Question

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

• A. 59
• B. 60
• C. 61
• D. 58

Answer 1

The Correct Option is B

We are given:

• \( f(4) = 2 \cdot 4^n + \lambda = 133 \) (1)
• \( f(5) = 2 \cdot 5^n + \lambda = 255 \) (2)

Subtract equation (1) from equation (2):

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

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

\( 5^n - 4^n = 61 \)

Testing integral values of \( n \):

For \( n = 3 \), we have:

\( 5^3 - 4^3 = 125 - 64 = 61 \)

Thus, \( n = 3 \). Substituting back into equation (1) to find \( \lambda \):

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

\( 128 + \lambda = 133 \)

\( \lambda = 5 \)

Now, compute \( f(3) - f(2) \):

\( f(3) = 2 \cdot 3^3 + 5 = 2 \cdot 27 + 5 = 54 + 5 = 59 \)

\( f(2) = 2 \cdot 2^3 + 5 = 2 \cdot 8 + 5 = 16 + 5 = 21 \)

\( f(3) - f(2) = 59 - 21 = 38 \)

Find the positive integer divisors of 38:

• 1, 2, 19, 38

Sum of divisors:

\( 1 + 2 + 19 + 38 = 60 \)

Thus, the correct answer is option (2).

Answer 2

The correct answer is (B) : 60
$f(x)=2x^n+\lambda$
$f(4)=133$
$f(5)=255$
$133=2\times 4^n+\lambda ...$(1)
$255=2\times 5^n+\lambda ....$(2)
$(2)-(1)$
$122=2\left(5^n-4^n\right)$
$\Rightarrow 5^n-4^n=61$
$\therefore n=3\&\lambda =5$
Now, $f(3)-f(2)=2\left(3^3-2^3\right)=38$
Number of Divisors is $1,2,19,38$; & their sum is $60$