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: $\frac{15}{16}$
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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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Question : If $\cos^2x+\cos^4x=1$, then $\tan^2x+\tan^4x$?
Option 1: $0$
Option 2: $1$
Option 3: $2 \tan^2x$
Option 4: $2\tan^4x$
Question : If $(4 y-\frac{4}{y})=13$, find the value of $(y^2+\frac{1}{y^2})$.
Option 1: $12 \frac{11}{16}$
Option 2: $10 \frac{9}{16}$
Option 3: $12 \frac{9}{16}$
Option 4: $8 \frac{9}{16}$
Question : Which of the following is correct?
Option 1: $\frac{2}{3}< \frac{3}{5}< \frac{11}{15}$
Option 2: $\frac{3}{5}< \frac{2}{3}< \frac{11}{15}$
Option 3: $\frac{11}{15}< \frac{3}{5}< \frac{2}{3}$
Option 4: $\frac{3}{5}< \frac{11}{15}< \frac{2}{3}$
Question : If $2x+\frac{1}{3x}$ = 5, then the value of $\frac{5x}{6x^{2}+20x+1}$ is:
Option 1: $\frac{1}{4}$
Option 2: $\frac{1}{6}$
Option 3: $\frac{1}{5}$
Option 4: $\frac{1}{7}$
Question : Simplify the following expression. 120 ÷ 15 of 4 + [11 × 4 ÷ 4 of {4 × 2 – (8 – 11)}]
Option 1: $\frac{120}{61}$
Option 2: $\frac{61}{120}$
Option 3: $3$
Option 4: $-3$
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