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Bunuel
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When solving case one, why are we ruling out option B? It's also greater than 7.
chetan2u
Bunuel
If \( x = \sqrt{2} - 1\), then the value of \(x^4+4x^3+6x^2+4x+7\) is


(A) \((\sqrt{2} -1)^4\)

(B) \(4 + 4\sqrt{2}\)

(C) 6

(D) 10

(E) 12


Are You Up For the Challenge: 700 Level Questions


Two ways I can think of immediately are

(I) If the choices are so wide and I am in need of saving time, I will go for approximation.

Now \(x=\sqrt{2}-1=1.4-1=0.4\)

\(x^4+4x^3+6x^2+4x+7\)
As x is positive, the answer is surely greater than 7. Only D and E are left.
\(x^4+4x^3\) will make too less an effect on overall sum.
\(6x^2+4x+7=6*(0.4)^2+4*0.4+7= ~1+1.5+7=~9.5\)
So D is the answer.


(I) If the choices were close, I will analyse the terms.

Now, \(x=\sqrt{2}-1\) or \(x+1=\sqrt{2}\), so let us see if we can get the expression in same format.
\(x^4+4x^3+6x^2+4x+7\)
\(x^4+x^3+3x^3+6x^2+4x+7\)
\(x^3(x+1)+3x^3+3x^2+3x^2+4x+7\)
\(x^3(x+1)+3x^2(x+1)+3x^2+3x+x+7\)
\(x^3(x+1)+3x^2(x+1)+3x(x+1)+(x+1)+6\)
\((x^3+3x^2+3x+1)(x+1)+6\)
\((x(x^2+2x^2+1)+x^2+2x+1)(x+1)+6\)
\((x(x+1)^2+(x+1)^2)(x+1)+6\)
\((x+1)^2(x+1)(x+1)+6=(x+1)^4+6=(\sqrt{2})^4+6=4+6=10\)


D
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