Question

Let $f\left( x \right)$ be a polynomial function of second degree. If $f\left( 1 \right) = f\left( { - 1} \right)$ and a, b, c are in AP, then $f'\left( a \right),f'\left( b \right),f'\left( c \right)$ are in
A) $GP$
B) $HP$
C) $AGP$
D) $AP$

Answer 1

Hint: If a, b, c are in AP, then $2b = a + c$. Calculate the values of $f'\left( a \right),f'\left( b \right),f'\left( c \right)$ and check whether they satisfy the condition of GP, HP, AGP or AP.

Complete step-by-step answer:
Let the polynomial be $f\left( x \right) = p{x^2} + qx + r$
Given that, $f\left( 1 \right) = f\left( { - 1} \right)$
Substituting $1$ and $ - 1$ in $f\left( x \right)$ in place of $x$ and equating them,
$
  p{\left( 1 \right)^2} + q\left( 1 \right) + r = p{\left( { - 1} \right)^2} + q\left( { - 1} \right) + r \\
  p + q + r = p - q + r \\
  2q = 0 \\
  q = 0 \\
 $
Therefore, on substituting $q = 0$ in $f\left( x \right)$ ;
$f\left( x \right) = p{x^2} + r$
To need to be able to calculate $f'\left( a \right)$ we need to find $f'\left( x \right)$ , that is
$\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}}$
$f'\left( x \right) = 2px$ (1)
($\because \dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$ and $d\left( {{\text{constant}}} \right) = 0$)
We are given that a, b and c are in AP.
That is,
$2b = a + c$ (2)
To obtain the values of $f'\left( a \right){\text{ , }}f'\left( b \right){\text{ , }}f'\left( c \right)$ , replace $x$ by $a{\text{ , }}b{\text{ and }}c$ respectively.
$
  f'\left( a \right) = 2pa \\
  f'\left( b \right) = 2pb \\
  f'\left( c \right) = 2pc \\
 $ (3)
Here we can see that;
$
  f'\left( a \right) + f'\left( c \right) = 2pa + 2pc \\
   = 2p\left( {a + c} \right) \\
   = 2p\left( {2b} \right) \\
 $
(using (2))
$
   = 4pb \\
   = 2\left( {2pb} \right) \\
   = 2\left( {f'\left( b \right)} \right) \\
 $
(using (3))
Hence, $f'\left( a \right) + f'\left( c \right) = 2f'\left( b \right)$
Hence, this confirms that $f'\left( a \right),f'\left( b \right),f'\left( c \right)$ are in AP.

Option D is the correct answer.

Note: The condition of GP, that is, when a, b, c are in GP, they must satisfy the relation, ${b^2} = ac$. And if a , b, c are in HP, they must satisfy $\dfrac{1}{a} + \dfrac{1}{c} = \dfrac{2}{b}$. Here it should be noted that a, b, c being in HP means the reciprocal of these numbers are in AP: $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$are in AP.
Similarly if $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$are in AP, then a,b,c are in HP.
AGP is the most complex progression, here the numbers are in AP as well as GP at the same time. Here the numbers are : $a,\left( {a + d} \right)r,\left( {a + 2d} \right){r^2}$------. d is the common difference and r is the common ratio.