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JEE Mathematics 2023 Question with Solution

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

  • A

    6161

  • B

    6060

  • C

    5858

  • D

    5959

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: f(x)=2xn+λf(x) = 2x^n + \lambda, f(4)=133f(4) = 133, and f(5)=255f(5) = 255.

Find: The sum of all positive integer divisors of f(3)−f(2)f(3) - f(2).

From the given values,

2⋅4n+λ=1332 \cdot 4^n + \lambda = 133

and

2⋅5n+λ=2552 \cdot 5^n + \lambda = 255

Subtract the first equation from the second:

2⋅5n+λ−(2⋅4n+λ)=255−1332 \cdot 5^n + \lambda - (2 \cdot 4^n + \lambda) = 255 - 133 2(5n−4n)=1222(5^n - 4^n) = 122 5n−4n=615^n - 4^n = 61

Testing integral values of nn, for n=3n = 3,

53−43=125−64=615^3 - 4^3 = 125 - 64 = 61

Thus, n=3n = 3.

Substitute n=3n = 3 into

2⋅43+λ=1332 \cdot 4^3 + \lambda = 133 128+λ=133128 + \lambda = 133 λ=5\lambda = 5

Now compute:

f(3)=2⋅33+5=2⋅27+5=54+5=59f(3) = 2 \cdot 3^3 + 5 = 2 \cdot 27 + 5 = 54 + 5 = 59 f(2)=2⋅23+5=2⋅8+5=16+5=21f(2) = 2 \cdot 2^3 + 5 = 2 \cdot 8 + 5 = 16 + 5 = 21

Therefore,

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

The positive divisors of 3838 are 1,2,19,381, 2, 19, 38. Their sum is

1+2+19+38=601 + 2 + 19 + 38 = 60

Therefore, the correct option is B.

Direct Difference Method

Given: f(x)=2xn+λf(x) = 2x^n + \lambda, f(4)=133f(4) = 133, and f(5)=255f(5) = 255.

Find: The sum of all positive integer divisors of f(3)−f(2)f(3) - f(2).

Using the two given values directly,

f(5)−f(4)=255−133=122f(5) - f(4) = 255 - 133 = 122

But

f(5)−f(4)=2(5n−4n)f(5) - f(4) = 2(5^n - 4^n)

So,

5n−4n=615^n - 4^n = 61

Since 125−64=61125 - 64 = 61, we immediately get n=3n = 3.

Then from f(4)=133f(4) = 133,

2⋅43+λ=1332 \cdot 4^3 + \lambda = 133 128+λ=133128 + \lambda = 133 λ=5\lambda = 5

Now,

f(3)−f(2)=2(33−23)=2(27−8)=38f(3) - f(2) = 2(3^3 - 2^3) = 2(27 - 8) = 38

The divisors of 3838 are 1,2,19,381, 2, 19, 38, and their sum is 6060.

Therefore, the correct option is B.

Common mistakes

  • A common mistake is forgetting to subtract the two given equations to eliminate λ\lambda. That makes the problem longer and more error-prone. Instead, first use f(5)−f(4)f(5) - f(4) to isolate an equation involving only nn.

  • Some students test the wrong value of nn in 5n−4n=615^n - 4^n = 61. This is incorrect because nn must be a natural number and should satisfy the equation exactly. Check small integer values carefully until the equality holds.

  • Another mistake is computing f(3)−f(2)f(3) - f(2) as f(3)+f(2)f(3) + f(2) or substituting the wrong powers. Use n=3n = 3 and evaluate 333^3 and 232^3 correctly before subtracting.

  • Some students list the divisors of 3838 incorrectly by missing 1919. Since 38=2×1938 = 2 \times 19, its positive divisors are exactly 1,2,19,381, 2, 19, 38. Factor the number first before summing divisors.

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