MCQ(1M)

MCQ 11 Mark

$4 a^2+b^2+4 a b+8 a+4 b+4=?$

✓
$(2 a+b+2)^2$
B
$(2 a-b+2)^2$
C
$(a+2 b+2)^2$
D
None of these

Answer

Correct option: A.

$(2 a+b+2)^2$

$4 a^2+b^2+4 a b+8 a+4 b+4$
$=4 a^2+b^2+4+4 a b+4 b+8 a$
$=(2 a)^2+b^2+2^2+2 \times 2 a \times b+2 \times b \times 2+2 \times 2 a \times 2$
$=(2 a+b+2)^2$

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MCQ 21 Mark

If $\frac{a}{b}+\frac{b}{a}=-1$ then $\left(a^3-b^3\right)= ?$

A
$-3$
B
$-2$
C
$-1$
✓
$0$

Answer

Correct option: D.

$0$

$\frac{a}{b}+\frac{b}{a}=-1$
$\Rightarrow \frac{a^2+b^2}{a b}=-1$
$\Rightarrow a^2+b^2=-a b$
$\Rightarrow a^2+b^2+a b=0$
Thus, we have:
$\left(a^3-b^3\right)=(a-b)\left(a^2+b^2+a b\right)$
$=(a-b) \times 0$
$=0$

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MCQ 31 Mark

If $x+y+z=9$ and $x y+y z+z x=23$, the value of $\left(x^3+y^3+z^3-3 x y z\right)=?$

✓
$108$
B
$207$
C
$669$
D
$729$

Answer

Correct option: A.

$108$

$x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$
$=(x+y+z)\left[(x+y+z)^2-3(x y+y z+z x)\right]$
$=9 \times(81-3 \times 23)$
$=9 \times 12$
$=108$

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MCQ 41 Mark

$\left(x^2-4 x-21\right)=?$

A
$(x-7)(x-3)$
B
$(x+7)(x-3)$
✓
$(x-7)(x+3)$
D
None of these

Answer

Correct option: C.

$(x-7)(x+3)$

$x^2-4 x-21$
$x^2-7 x+3 x-21$
$=x(x-7)+3(x-7)$
$=(x-7)(x+3)$

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MCQ 51 Mark

$3 x^3+2 x^2+3 x+2=?$

A
$(3 x-2)\left(x^2-1\right)$
B
$(3 x-2)\left(x^2+1\right)$
C
$(3 x+2)\left(x^2-1\right)$
✓
$(3 x+2)\left(x^2+1\right)$

Answer

Correct option: D.

$(3 x+2)\left(x^2+1\right)$

$(3 x+2)\left(x^2+1\right)$
$3 x^3+2 x^2+3 x+2$
$=x^2(3 x+2)+1(3 x+2)$
$=(3 x+2)\left(x^2+1\right)$

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MCQ 61 Mark

The coefficient of $x$ in the expansion of $(x+3)^3$ is:

A
$1$
B
$9$
C
$18$
✓
$27$

Answer

Correct option: D.

$27$

$(x+3)^3$
$=x^3+3^3+9 x(x+3)$
$=x^3+27+9 x^2+27 x$
So, the coefficient of $x$ in $(x+3)^3$ is $27 $.

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MCQ 71 Mark

$6 x^2+17 x+5=?$

A
$(2 x+1)(3 x+5)$
✓
$(2 x+5)(3 x+1)$
C
$(6 x+5)(x+1)$
D
None of these

Answer

Correct option: B.

$(2 x+5)(3 x+1)$

$(2 x+5)(3 x+1)$
$6 x^2+17 x+5$
$=6 x^2+15 x+2 x+5$
$=3 x(2 x+5)+1(2 x+5)$
$=(2 x+5)(3 x+1)$

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MCQ 81 Mark

$\left(4 x^2+4 x-3\right)=?$

A
$(2 x-1)(2 x-3)$
B
$(2 x+1)(2 x-3)$
✓
$(2 x+3)(2 x-1)$
D
None of these

Answer

Correct option: C.

$(2 x+3)(2 x-1)$

$(2 x+3)(2 x-1)$
$4 x^2+4 x-3$
$=4 x^2+6 x-2 x-3$
$=2 x(2 x+3)-1(2 x+3)$
$=(2 x+3)(2 x-1)$

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MCQ 91 Mark

If $a + b + c = 0$ then $\Big(\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}\Big)=?$

A
$1$
B
$0$
C
$-1$
✓
$3$

Answer

Correct option: D.

