(solved)Question 16 SSC-CGL 2018 June 4 Shift 1

If $\displaystyle x^{4} + x^{-4} = 194, x \gt 0$, then what would be the value of $\displaystyle (x - 2)^2 ?$

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Parveen,

Question 16
SSC-CGL 2018 June 4 Shift 1

If $\displaystyle x^{4} + x^{-4} = 194, x \gt 0$, then what would be the value of $\displaystyle (x - 2)^2 ?$

  1. 6
  2. 2
  3. 1
  4. 3

Solution 1

every year one question is always there on this standard form

Standard form: $\displaystyle \bigg(x^{n} + \frac{1}{x^{n}}\bigg)^2$ = $\displaystyle x^{2n} + \frac{1}{x^{2n}} + 2$

So $\displaystyle \bigg(x^2 + \frac{1}{x^2}\bigg)^2 = 194 + 2 = 14^2$

We get $\displaystyle x^2 + \frac{1}{x^2} = 14$, likewise we can obtain $\displaystyle x + \frac{1}{x} = 4$

Which gives $\displaystyle x^2 - 4x = -1 \text{. . . (1)}$

Now $\displaystyle (x - 2)^2 = x^2 -4x + 4 = -1 + 4 = 3$ answer.

Solution 2

If nothing works, then approximations can be tried. But some risk may be there.

Observe that as $\displaystyle x$ becomes larger $\displaystyle x^{-4}$ becomes smaller and smaller.

So let us approximate $\displaystyle x^{4} + x^{-4} \approx x^4 \approx 194$

Since $\displaystyle 3^4 \lt 194 \lt 4^4$, we can say that $\displaystyle x$ is between $\displaystyle 3$ and $\displaystyle 4$

Hence, $\displaystyle (x - 2)^2$ should be between $\displaystyle (3 - 2)^2$ and $\displaystyle (4 - 2)^2$, i.e., between $\displaystyle 1$ and $\displaystyle 4$, which leads to $\displaystyle 3$ as the answer!

Solution 3

Cycle through the options: If $\displaystyle (x - 2)^2 = 6$, then $\displaystyle x = \sqrt 6 + 2 = 2.23 + 2 = 4.45$, but $\displaystyle (4.45)^4 + \frac{1}{(4.45)^4}$ will be too big. So can be rejected.

If $\displaystyle (x - 2)^2 = 6$, then $\displaystyle x = \sqrt 2 + 2 = 1.41 + 2 = 3.41$, but $\displaystyle (3.41)^4 + \frac{1}{(3.41)^4}$ will be too small for 194. So can be rejected.

If $\displaystyle (x - 2)^2 = 1$, then $\displaystyle x = 3$, but $\displaystyle 3^4 + \frac{1}{3^4}$ is even smaller. So can be rejected.

The last option, i.e., 3 can be marked as the answer!

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