Question

Let $f(x)$ be a second degree polynomial. If $f(1) = f(-1)$ and $p, q, r$ are in A.P., then $f'(p), f'(q), f'(r)$ are

• A. in A.P.
• B. in G.P.
• C. in H.P.
• D. neither in A.P. or G.P. or H.P.

Hint:

Whenever you are given conditions like $f(1) = f(-1)$ for a polynomial, check if it forces symmetry (like eliminating the $x$ term). For A.P. checks, remember the middle term must be the average of the two outer terms!

Answer 1

The Correct Option is A

Step 1: Define the general form of \(f(x)\)

Since \(f(x)\) is a second-degree polynomial, we express it as:

f(x) = ax^2 + bx + c

where \(a\), \(b\), and \(c\) are constants.

Step 2: Utilize the condition \(f(1) = f(-1)\)

Calculate \(f(1)\) and \(f(-1)\):

f(1) = a(1)^2 + b(1) + c = a + b + c

f(-1) = a(-1)^2 + b(-1) + c = a - b + c

Given \(f(1) = f(-1)\), equate the expressions:

a + b + c = a - b + c

2b = 0

b = 0

Therefore, the polynomial simplifies to:

f(x) = ax^2 + c

Step 3: Determine the derivative \(f'(x)\)

Differentiate \(f(x)\) with respect to \(x\):

f'(x) = d/dx (ax^2 + c) = 2ax

Thus,

f'(x) = 2ax

Step 4: Analyze \(f'(p)\), \(f'(q)\), \(f'(r)\)

Given that \(p\), \(q\), and \(r\) are in A.P., then:

2q = p + r

Now:

f'(p) = 2ap, f'(q) = 2aq, f'(r) = 2ar

Since \(p\), \(q\), \(r\) are in A.P., \(q\) is the average of \(p\) and \(r\):

q = (p + r)/2

Thus:

2aq = 2a((p + r)/2) = a(p + r)

Now check:

f'(q) = (f'(p) + f'(r))/2

Substituting:

f'(q) = (2ap + 2ar)/2 = a(p + r)

which matches with the earlier expression for \(f'(q)\).

Step 5: Conclusion

Since \(f'(q)\) is the average of \(f'(p)\) and \(f'(r)\), \(f'(p)\), \(f'(q)\), and \(f'(r)\) are in Arithmetic Progression (A.P.).