Given- x-20 is a root of the equation lnx2-20 a2-ln a-4-ln9 a x-a-4- Then the sum of all distinct roots of the equation isA- 36B- 41C- 45D- 61

The correct option is C 45 nbsp; nbsp;ln(x^2+20a^2) ln(a 4)=ln(9axa 4) ⇒ln(x^2+20a^2a 4)=ln(9axa 4) ⇒x^2+20a^2a 4=9axa 4, a 4 ⋯(1) ⇒ x^2+20a^2=9ax ⇒ ( ...

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Standard XII

Mathematics

Descartes' Rule

Given, x=20 i...

Question

Given, $x = 20$ is a root of the equation $ln (x^{2} + 20 a^{2}) - ln (a - 4) = ln (\frac{9 a x}{a - 4}) .$ Then the sum of all distinct roots of the equation is

36

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41

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45

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61

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Solution

The correct option is C $45$
$ln (x^{2} + 20 a^{2}) - ln (a - 4) = ln (\frac{9 a x}{a - 4})$
$\Rightarrow ln (\frac{x^{2} + 20 a^{2}}{a - 4}) = ln (\frac{9 a x}{a - 4})$
$\Rightarrow \frac{x^{2} + 20 a^{2}}{a - 4} = \frac{9 a x}{a - 4}, a \neq 4 \dots (1)$
$\Rightarrow x^{2} + 20 a^{2} = 9 a x$
$\Rightarrow (x - 5 a) (x - 4 a) = 0$
$\Rightarrow x = 5 a or x = 4 a$

Now, if $5 a = 20 \Rightarrow a = 4$
which is not possible from eqn $(1)$
If $4 a = 20 \Rightarrow a = 5$
Therefore, other root is, $5 a = 25$

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Given, x=20 is a root of the equation ln(x2+20a2)−ln(a−4)=ln(9axa−4). Then the sum of all distinct roots of the equation is

The correct option is C 45 ln(x2+20a2)−ln(a−4)=ln(9axa−4) ⇒ln(x2+20a2a−4)=ln(9axa−4) ⇒x2+20a2a−4=9axa−4, a≠4 ⋯(1) ⇒x2+20a2=9ax ⇒(x−5a)(x−4a)=0 ⇒x=5a or x=4a Now, if 5a=20⇒a=4 which is not possible from eqn(1) If 4a=20⇒a=5 Therefore, other root is, 5a=25

Given, $x = 20$ is a root of the equation $ln (x^{2} + 20 a^{2}) - ln (a - 4) = ln (\frac{9 a x}{a - 4}) .$ Then the sum of all distinct roots of the equation is