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: Assume general form of $f(x)$

Since $f(x)$ is a second degree polynomial, we can write:

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

where $a$, $b$, and $c$ are constants.

Step 2: Use the given condition $f(1) = f(-1)$

We 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

Since $f(1) = f(-1)$, equating:

a + b + c = a - b + c

2b = 0

b = 0

Thus, the polynomial simplifies to:

f(x) = ax^2 + c

Step 3: Find the derivative $f'(x)$

Differentiating $f(x)$ 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., thus:

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 $f'(r)$, it means that $f'(p)$, $f'(q)$, and $f'(r)$ are in Arithmetic Progression (A.P.).