ISS 2016 Statistics Paper-1 Solution: Question 56 (Finite Differences of a Cubic Polynomial)

Continuing our question-by-question walk through the ISS Statistics Paper-1 previous year papers, we now arrive at Question 56 of the 2016 paper. This one sits in the Numerical Analysis portion of the syllabus and tests whether you really understand what the finite difference operator Δdoes to a polynomial, rather than just memorising a formula for it.

Quick Summary

• Topic:Finite Differences (first, second and third order differences of a cubic polynomial)

• Question Reference:ISS 2016, Statistics Paper-1, Question 56

• Correct Answer:Option (a) — Statements 1 and 2 only

The Question

Let f(x) = 5x3 − 13x2 + 11x + 4. If the interval of differencing is 1, then consider the following statements:

1. Δ3 f(x) = 30 for all x

2. Δ2 f(x) = 34 for x = 1

3. Δ f(x) = 4 for x = 1

Which of the above statements are correct?

• (a) 1 and 2 only

• (b) 1 and 3 only

• (c) 2 and 3 only

• (d) 1, 2 and 3

Step 1: Remember what Δ actually means

The forward difference operator Δ is defined, for an interval of differencing h, as:

Δ f(x) = f(x + h) − f(x)

Here the interval of differencing h = 1, so Δ f(x) = f(x + 1) − f(x). This is the whole idea — nothing fancier than "subtract the current value from the next value." Once you have Δf(x) as a new function of x, you apply the same operator to it again to get Δ²f(x), and again to get Δ³f(x).

Step 2: Compute Δ f(x) algebraically

We need f(x + 1). Substitute (x + 1) in place of x in f(x) = 5x3 − 13x2 + 11x + 4.

First expand (x + 1)3 = x3 + 3x2 + 3x + 1, and (x + 1)2 = x2 + 2x + 1.

So:

5(x + 1)3 = 5x3 + 15x2 + 15x + 5

−13(x + 1)2 = −13x2 − 26x − 13

11(x + 1) = 11x + 11

+4 stays as +4

Adding all four pieces together:

f(x + 1) = 5x3 + (15 − 13)x2 + (15 − 26 + 11)x + (5 − 13 + 11 + 4) = 5x3 + 2x2 + 0·x + 7

Now subtract the original f(x) = 5x3 − 13x2 + 11x + 4:

Δf(x) = f(x + 1) − f(x) = (5x3 + 2x2 + 7) − (5x3 − 13x2 + 11x + 4)

The x3 terms cancel (5x3 − 5x3 = 0), which is expected — differencing always drops the degree of a polynomial by exactly one. What remains is:

Δf(x) = (2 − (−13))x2 + (0 − 11)x + (7 − 4) = 15x2 − 11x + 3

Step 3: Check Statement 3 using Δf(x)

Statement 3 claims Δf(x) = 4 for x = 1. Substitute x = 1 into Δf(x) = 15x2 − 11x + 3:

Δf(1) = 15(1) − 11(1) + 3 = 15 − 11 + 3 = 7

So Δf(1) = 7, not 4.Statement 3 is false.

You can sanity-check this directly from the raw function values instead of the algebra, which is a good habit whenever you're unsure of an expansion: f(1) = 5 − 13 + 11 + 4 = 7, and f(2) = 5(8) − 13(4) + 11(2) + 4 = 40 − 52 + 22 + 4 = 14. Then Δf(1) = f(2) − f(1) = 14 − 7 = 7, matching our formula exactly.

Step 4: Find Δ²f(x) and check Statement 2

Δ²f(x) is simply Δ applied to Δf(x) = 15x2 − 11x + 3. Using the same substitution rule:

Δ(15x2) = 15[(x+1)2 − x2] = 15(2x + 1) = 30x + 15

Δ(−11x) = −11[(x+1) − x] = −11

Δ(3) = 0 (a constant never changes under differencing)

Adding these: Δ²f(x) = 30x + 15 − 11 = 30x + 4.

