ISS 2016 Statistics Paper-1 Solution: Question 63 (Symmetry Trick in Lagrange Interpolation)

Continuing our question-by-question walk through the ISS 2016 Statistics Paper-1, today's post tackles a numerical analysis item that looks intimidating at first glance but hides a very elegant shortcut once you notice the pattern in the data. If you've been following this series, you already know the drill: full working, no skipped steps, and a direct check against the official answer key at the end.

Quick Summary

• Topic:Numerical Analysis — Lagrange's Interpolation Formula and locating extrema of the interpolating polynomial

• Question Reference:ISS 2016, Statistics Paper-1, Question 63

• Correct Answer:(a) f(x) is maximum when x = 4.5

The Question As Asked

The values of the function f(x) for values of x are given as f(1) = 4, f(2) = 5, f(7) = 5, f(8) = 4. Which one of the following is correct in respect of f(x) when it is computed by using Lagrange's formula?

• (a) f(x) is maximum when x = 4.5

• (b) f(x) is minimum when x = 4.5

• (c) f(x) is maximum when x = 4

• (d) f(x) is minimum when x = 4

Step 1: Write Down What We Actually Have

We're given four data points: (x0, f0) = (1, 4), (x1, f1) = (2, 5), (x2, f2) = (7, 5), and (x3, f3) = (8, 4). With four points, Lagrange's formula will give us a unique interpolating polynomial of degree at most 3 (a cubic). The question is really asking: where on this cubic does f(x) reach a maximum or minimum, and is that point x = 4.5 or x = 4?

Before jumping into the full Lagrange expansion, it's worth pausing and simply looking at the numbers. f(1) = 4 and f(8) = 4 are equal. f(2) = 5 and f(7) = 5 are equal. That is not a coincidence the examiner put there by accident — it is the entire question.

Step 2: Spot the Symmetry

Look at the midpoint of the x-values: (1 + 8) / 2 = 4.5, and (2 + 7) / 2 = 4.5 as well. So all four points are symmetric about the vertical line x = 4.5. If we shift our variable so that this midpoint becomes the origin, the symmetry becomes very easy to exploit algebraically.

Let u = x − 4.5. Then the four points become:

• x = 1 → u = −3.5, f = 4

• x = 2 → u = −2.5, f = 5

• x = 7 → u = 2.5, f = 5

• x = 8 → u = 3.5, f = 4

Notice that f(−3.5) = f(3.5) = 4 and f(−2.5) = f(2.5) = 5. In terms of u, the function value at −u always equals the function value at +u. That means the data itself is an even function of u.

Step 3: Why This Forces the Polynomial to Be Even in u

Here is the key theoretical point, spelled out in full rather than assumed. The Lagrange interpolating polynomial through four distinct points is unique — there is only one polynomial of degree ≤ 3 that passes through all four given points. Now suppose P(u) is that unique polynomial, written in terms of u instead of x. Since the data satisfies f(−u) = f(u) at every one of the four points we were given, the polynomial P(−u) also passes through exactly the same four points (because plugging in −u where u was one of our four u-values just lands you back on the matching point, by the symmetry we just checked).

But since the interpolating polynomial through four points is unique, there can only be one such polynomial. So P(u) and P(−u) must be the same polynomial: P(u) = P(−u) for all u, not just at the four data points. A polynomial that satisfies P(u) = P(−u) identically can only contain even powers of u — any odd-power term (like a u or a u3 term) would flip sign under u → −u and break the equality. Since P(u) has degree at most 3, the only even powers available are u0 and u2. So we can write:

P(u) = A + B·u2

Pro Tip: This is the real trick of the question, and it's the step most students skip past — they try to grind out all four Lagrange basis polynomials L0, L1, L2, L3 in full and then differentiate the resulting cubic, which works but takes four or five times as long. The way this kind of question is usually taught at Sunrise Classes is to train yourself to check the x-values for symmetry about a midpoint

Step 4: Solve for A and B

We only need two of the four points now, since P(u) = A + B·u2 has just two unknowns. Use u = 2.5 (f = 5) and u = 3.5 (f = 4):

A + B(2.5)2 = 5 → A + 6.25B = 5

A + B(3.5)2 = 4 → A + 12.25B = 4

Subtract the first equation from the second:

(12.25B − 6.25B) = (4 − 5)

6B = −1

B = −1/6

Substitute back into A + 6.25B = 5:

A = 5 − 6.25 × (−1/6) = 5 + 6.25/6 = 5 + 1.041666... = 6.041666...

