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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 A.P., then ${f}'\left ( a \right ),{f}'\left ( b \right ),{f}'\left ( c \right )$ are in

Option 1)
G.P.

Option 2)
H.P.

Option 3)
ArithmeticGeometric Progression

Option 4)
A.P.

Answers (1)

As we learnt in

Properties of an A.P. -

If each term of an AP is multiplied by a fixed constant (or divided by a constant),then resultant is also an AP.

- wherein

If $a_{1},a_{2},a_{3}------is \: AP$

Then $Ka_{1},Ka_{2},Ka_{3}------is \: AP$

and $a_{1}/K,a_{2}/K,a_{3}/K------is \: also \: an \: AP$

Let the polynomial be

$f\left ( x \right )= Ax^{2}+Bx+C$

$f\left ( 1 \right )= A+B+C$

$f\left ( -1 \right )= A-B+C$

$\therefore A+B+C= A-B+C$

$B=0$

$f\left ( x \right )= Ax^{2}+c$

$f'\left ( x \right )= 2Ax$

$f'\left ( a \right )= 2Aa$

$f'\left ( b \right )= 2Ab$

$f'\left ( c \right )= 2Ac$

Now, a, b, c in AP

So 2Aa, 2Ab, 2Ac in AP

So $f'\left ( a \right ), f'\left ( b \right ), f'\left ( c \right )$ in AP

Option 1)

G.P.

This option is incorrect.

Option 2)

H.P.

This option is incorrect.

Option 3)

ArithmeticGeometric Progression

This option is incorrect.

Option 4)

A.P.

This option is correct.

Posted by

Sabhrant Ambastha

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