You visited us 1 times! Enjoying our articles? Unlock Full Access!

Determinant

Let fx be a p...

Question

Let $f (x)$ be a polynomial of the second degree. If $f (1) = f (- 1)$ and $a_{1} : a_{2} : a_{3}$ are in AP then $f' (a_{1}) : f' (a_{2}) : f' (a_{3})$ are in

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
AP

Explanation for the correct option:
let $f (x) = a x^{2} + b x + c$
$f (1) = a + b + c f (- 1) = a - b + c$
$a + b + c = a - b + c \Rightarrow b = 0$
$f (x) = a x^{2} + c$
$f' (x) = 2 a x$
$f' (a_{1}) = 2 a a_{1} f' (a_{2}) = 2 a a_{2} f' (a_{3}) = 2 a a_{3}$
They are in AP since $a_{1}, a_{2}, a_{3}$ are in AP
Hence, option(A) is the correct answer

Suggest Corrections

Similar questions

a_{1}, a_{2}, a_{3}

a_{p}, a_{q}, a_{r}

\frac{a_{q}}{a_{p}} =

a_{1}, a_{2}, a_{3}, a_{4}

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{1}{\sqrt{a_{3}} + \sqrt{a_{4}}}

a_{1}

a_{2}

a_{3}

a_{2}

a_{3}

a_{4}

a_{3}

a_{4}

a_{5}

a_{1}

a_{3}

a_{5}

a_{1}, a_{2}, a_{3}, . . . .

a_{p}, a_{q}, a_{r}

a_{q} : a_{p}

a_{1}

a_{2}

a_{3}

a_{p}

a_{q}

a_{r}

a_{q} : a_{p}

Join BYJU'S Learning Program