If - - are the roots of equation a x 2 -bx-c-0- then log e a -2log e x - 1 x - - - 1 2 x 2 - 2 - 2 - 1 3 x 2 - 3 - 3 - is equal to

The correct option is C log _ e ( a x ^ 2 +bx+c ) Since, α ,β are the roots of a x ^ 2 +bx+c=0 . #160;Also, we need a series in the descendin ...

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Standard XII

Mathematics

Algebra of Limits

If α ,β are...

Question

If $α, β$ are the roots of equation $a x^{2} + b x + c = 0$ , then ${log}_{e} a + 2 {log}_{e} x - \frac{1}{x} (α + β) - \frac{1}{2 x^{2}} (α^{2} + β^{2}) - \frac{1}{3 x^{2}} (α^{3} + β^{3}) . . . .$ is equal to

{log}_{e} a

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{log}_{e} β

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{log}_{e} (a x^{2} + b x + c)

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{log}_{e} \frac{β}{α}

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Solution

The correct option is C ${log}_{e} (a x^{2} + b x + c)$
Since, $α, β$ are the roots of $a x^{2} + b x + c = 0$ .
Also, we need a series in the descending powers of $x$
Therefore, $a x^{2} + b x + c = a (x - α) (x - β) = a x^{2} (1 - \frac{α}{x}) (1 - \frac{β}{x})$
Let $E = {log}_{e} ({a x}^{2} + b x + c)$
$\Rightarrow E = {log}_{e} a + 2 {log}_{e} x + {log}_{e} (1 - \frac{α}{x}) + {log}_{e} (1 - \frac{β}{x})$
$\Rightarrow E = {log}_{e} a + 2 {log}_{e} x - (\frac{α}{x} + \frac{1}{2} . \frac{α^{2}}{x^{2}} + \frac{1}{3} . \frac{α^{3}}{x^{3}}) - (\frac{β}{x} + \frac{1}{2} . \frac{β^{2}}{x^{2}} + \frac{1}{3} . \frac{β^{3}}{x^{3}})$
$\Rightarrow E = {log}_{e} a + 2 {log}_{e} x - \frac{1}{x} (α + β) - \frac{1}{2 x^{2}} (α^{2} + β^{2}) - \frac{1}{3 x^{3}} (α^{3} + β^{3})$
$∴$ Ans is Opt: $[C]$

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If $α, β$ are the roots of equation $a x^{2} + b x + c = 0$ , then ${log}_{e} a + 2 {log}_{e} x - \frac{1}{x} (α + β) - \frac{1}{2 x^{2}} (α^{2} + β^{2}) - \frac{1}{3 x^{2}} (α^{3} + β^{3}) . . . .$ is equal to