Let f(x) = ax2 + bx + c, where a, b, cR andf 2(1) + f 2(2) + f 2(3) – 2 f (1) – 4 f (2) – 6 f (3) + 14 = 0 and If g(x) = then find g' (1) + g (1).

Engineering

Mathematics

Derivative of Functions Expressed in The Determinant From

Question

Let f(x) = ax² + bx + c, where a, b, c $\in$ R and
f ²(1) + f ²(2) + f ²(3) – 2 f (1) – 4 f (2) – 6 f (3) + 14 = 0 and If g(x) = $| \begin{matrix} f (x) & f (x + 1) & f (x + 2) \\ f' (x) & 2 f' (x + 1) & 3 f' (x + 2) \\ f^{2} (x) & f^{2} (x + 1) & f^{2} (x + 2) \end{matrix} |$ then find g' (1) + g (1).

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$∵$ (f (1) – 1)² + (f(2) – 2)² + (f (3) – 3)² = 0
⇒ f (1) = 1 , f (2) = 2 & f (3) = 3
$∴$ f (x) = x ⇒ ax² + (b – 1) x + c = 0 is satisfied by x = 1 , 2 , 3
$∴$ a = 0, b = 1 , c = 0
$∴$ f (x) = x
$∴$ g (x) = $| \begin{matrix} x & (x + 1) & (x + 2) \\ 1 & 2 & 3 \\ x^{2} & {(x + 1)}^{2} & {(x + 2)}^{2} \end{matrix} |$
Applying c₂ → c₂ – c₁ & c₃ → c₃ – c₂, we get
g (x) = $| \begin{matrix} x & 1 & 1 \\ 1 & 1 & 1 \\ x^{2} & 2 x + 1 & 2 x + 3 \end{matrix} |$
Again, c₃ → c₃ –c₂ , we get
g (x) = $| \begin{matrix} x & 1 & 0 \\ 1 & 1 & 0 \\ x^{2} & 2 x + 1 & 2 \end{matrix} |$ = 2 (x – 1)
$∴$ g'(x) = 2
$∴$ g' (1) + g(1) = 2