Question 17
SSC-CGL 2018 June 4 Shift 1
If $\displaystyle \frac{5\sqrt 5 x^3 - 81\sqrt 3 y^3}{\sqrt 5 x - 3\sqrt 3 y} = $ $\displaystyle Ax^2 + By^2 +Cxy, $ then $\displaystyle 6A + B - \sqrt {15}C = ?$
- 12
- 15
- 10
- 9
Solution 1
Factorize the numerator as
$\displaystyle (\sqrt 5 x - 3\sqrt 3y) (5x^2 + 3\sqrt {15}xy + 27y^2)$
So, $\displaystyle (5x^2 + 3\sqrt {15}xy + 27y^2) = Ax^2 + By^2 + Cxy$
Equating the coefficients, $\displaystyle A = 5, B = 27, C = 3\sqrt 15$
Putting in $\displaystyle 6A + B - \sqrt {15}C$, we get 12 Ans.
Solution 2
To determine A, B and C first put x= 1 and y = 0 in the given equation. By this choice, B and C will disappear on the RHS, and we shall get A as
$ \begin{aligned} & = \frac{5\sqrt 5 \times 1^3 - 81\sqrt 3 \times 0^3}{\sqrt 5 \times 1 - 3\sqrt 3 \times 0} \\ & = 5\\ \end{aligned} $
Similarly, putting x = 0, and y = 1, we obtain B = 27
Mentally take the denominator to the RHS and calculate the coefficient of, say, $\displaystyle x^2y$ as $\displaystyle -3\sqrt3 A + \sqrt 5C$. This has to be zero because there is no $\displaystyle x^2y$ on the LHS. We get
$ \begin{aligned} -3 \sqrt 3 \times 5 + \sqrt 5 C &= 0\\ \implies C &= 3 \sqrt {15} \end{aligned} $
Putting in $\displaystyle 6A + B - \sqrt {15}C$, we get 12 Ans.
Solution 3
We can totally avoid the calculation for C. It requires a bit of imagination to start with $\displaystyle x = \sqrt 3, y = -\sqrt 5$ to get
$ \begin{aligned} &A (\sqrt 3)^2 + B (-\sqrt 5)^2 + C\sqrt 3 \times (-\sqrt 5) \\ & = \frac{5\sqrt 5 \times (\sqrt 3)^3 - 81\sqrt 3 \times (-\sqrt 5)^3}{\sqrt 5 \times \sqrt 3 - 3\sqrt 3 \times (-\sqrt 5)} \\ & = \frac{15\sqrt {15} + 405 \sqrt {15}}{4\sqrt 15}\\ & = 105\\ &\text{i.e., } 3A + 5B - \sqrt {15}C = 105\\ \end{aligned} $
$\displaystyle\text{add } 3A - 4B \text{ i.e., } -93 \text{ to both sides }$
$ \begin{aligned} 6A + B - \sqrt {15}C &= 105 + (-93)\\ & = 12\:\underline{Ans}\\ \end{aligned} $
This Blog Post/Article "(solved)Question 17 SSC-CGL 2018 June 4 Shift 1" by Parveen is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.