Question 20

If $$x^3 - 4x^2 + 19 = 6(x - 1)$$, then what is the value of $$\left[x^2 + \left(\frac{1}{x - 4}\right)\right]$$?

Solution

Given that :

$$x^3 - 4x^2 + 19 = 6(x - 1)$$

or, $$x^3-4x^2+1=\left(6x-6-18\ \right).$$

So,

$$\left[x^2+\left(\frac{1}{x-4}\right)\right]=\frac{x^3-4x^2+1}{\left(x-4\right)}=\frac{6x-6-18}{\left(x-4\right)}=\frac{6\left(x-4\right)}{\left(x-4\right)}=6.$$

C is correct choice.

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SSC CGL Tier-2 21-February-2018 Maths

If $$x^3 - 4x^2 + 19 = 6(x - 1)$$, then what is the value of $$\left[x^2 + \left(\frac{1}{x - 4}\right)\right]$$?

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