MCQ

MCQ 11 Mark

If $x+1$ is a factor of $3 x^3+k x^2+7 x+4$, then the value of $k$ is

A
$-1$
B
$0$
✓
6
D
10

Answer

Correct option: C.

$f(x)=3 x^3+k x^2+7 x+4$
$g(x)=x+1$
$\text { Remainder }=0$
$\text { Let } x+1=0$
$\text { then } x=-1$
$f(-1)=3(-1) 3+k(-1)^2+7(-1)+4$
$=-3+k-7+4$
$=k-6$
$\therefore \text { Remainder }=0$
$\therefore k-6=0$
$\Rightarrow k=6$

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MCQ 21 Mark

If on dividing $2 x^3+6 x^2-(2 k-7) x+5$ by $x+3$, the remainder is $k-1$ then the value of $k$ is

A
2
B
$-2$
C
$-3$
✓
3

Answer

Correct option: D.

$
\begin{aligned}
& f(x)=2 x^3+6 x^2-(2 k-7) x+5 \\
& g(x)=x+3
\end{aligned}
$
Remainder $=k-1$
$
\text { If } x+3=0 \text {, }
$
then $x=-3$
$\therefore$ Remainder will be
$
\begin{aligned}
& f(-3)=2(-3)^2+6(-3)^2-(2 k-7)(-3)+5 \\
& =-54+54+3(2 k-7)+5 \\
& =-54+54+6 k-21+5 \\
& =6 k-16 \\
& \therefore 6 k-16=k-1 \\
& 6 k-k=-1+16 \\
& \Rightarrow 5 k-15 \\
& k=\frac{15}{5}=3 \\
& \therefore k=3 .
\end{aligned}
$

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MCQ 31 Mark

If on dividing $4 x^2-3 k x+5$ by $x+2$, the remainder is -3 then the value of $k$ is

A
4
✓
$-4$
C
3
D
$-3$

Answer

Correct option: B.

$-4$

$
\begin{aligned}
& f(x)=4 x^2-3 k x+5 \\
& g(x)=x+2 \\
& \text { Remainder }=-3
\end{aligned}
$
Remainder $=-3$
Let $x+2=0$, then $x=-2$
Now remainder will be
$
\begin{aligned}
& f(-2)=4(-2)^2-3 k(-2)+5 \\
& =16+6 k+5 \\
& =21+6 k \\
& \therefore 21+6 k=-3 \\
& \Rightarrow 6 k=-3-21 \\
& =-24 \\
& \Rightarrow k=\frac{-24}{6}=-4 \\
& \therefore k=-4
\end{aligned}
$

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MCQ 41 Mark

When $2 x^3-x^2-3 x+5$ is divided by $2 x+1$, then the remainder is

✓
6
B
$-6$
C
$-3$
D
$0$

Answer

Correct option: A.

$
\begin{aligned}
& f(x)=2 x^3-x^2-3 x+5 \\
& g(x)=2 x+1 \\
& \text { Let } 2 x+1=0 \\
& \text { then } x=\frac{-1}{2}
\end{aligned}
$
Then remainder will be
$
\begin{aligned}
& f\left(\frac{-1}{2}\right)=2\left(\frac{-1}{2}\right)^3-\left(\frac{-1}{2}\right)^2-3\left(\frac{-1}{2}\right)+5 \\
& =2 \times \frac{-1}{8}-\frac{1}{4}+\frac{3}{2}+5 \\
& =\frac{-1}{4}-\frac{1}{4}+\frac{3}{2}+5 \\
& =\frac{-1-1^4+6+20}{4} \\
& =\frac{24}{4} \\
& =6 \\
& \therefore \text { Remainder }=6 .
\end{aligned}
$

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MCQ 51 Mark

When $x^3-3 x^2+5 x-7$ is divided by $x-2$, then the remainder is

A
$0$
B
1
C
2
✓
$-1$

Answer

Correct option: D.

$-1$

$f(x)=x^3-3 x^2+5 x-7$
$g(x)=x-2$, if $x-2=0$, then $x=2$
Remainder will be
$\therefore f(2)=(2)^3-3(2)^3+5 \times 2-7$
$=8-12+10-7$
$=18-19$
$=-1$
$\therefore \text { Remainder }=-1$

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