At x = 1: Δ²f(1) = 30(1) + 4 = 34. This matches Statement 2 exactly, so Statement 2 is true.

Again, you can verify this with raw values: f(3) = 5(27) − 13(9) + 11(3) + 4 = 135 − 117 + 33 + 4 = 55. So Δf(2) = f(3) − f(2) = 55 − 14 = 41, and Δ²f(1) = Δf(2) − Δf(1) = 41 − 7 = 34. Confirmed.

Step 5: Find Δ³f(x) and check Statement 1

Differencing Δ²f(x) = 30x + 4 one more time:

Δ³f(x) = Δ(30x + 4) = 30[(x+1) − x] + 0 = 30

This is a constant — it does not depend on x at all. So Δ³f(x) = 30 for every value of x, exactly as Statement 1 claims.Statement 1 is true.

Pro Tip: There is a shortcut that saves all of this algebra once you trust it. For any polynomial of degree n with leading coefficient a and interval of differencing h, the n-th order difference is always the constant

Step 6: Put it together and confirm the answer

Statement 1 — true. Statement 2 — true. Statement 3 — false. So the statements that are correct are 1 and 2 only, which is option (a).

This matches the official ISS 2016 Statistics Paper-1 answer key (Series A), which also lists option (a) as correct for Question 56. Our independent derivation and the official key agree.

Why This Question Matters

Finite difference questions like this one are a recurring theme across ISS papers because they test two things at once: whether you can mechanically expand and simplify polynomial expressions without error, and whether you understand the deeper structural fact that differencing a degree-n polynomial n times always collapses it to a constant, and differencing it further always gives zero. Students who drill this kind of derivation by hand — rather than jumping straight to the shortcut formula — tend to catch sign errors and off-by-one mistakes much faster in the exam hall, because they can mentally verify an answer using a couple of raw function values the way we did in Steps 3 and 4.

Frequently Asked Questions

What is the interval of differencing, and why does it matter here?

The interval of differencing (usually called h) is the fixed gap between successive x-values at which the function is evaluated. In this question h = 1, meaning Δf(x) = f(x+1) − f(x). If h had been 2, every formula above would change, since Δf(x) would instead be f(x+2) − f(x), so always check the value of h before applying any difference formula.

Why does the degree of the polynomial drop by one every time we apply Δ?

Each time you expand f(x+h) and subtract f(x), the highest-degree terms are identical in both expressions and cancel out exactly, leaving only lower-degree terms. This is why a cubic becomes a quadratic after one differencing, a linear function after two, and a constant after three.

Is there a faster way to get Δ³f(x) without expanding everything?

Yes — for a degree-n polynomial with leading coefficient a, the n-th difference is always a·n!·hn, as shown in the Pro Tip above. It's worth deriving this formula yourself once so you understand where the n! and hn come from, rather than memorising it blindly.

How is this different from a derivative?

A derivative measures instantaneous rate of change over an infinitesimally small interval, while a finite difference measures the actual change over a fixed, finite interval h. As h approaches zero, Δf(x)/h approaches f'(x), which is why finite differences are the backbone of numerical differentiation methods.

What is the most common mistake students make on this type of question?

The most frequent error is an arithmetic slip while expanding (x+1)2 or (x+1)3, which then propagates through every subsequent step. Always double-check your expanded polynomial by plugging in a simple value like x = 0 on both sides before moving forward.

Why does statement 3 look plausible but turn out to be wrong?

Statement 3 gives a value, 4, that is close to the constant term of f(x) itself (which is also 4), making it an easy trap for anyone who confuses f(x) with Δf(x). Carefully separating what each operator actually produces — as we did by computing Δf(x) as its own standalone function first — avoids this confusion entirely.

If anything in this derivation felt unclear, drop a comment below with the exact step you'd like explained further, and do share this with a fellow ISS aspirant who is also working through finite differences — explaining it to someone else is often the fastest way to make it stick.