Question

If \(f(x^{2}-1)=x^{4-7x^{2}+k_{1}\) and \(f(x^{3}-2)=x^{6}-9x^{3}+k_{2}\) then the value of \((k_{2}-k_{1})\) is}

• A. 6
• B. 7
• C. 8
• D. 9
• E. None of the above

Hint:

When \(f(\cdot)\) is evaluated at linear or polynomial shifts like \(x^{2}-1\), \(x^{3}-2\), try a low-degree ansatz for \(f\) and match coefficients.

Answer 1

The Correct Option is C

Step 1: Assume a quadratic form for \(f\).
Since the outputs are quartic and sextic with only even/odd middle terms, try \(f(u)=au^{2}+bu+c\).
Then \[ f(x^{2}-1)=a(x^{2}-1)^{2}+b(x^{2}-1)+c = ax^{4}+(-2a+b)x^{2}+(a-b+c). \] Match with \(x^{4}-7x^{2}+k_{1}\) \(\Rightarrow\) \[ a=1,\quad -2a+b=-7 \Rightarrow b=-5,\quad a-b+c=k_{1}\Rightarrow c=k_{1}-6. \] Step 2: Use the second condition.
\[ f(x^{3}-2)=a(x^{3}-2)^{2}+b(x^{3}-2)+c = ax^{6}+(-4a+b)x^{3}+(4a-2b+c). \] With \(a=1,\ b=-5,\ c=k_{1}-6\), \[ x^{6}\text{-coeff}=1\ \checkmark,\quad x^{3}\text{-coeff}=-4-5=-9\ \checkmark, \] \[ \text{constant}=4-2(-5)+(k_{1}-6)=14+k_{1}-6=8+k_{1}. \] Hence \(k_{2}=k_{1}+8 \Rightarrow k_{2}-k_{1}=\boxed{8}\).