Limits, Continuity & Differentiability
Methods of Differentiation
Grade 12
Question:
<p>Let $f(x) = \begin{pmatrix} a & -1 & 0 \\ ax & a & -1 \\ ax^2 & ax & a \end{pmatrix}$, $a\in\mathbb{R}$. Then the sum of the squares of all the values of $a$ for which $2f'(10)-f'(5)+100=0$ is:</p>
<p>$117$</p>
<p>$106$</p>
<p>$125$</p>
<p>$136$</p>
Step-by-Step Solution
Key Concept: General
<b>Derivative of Determinant Function</b><br>
Expand the $3\times3$ determinant:<br>
$f(x)=a(a^2-(-1)(ax))-(-1)(ax\cdot a-(-1)\cdot ax^2)+0$<br>
$=a(a^2+ax)+(a^2x+ax^2) = a^3+a^2x+a^2x+ax^2 = a^3+2a^2x+ax^2$.<br>
$f'(x)=2a^2+2ax$, so $f'(10)=2a^2+20a$ and $f'(5)=2a^2+10a$.<br>
Equation: $2(2a^2+20a)-(2a^2+10a)+100=0$<br>
$4a^2+40a-2a^2-10a+100=0$<br>
$2a^2+30a+100=0$<br>
$a^2+15a+50=0$<br>
$(a+5)(a+10)=0\Rightarrow a=-5$ or $a=-10$.<br>
Sum of squares $=25+100=125$. <b>Answer: 3 ($=125$)</b><br>
<b>Key concept:</b> Expand the determinant first (it's a polynomial in $x$), then differentiate that polynomial.<br>
<b>Trap:</b> Using the row-differentiation formula for determinants — direct expansion is much simpler here.
Correct Answer: 3