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TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Section: Mathematics

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Question : 13 of 160

Marks: +1, -0

Let

\alpha,\beta,\gamma(\alpha < \beta < \gamma)

ax^3+bx^2+cx+d

0

u,v,w(u < v < w)

ak^3x^3+bk^2x^2+ckx+d

0

\beta^2

\alpha\gamma

, then

$v$ = $\frac{2vw}{u+w}$
$2v$ = $u+w$
$v^2$ = $uw$
$v^2$ = $2uw$

Solution:

\because \alpha, \beta

and

\gamma

are the roots of

a x^{3}+b x^{2}+c x+d=0.

and given

\beta^{2}=\alpha \gamma

i.e.

\alpha, \beta

and

\gamma

are in G.P.

So,

\; \frac{\alpha}{K}, \; \frac{\beta}{K}

and

\; \frac{\gamma}{K}

are also in G.P.

Now,

u, v

and

w

are roots of

a k^{3} x^{3}+b k^{2} x^{2}+c k x+d \;\; =0

\;\text{ i.e. }\; \;\; a(k x)^{3}+b(k x)^{2}+c(k x)+d \;\; =0 \;\; . . .(i)

So, roots of Eq. (i) are

\; \frac{\alpha}{K}, \; \frac{\beta}{K}

and

\; \frac{\gamma}{K}

i.e.

u=\; \frac{\alpha}{K}

,

v=\; \frac{\beta}{K}

and

w=\; \frac{\gamma}{K}

From above,

u, v

and

w

are also in G.P.

Hence,

v^{2}=u w

.

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