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The sum of the values of a for which $$\frac{x^3-6x^2+11x-6}{x^3+x^2-10x+8} + \frac a{30} = 0$$ does not have a real solution is

A. $1$

B. $12$

C. $13$

D. $2$

I tried to factorise the numerator and the denominator and got $$\frac{(x-1)(x-2)(x-3)}{(x-1)(x-2)(x+4)}$$ Then $f(x) = \frac{x-3}{x+4}$, but after this how to get the range of $f(x)$?

Any help will be appreciated ....thanks.

Sayan Dutta
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Neetu Yadav
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1 Answers1

0

Take $$y=f(x) = \frac{x-3}{x+4}$$ Then, $$\frac{y+1}{y-1}=\frac{2x+1}{-7}$$ or $$x=\frac 1 2 \left(\frac{7y+7}{1-y}-1\right)$$ So, the range is $\mathbb R \backslash \{1\}$ because $x$ is defined iff $y\neq 1$.

Now, check that $y=-\frac{12}{30}$ gives $x=1$, for which the original fraction takes $\frac 0 0$ form. So, $a\neq 12$. Also, we can see that $a\neq -30$ because $y\neq 1$. Again, $a\neq 5$ because $y\neq -\frac 1 6$ (as that would imply $x=2$ which is not possible). Summing these up, we aren't getting $13$, we are getting $-13$. So, I think the option has a sign flip.

Does that help?

Sayan Dutta
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  • Answer is not 12 it is 13 – Neetu Yadav Jul 20 '21 at 06:43
  • you have not find out the range values – Neetu Yadav Jul 20 '21 at 06:44
  • @NeetuYadav You don't need the range of $f(x)$. You only need the value of $a$ for which $\frac{x^3-6x^2+11x-6}{x^3+x^2-10x+8} + \frac a{30} = 0$ does not have a real solution. And, if the answer given is C, then it's wrong. You can check that $x=\frac{38}{43}$ is a solution of $\frac{x^3-6x^2+11x-6}{x^3+x^2-10x+8} + \frac {13}{30} = 0$. Also, I proved B in my solution above. – Sayan Dutta Jul 20 '21 at 07:14
  • $a=12$ appears to work, yes. But the problem statement asks for "the sum of the values of $a$ for which...". Is $a=12$ the only one possibility? – user3733558 Jul 20 '21 at 08:43
  • @SayanDutta it is not asking value of a it is asking sum of values of a and it is 13 – Neetu Yadav Jul 20 '21 at 09:09
  • @user3733558 ur right ...it is asking sum of values of a......and it is 13 – Neetu Yadav Jul 20 '21 at 09:09
  • @SayanDutta you will get 3 three different of value of a and then you have to add three value and that is a result – Neetu Yadav Jul 20 '21 at 09:44
  • @user3733558 Oh! It was sum? I didn't notice that. Just a moment. – Sayan Dutta Jul 20 '21 at 09:53
  • @NeetuYadav The range as you can see is $\mathbb R \backslash {1}$ because look at the expression of $x$ as a function of $y$ and note that $x$ is defined iff $y\neq 1$. – Sayan Dutta Jul 20 '21 at 09:55
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    @NeetuYadav I can see that $a\neq -30$ because $y\neq 1$. Also, $a\neq 5$ because $y\neq -\frac 1 6$ (as that would imply $x=2$ which is not possible). Still, summing these up, you aren't getting $13$, you are getting $-13$. So, I think the option has a sign flip. – Sayan Dutta Jul 20 '21 at 09:59
  • I did the edits. Thanks for pointing the mistake out. Please check whether it's okay now. – Sayan Dutta Jul 20 '21 at 10:02
  • @SayanDutta can u tell me how 1 is excluded from range? range = R- {1,-2,5 ,-1/6} – Neetu Yadav Jul 20 '21 at 10:04
  • @SayanDutta why 1 is excluded from Real ? – Neetu Yadav Jul 20 '21 at 10:06
  • @NeetuYadav You have $x=\frac 1 2 \left(\frac{7y+7}{1-y}-1\right)$. See that $y=1$ is the only value for which $\frac 1 2 \left(\frac{7y+7}{1-y}-1\right)$ is not defined. So, there is no $x$ for which $y=1$. That is why, ${1}$ is not in the range. Is that clear now? – Sayan Dutta Jul 20 '21 at 10:08
  • @SayanDutta okk....understood – Neetu Yadav Jul 20 '21 at 10:18
  • @SayanDutta how did you got y+1/y-1 = 2x+1/-7 ? – Neetu Yadav Jul 20 '21 at 10:31
  • It's componendo dividendo. Check this out - https://brilliant.org/wiki/componendo-and-dividendo/ – Sayan Dutta Jul 20 '21 at 10:36
  • @SayanDutta okkkk – Neetu Yadav Jul 20 '21 at 10:59