JEE Main 2022 · 27 Jun · Shift 2 · Mathematics Q2

← JEE Main 2022 · 27 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 2 of 90

Let $f(x)=\begin{vmatrix}a&-1&0\\ax&a&-1\\ax^2&ax&a\end{vmatrix}$ , $a\in\mathbb R$ . Then the sum of the squares of all the values of $a$ , for which $2f'(10)-f'(5)+100=0$ , is

117

106

125

136

Answer and worked solution

Official answer: C

Expand. $f(x)=a(a^2+ax)+(a^2x+ax^2)=a^3+2a^2x+ax^2$ , so $f'(x)=2a^2+2ax$ .

Condition. $2(2a^2+20a)-(2a^2+10a)+100=2a^2+30a+100=0$ , i.e. $a^2+15a+50=0$ : $a=-5,-10$ .

Sum of squares. $25+100=125$ .

(C)

Topics:Determinants and properties Derivatives of standard functions

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JEE Main 2022 · 27 Jun · Shift 2, Mathematics question 2