As an exact fraction, A = 145/24. So the interpolating polynomial, in terms of u, is:

P(u) = 145/24 − (1/6)u2

Step 5: Identify the Maximum

This is now just a downward-opening parabola in u, since the coefficient of u2 is negative (B = −1/6 < 0). A downward parabola A + Bu2 with B < 0 has its single highest point exactly at u = 0, where the squared term vanishes and contributes nothing to subtract from A. Moving away from u = 0 in either direction only decreases P(u), because u2 grows and we're subtracting a positive multiple of it.

Recall u = x − 4.5, so u = 0 corresponds to x = 4.5. Therefore f(x) reaches its maximum value exactly at x = 4.5, and that maximum value is P(0) = A = 145/24 ≈ 6.0417.

Step 6: Rule Out x = 4 Directly

To be thorough, let's also compute P at x = 4, since options (c) and (d) tempt you toward that point. Here u = 4 − 4.5 = −0.5, so u2 = 0.25:

P(4) = 145/24 − (1/6)(0.25) = 145/24 − 1/24 = 144/24 = 6

So P(4) = 6, while the actual maximum P(4.5) ≈ 6.0417 is strictly larger. This confirms that x = 4 is not where the extremum occurs — it's simply a nearby point with a slightly smaller function value, included in the options precisely to test whether you've correctly located the turning point.

As a full cross-check, you can verify P(4) using the original Lagrange basis polynomials directly in x (without the u-substitution): L0(4) = −4/7, L1(4) = 6/5, L2(4) = 4/5, L3(4) = −3/7, giving P(4) = 4(−4/7) + 5(6/5) + 5(4/5) + 4(−3/7) = −16/7 + 6 + 4 − 12/7 = 10 − 4 = 6, which matches exactly.

Final Answer

f(x) is maximum when x = 4.5 — this is option (a), which matches the official ISS 2016 Statistics Paper-1 answer key.

Why This Question Matters

On the surface this looks like a routine "plug into Lagrange's formula" question, but it's really testing whether you can recognize structure in data before committing to brute-force computation. In a two-hour, 80-question paper, the difference between solving this in under two minutes (via symmetry) versus eight or nine minutes (via full cubic expansion and calculus) can meaningfully affect how much time you have left for the rest of the paper. Questions like this reward candidates who have drilled pattern recognition in numerical methods — symmetric nodes, equally spaced arguments, repeated differences — until spotting them becomes close to automatic, which is exactly the kind of habit that disciplined, repeated practice on past papers builds over time.

Frequently Asked Questions

Why does symmetric data guarantee an even polynomial?

Because the interpolating polynomial through a fixed set of points is unique. If the data is unchanged when you replace u with −u, then the polynomial you get by substituting −u into P must still pass through all the same points, and since there's only one polynomial of that degree that can do so, P(u) and P(−u) have to be identical. A polynomial that equals its own mirror image can only have even-power terms.

What if the data points aren't exactly symmetric about a midpoint?

Then this shortcut doesn't apply, and you'd need to construct the Lagrange polynomial in full using all four basis functions L0 through L3, then differentiate the resulting cubic with respect to x, set the derivative to zero, and check the second derivative to classify the critical point as a maximum or minimum.

Could this interpolating polynomial have more than one maximum?

No. Once we established that P(u) = A + Bu2 with B negative, it's a simple parabola in u, and a parabola has exactly one turning point. The reduction to just two terms (instead of a general cubic with up to two turning points) is itself a direct consequence of the symmetry.

Is Lagrange's formula the only way to build this interpolating polynomial?

No — Newton's divided difference formula or Newton's forward/backward difference formulas would give the same unique polynomial, just arrived at through a different computational route. The question specifically asks for Lagrange's formula, but the underlying polynomial, and therefore the answer, doesn't change with the method used.

Why is there a penalty structure mentioned in the paper's instructions?

ISS Paper-1 and similar objective papers use negative marking (typically one-third of the marks for that question) to discourage random guessing, which is exactly why a two-minute symmetry-based solution is so valuable here — it leaves you confident enough in the answer to mark it rather than guess among four close-looking options.

How can I get faster at recognizing symmetry tricks like this under exam time pressure?

The most reliable way is repetition: work through enough numerical analysis PYQs that checking "are the x-values equally spaced from a common midpoint, and do the matching f-values repeat?" becomes the very first thing you do before writing any formula, rather than an afterthought.

If anything in this derivation felt unclear, or you solved it a different way, drop a comment below — working through the disagreement is often the fastest way to actually lock in the concept. And if you know a fellow ISS aspirant grinding through the same Paper-1 PYQs, sharing this one with them costs you nothing and might save them several minutes on exam day.