Let $f(x)=\left|\begin{array}{ccc}a & -1 & 0 \\ a x & a & -1 \\ a x^{2} & a x & a\end{array}\right|, a \in R$. Then the sum of the squares of all the values of $a$ for which $2 f^{\prime}(10)-f^{\prime}(5)+100=0$ is :

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Determinants

Let

f (x) = a a x a x^{2} - 1 a a x 0 - 1 a, a \in R

. Then the sum of the squares of all the values of

a

for which

2 f^{'} (10) - f^{'} (5) + 100 = 0

is :

Answer: (C)

Expanding the determinant:

f (x) = a 1 x x^{2} - 1 a a x 0 - 1 a = a [a^{2} + a x + a x + x^{2}] = a (x + a)^{2}

.

Differentiating:

f^{'} (x) = 2 a (x + a)

.

Substituting into the given condition:

2 f^{'} (10) - f^{'} (5) + 100 = 0

4 a (10 + a) - 2 a (5 + a) + 100 = 0

40 a + 4 a^{2} - 10 a - 2 a^{2} + 100 = 0

2 a^{2} + 30 a + 100 = 0

a^{2} + 15 a + 50 = 0

(a + 10) (a + 5) = 0

a = - 10

a = - 5

. The sum of squares is

(- 10)^{2} + (- 5)^{2} = 100 + 25 = 125

.

Answer: (C)

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Single correct option