Question : Find the remainder when we divide $3x^4-2x^2+4x-1$ by $(2x-1)$.
Option 1: $2$
Option 2: $3$
Option 3: $\frac{11}{16}$
Option 4: $\frac{15}{16}$
Correct Answer: $\frac{11}{16}$
Solution :
Assume, $2x-1$ = 0
⇒ $x$ = $\frac{1}{2}$
Let $f(x)=3x^4-2x^2+4x-1$
So, the remainder when $3 x^4-2 x^2+4 x-1$ is divided by $(2x-1)$ is $f(\frac{1}{2})$
= $3 (\frac{1}{2})^4-2 (\frac{1}{2})^2+4(\frac{1}{2})-1$
= $\frac{3}{16} - \frac{1}{2} + 2 - 1$
= $\frac{3}{16} + \frac{1}{2}$
= $\frac{11}{16}$
Hence, the correct answer is $\frac{11}{16}$.
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