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.

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