$3$

$a + b + c = 0 \Rightarrow a^3 + b^3 + c^3 = 3abc$
Thus, we have:
$\Big(\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}\Big)=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}$
$=\frac{3\text{abc}}{\text{abc}}$
$=3$

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MCQ 101 Mark

If $(x + 1)$ is a factor of the polynomial $(2x^2 + kx)$ then the value of $k$ is:

A
$-2$
B
$-3$
✓
$2$
D
$3$

Answer

Correct option: C.

$2$

$(x+1)$ is a factor of $2 x^2+k x$
So, $-1$ is a zero of $2 x^2+k x$
Thus, we have:
$2 \times(-1)^2+k \times(-1)=0$
$\Rightarrow 2-k=0$
$\Rightarrow k=2$

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MCQ 111 Mark

The value of $(249)^2-(248)^2$ is:

A
$1^2$
B
$477$
C
$487$
✓
$497$

Answer

Correct option: D.

$497$

$(249)^2 - (248)^2$
We know
$a^2-b^2=(a+b)(a-b)$
So,
$(249)^2-(248)^2$
$(249-248)(249+248)$
$=497$

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MCQ 121 Mark

$(207 \times 193)=?$

A
$39851$
✓
$39951$
C
$39961$
D
$38951$

Answer

Correct option: B.

$39951$

$207 \times 193$
$=(200+7)(200-7)$
$=(200)^2-(7)^2$
$=40000-49$
$=39951$

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MCQ 131 Mark

Which of the following is a factor of $(x+y)^3-\left(x^3+y^3\right) ?$

A
$x^2+y^2+2 x y$
B
$x^2+y^2-x y$
C
$x y^2$
✓
$3 x y$

Answer

Correct option: D.

$3 x y$

$(x+y)^3-\left(x^3+y^3\right)$
$=x^3+y^3+3 x y(x+y)-\left(x^3+y^3\right)$
$=3 x y(x+y)$
Thus, the factors of $(x+y)^3-\left(x^3+y^3\right)$ are $3 x y$ and $(x+y)$

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MCQ 141 Mark

$(x+1)$ is a factor of the polynomial:

A
$x^3-2 x^2+x+2$
B
$x^3+2 x^2+x-2$
✓
$x^3+2 x^2-x-2$
D
$x^3+2 x^2-x+2$

Answer

Correct option: C.

$x^3+2 x^2-x-2$

Let: $f(x)=x^3-2 x^2+x+2$
By the factor theorem, $(x+1)$ will be a factor of $f(x)$ if $f(-1)=0$.
We have:
$f(-1)=(-1)^3-2 \times(-1)^2+(-1)+2$
$=-1-2-1+2$
$=-2 \neq 0$
Hence, $(x+1)$ is not a factor of $f(x)=x^3-2 x^2+x+2$.
Now,
Let:
$f(x)=x^3+2 x^2+x-2$
By the factor theorem, $(x+1)$ will be a factor of $f ( x )$ if $f (-1)=0$.
We have:
$f(-1)=(-1)^3+2 \times(-1)^2+(-1)-2$
$=-1+2-1-2$
$=-2 \neq 0$
Hence, $(x+1)$ is not factor of $f(x)=x^3+2 x^2+x-2$.
Now,
Let:
$f(x)=x^3+2 x^2-x-2$
By the factor theorem, $(x+1)$ will be a factor of $f(x)$ if $f(-1)=0$.
We have:
$f(-1)=(-1)^3+2 \times(-1)^2-(-1)-2$
$=-1+2+1-2$
$=0$
Hence, $(x+1)$ is a factor of $f(x)=x^3+2 x^2-x-2$.

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MCQ 151 Mark

If $a+b+c=0$ then $\left(a^3+b^3+c^3\right)$ is:

A
$0$
B
$a b c$
C
$2 a b c$
✓
$3 a b c$

Answer

Correct option: D.

$3 a b c$

$a+b+c=0$
$\Rightarrow a+b=-c$
$\Rightarrow(a+b)^3=(-c)^3$
$\Rightarrow a^3+b^3+3 a b(a+b)=-c^3$
$\Rightarrow a^3+b^3+3 a b(-c)=-c^3$
$\Rightarrow a^3+b^3+c^3=3 a b c$

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MCQ 161 Mark

If $\frac{x}{y}+\frac{y}{x}=+1$, where $x, y \neq 0$ then the value of $\left(x^3-y^3\right)$ is:

A
$1$
B
$-1$
✓
$0$
D
$\frac{1}{2}$

Answer

Correct option: C.

