is the value of $f(3) - f(2)$ ?
$38$
What is the sum of all positive integer divisors of $38$ ?
$60$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do we not need to find the value of lambda?

Because lambda cancels out in differences of the form $f(a) - f(b) = 2(a^n - b^n)$ .

Answer

The value of $f(3) - f(2)$ is $38$ , and the sum of all its positive divisors is $60$ .

Why each option works or fails

A · correct: $60$ - None. The student solved for $n=3$ , found $f(3) - f(2) = 2(27 - 8) = 38$ , and correctly summed its divisors $1 + 2 + 19 + 38 = 60$ . Correctly determined $n=3$ , evaluated the difference, and summed all positive divisors.
B: $59$ - The student computed the sum of proper divisors (excluding $1$ or forgetting to include $1$ ) instead of all positive divisors. Remember that the set of positive integer divisors of $N$ includes $1$ and $N$ itself. For $38$ , the divisors are $1, 2, 19, 38$ , which sum to $60$ , not $59$ .
C: $61$ - The student made an arithmetic error when adding the divisors, adding $1$ extra or misidentifying a divisor. Check the divisor sum carefully: $\sigma(38) = (1 + 2)(1 + 19) = 3 \times 20 = 60$ .
D: $58$ - The student made an arithmetic slip in evaluating $f(3) - f(2)$ or when summing the divisors. Ensure $f(3) - f(2) = 2(3^3 - 2^3) = 2(27 - 8) = 38$ , whose prime factorization is $2 \times 19$ .

Step-by-step solution

given: $f(x) = 2x^n + \lambda$ where $\lambda \in \mathbb{R}, n \in \mathbb{N}$ , $f(4) = 133$ , and $f(5) = 255$ .
goal: Find the sum of all positive integer divisors of $f(3) - f(2)$ .
approach: Subtract $f(4)$ from $f(5)$ to eliminate $\lambda$ and solve for $n \in \mathbb{N}$ . Then calculate $f(3) - f(2) = 2(3^n - 2^n)$ and compute the divisor sum formula $\sigma_1(N)$ .
execute: Subtracting the equations gives $f(5) - f(4) = 2(5^n - 4^n) = 255 - 133 = 122$ , hence $5^n - 4^n = 61$ . For $n=3$ , $5^3 - 4^3 = 125 - 64 = 61$ , so $n = 3$ .
execute: Evaluate $f(3) - f(2) = 2(3^3 - 2^3) = 2(27 - 8) = 2 \times 19 = 38$ .
execute: The prime factorization of 38 is $2^1 \times 19^1$ . The sum of divisors is $\sigma(38) = (1 + 2)(1 + 19) = 3 \times 20 = 60$ .
verify: Divisors of 38 are 1, 2, 19, 38. Their direct sum is $1 + 2 + 19 + 38 = 60$ , matching the result.

Sets, Relations and Functions: Mathematics | JEE Main

What feels right?

The value of $f(3) - f(2)$ is $38$ , and the sum of all its positive divisors is $60$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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