WBJEE · Maths · Quadratic Equation

Let \(f(x)=2 x^{2}+5 x+1\). If we write \(f(x)\) as \(f(x)=a(x+1)(x-2)+b(x-2)(x-1)+c(x-1)(x+1)\)
for real numbers \(a, b, c,\) then

A there are infinite number of choices for \(a, b, c\)
B only one choice for a but infinite number of choices for \(b\) and \(c\)
C exactly one choice for each of \(a, b, c\)
D more than one but finite number of choices for \(a, b, c\)

Verified Solution

Answer & Solution

Correct Answer

(C) exactly one choice for each of \(a, b, c\)

Step-by-step Solution

Detailed explanation

\begin{array}{l} \text { Given, } f(x)=2 x^{2}+5 x+1 \\ \text { Also, } f(x)=a(x+1)(x-2)+b(x-2)(x-1) \\ +c(x-1)(x+1) \\ \qquad \begin{array}{r} =a\left(x^{2}-x-2\right)+b\left(x^{2}-3 x+2\right) \\ +c\left(x^{2}-1\right) \\ f(x)=(a+b+c) x^{2}+(-a-3 b) x \\ +(-2 a+2 b-c)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Let \(f(x)=2 x^{2}+5 x+1\). If we write \(f(x)\) as \(f(x)=a(x+1)(x-2)+b(x-2)(x-1)+c(x-1)(x+1)\) for real numbers \(a, b, c,\) then

Step-by-step Solution

Let \(f(x)=2 x^{2}+5 x+1\). If we write \(f(x)\) as \(f(x)=a(x+1)(x-2)+b(x-2)(x-1)+c(x-1)(x+1)\)
for real numbers \(a, b, c,\) then