$0$

$\frac{x}{y}+\frac{y}{x}=-1$
$\Rightarrow \frac{x^2+y^2}{x y}=-1$
$\Rightarrow x^2+y^2=-x y$
$\Rightarrow x^2+y^2+x y=0$
Thus, we have:
$\left(x^3-y^3\right)=(x-y)\left(x^2+y^2+x y\right)$
$=(x-y) \times 0$
$=0$

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MCQ 171 Mark

One of the factors of $\left(25 x^2-1\right)+(1+5 x)^2$ is:

A
$5+x$
B
$5-x$
C
$5 x-1$
✓
$10 x$

Answer

Correct option: D.

$10 x$

$\left(25 x^2-1\right)+(1+5 x)^2$
$=(5 x-1)(5 x+1)+(1+5 x)^2$
$=(5 x+1)[(5 x-1)+(1+5 x)]$
$=(5 x+1)(10 x)$
So, the factors of $\left(25 x^2-1\right)+(1+5 x)^2$ are $(5 x+1)$ and $10 x$

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MCQ 181 Mark

$(305 \times 308)=?$

A
$94940$
B
$93840$
✓
$93940$
D
$94840$

Answer

Correct option: C.

$93940$

$305 \times 308=(300+5)(300+8)$
$=(300)^2+300 \times(5+8)+5 \times 8$
$=90000+3900+40$
$=93940$

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MCQ 191 Mark

$(104 \times 96) = ?$

A
$9894$
✓
$9984$
C
$9684$
D
$9884$

Answer

Correct option: B.

$9984$

$9984$
$104 \times 96 = (100 + 4)(100 - 4)$
$= 100^2 - 4^2$
$= (10000 - 16)$
$= 9984$

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MCQ 201 Mark

If $(x + 2)$ and $(x - 1)$ are factor of $(x^3 + 10x^2 + mx + n)$ then:

A
$m = 5, n = -3$
✓
$m = 7, n = -18$
C
$m = 17, n = -8$
D
$m = 23, n = -19$

Answer

Correct option: B.

$m = 7, n = -18$

$m = 7, n = -18$
Let:
$p(x)=x^3+10 x^2+m x+n$
Now,
$x+2=0 \Rightarrow x=-2$
$(x+2)$ is a factor of $p(x)$.
So, we have $p(-2)^2+m \times(-2)+n=0$
$\Rightarrow(-2)^3+10 \times(-2)^2+m \times(-2)+n=0$
$\Rightarrow-8+40-2 m+n=0$
$\Rightarrow 32-2 m+n=0$
$\Rightarrow 2 m-n=32 \ldots(i)$
Now,
$x-1=0 \Rightarrow x=1$
Also,
$(x-1)$ is a factor of $p(x)$
We have:
$p(1)=0$
$\Rightarrow 1^3+10 \times 1^2+m \times 1+n=0$
$\Rightarrow 1+10+m+n=0$
$\Rightarrow 11+m+n=0$
$\Rightarrow m+n=-11 \ldots \text {..(ii) }$
From (i) and (ii),
We get:
$3 m=21 \Rightarrow m=7$
By substituting the value of $m$ in (i), we get $n=-18$
$\therefore m=7 \text { and } n=-18$

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MCQ 211 Mark

If $(x+5)$ is a factor of $p(x)=x^3-20 x+5 k$ then $k=$ ?

A
$-5$
✓
$5$
C
$3$
D
$-3$

Answer

Correct option: B.

$5$

$(x+5)$ is a factor of $p(x)=x^3-20 x+5 k$
$\therefore p(-5)=0$
$\Rightarrow(-5)^3-20 \times(-5)+5 k=0$
$\Rightarrow-125+100+5 k=0$
$\Rightarrow 5 k=25$
$\Rightarrow k=5$

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MCQ 221 Mark

If $\Big(3\text{x}+\frac{1}{2}\Big)\Big(3\text{x}-\frac{1}{2}\Big)=9\text{x}^2-\text{p}$ then the value of $p$ is:

A
$0$
B
$-\frac{1}{4}$
✓
$\frac{1}{4}$
D
$\frac{1}{2}$

Answer

Correct option: C.

$\frac{1}{4}$

$\Big(3\text{x}+\frac{1}{2}\Big)\Big(3\text{x}-\frac{1}{2}\Big)=9\text{x}^2-\text{p}$
$9\text{x}^2-\frac{1}{4}$
$\Big(\therefore\ \big(\text{a}^2-\text{b}^2\big)=(\text{a}+\text{b})(\text{a}-\text{b})\Big)$
$=9\text{x}^2-\text{p}$
$\Rightarrow\text{p}=\frac{1}{4}$

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