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Derivative from First Principle

R1 and R2 a...

Question

$R_{1}$ and $R_{2}$ are the reminders when the polynomial $a x^{3} + 3 x^{2} - 3$ and $2 x^{3} - 5 x + 2 a$ are divided by (x-4) respectively. If $2 R_{1} - R_{2} = 0$ , then find the value of a

a = \frac{1}{6}

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a = \frac{1}{3}

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a = \frac{1}{7}

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a = \frac{1}{5}

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Solution

The correct option is C $a = \frac{1}{7}$
$a x^{3} - 5 x - 2 a$
Put $x = 4$
Then, $a (4)^{3} - 5 (4) - 3 = 64 a + 48 - 3 = 64 a + 45 \to R_{1}$
$2 x^{3} - 5 x + 2 a$
Put $x = 4$
Then, $2 (4)^{3} - 5 (4) + 2 a = 128 - 20 + 2 a = 2 a + 108 \to R_{2}$
Then, $2 R 1 - R 2 = 0$
$2 (64 a + 45) - (2 a + 108) = 0$
Or $128 a + 90 - 2 a - 108 = 0$
Or $126 a = 18$
Or $a = \frac{1}{7}$

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