M.C.Q - MATHS STD 9 Questions - Vidyadip

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MCQ 11 Mark

The expression $(a - b)^3 + (b - c)^3 + (c - a)^3$ can be factorized as:

A
$(a - b)(b - c)(c - a)$
✓
$3(a - b)(b - c)(c - a)$
C
$-3(a - b)(b - c)(c - a)$
D
$(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$

Answer

Correct option: B.

$3(a - b)(b - c)(c - a)$

By we know that $a^3 + b^3 + c^3 - 3abc$
$ = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
If $a + b + c = 0,$ then
$a^3 + b^3 + c^3 = 3abc$
In given expression,
Let $a - b = A, b - c = B, c - a = C$
Now, $a - b + b - c + c - a = 0$
$i.e. A + B + C = 0$
$\Rightarrow A^3 + B^3 + C^3 = 3 \text{ABC}$
$\Rightarrow (a - b)^3 + (b - c)^3 + (c - a)^3$
$ = 3(a - b)(b - c)(c - a)$

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MCQ 21 Mark

The expression $x^4 + 4$ can be factorized as:

✓
($x^2 + 2x + 2)(x^2 - 2x + 2)$
B
$(x^2 + 2x + 2)(x^2 + 2x - 2)$
C
$(x^2 - 2x - 2)(x^2{ }- 2x + 2)$
D
$(x^2 + 2)(x^2 - 2)$

Answer

Correct option: A.

($x^2 + 2x + 2)(x^2 - 2x + 2)$

$x^4 + 4$
$= x^4 + 4 + 4x^2 - 4x^2$
$= (x^4 + 4x^2 + 4) - 4x^2$
$= (x^2 + 2)^2 - (2x)^2$
$= (x^2 + 2 - 2x)(x^2 + 2 + 2x)$
$= (x^2 + 2x + 2)(x^2 - 2x + 2)$

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MCQ 31 Mark

The factors of $x^2 + 4y^2 + 4y - 4xy - 2x - 8,$ are:

✓
$(x - 2y - 4)(x - 2y + 2)$
B
$(x - y + 2)(x - 4y - 4)$
C
$(x + 2y - 4)(x + 2y + 2)$
D
None of these.

Answer

Correct option: A.

$(x - 2y - 4)(x - 2y + 2)$

$x^2 + 4y^2 + 4y - 4xy - 2x - 8$
$= x^2 +(2y)^2 - 2 \times x(2y) + 4y - 2x - 8$
$= (x - 2y)^2 + 4y - 2x - 8 ...(1)$
Now making $eq(1)$ a perfect square by adding $1$ and $-1$
$(x - 2y)^2 + 4y - 2x - 8 = (x - 2y)^2 + 4y - 2x - 8 + 1 - 1$
$= (x - 2y)^2 + (1)^2 - 2 \times (1) \times (x - 2y) - 9$
$= (x - 2y - 1)^2 - (3)^2$
$= [(x - 2y - 1) - 3][x - 2y - 1 + 3]$
$= (x - 2y - 4)(x - 2y + 2)$

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MCQ 41 Mark

The factors of $a^2 - 1 - 2x - x^2$ are:

✓
$(a - x + 1)(a - x - 1)$
B
$(a + x - 1)(a - x + 1)$
C
$(a + x +1)(a - x + 1)$
D
None of these.

Answer

Correct option: A.

$(a - x + 1)(a - x - 1)$

$a^2 - 1 - 2x - x^2$
$= a^2 - (1 + 2x + x^2)$
$= a^2 - (1 + x)^2$
$= [a - (1 + x)][a + (1 + x)]$
$= (a - x - 1)(a + x + 1)$

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MCQ 51 Mark

If $3x = a + b + c,$ then the value of $(x - a)^3 + (x - b)^3 + (x - c)^3 - 3(x - a) (x - b) (x - c)$ is:

A
$a + b + c$
B
$(a - b)(b - c)(c - a)$
✓
$0$
D
None of these.

Answer

Correct option: C.

$0$

$3x = a + b + c$
$\Rightarrow a + b + c - 3x = 0$
$\Rightarrow 3x - (a + b + c) = 0$
$\Rightarrow (x - a) + (x - b) + (x - c) = 0 ...(1)$
Using identity if $a + b + c = 0$ then, $a^3 + b^3 + c^3 - 3abc = 0$
If we take $x - a = A, x - b = B, x - c = C$ in equation $(1),$ we get
$A + B + C = 0$
$\Rightarrow A^3 + B^3 + C^3 - 3ABC= 0$
$\Rightarrow (x - a)^3 + (x - b)^3 + (x - c)^3 - 3(x - a) (x - b) (x - c) = 0$

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MCQ 61 Mark

The factors of $x^3 - 7x + 6$ are:

A
$x(x - 6)(x - 1)$
B
$(x^2 - 6)(x - 1)$
C
$(x + 1)(x + 2)(x + 3)$
✓
$(x - 1)(x + 3)(x - 2)$

Answer

Correct option: D.

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

$x^3 - 7x + 6 = x^3 - 7x + 6 + 1 - 1 ($by adding $+1 -1 $ to $\text{R.H.S})$
$= x^3 - 7x + 7 - 1$
$= (x^3 - 1) - 7(x - 1)$
Now by identity $a^3 - b^3 = (a - b)(a^2 + b^2 + ab),$ we get
$x^3 - 7x + 6 = (x^3 - 1) - 7(x - 1)$
$= (x - 1)(x^2 + x + 1) - 7(x - 1)$
$= (x - 1)(x^2 + x + 1 - 7)$
$= (x - 1)(x^2 + x - 6)$
$= (x - 1)(x + 3)(x - 2)$

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MCQ 71 Mark

The factors of $x^4 + x^2 + 25$ are:

✓
$(x^2 + 3x + 5)(x^2 - 3x + 5)$
B
$(x^2{ }+ 3x + 5)(x^2 + 3x − 5)$
C
$(x^2 + x +5)(x^2 - x + 5)$
D
None of these.

Answer

Correct option: A.

$(x^2 + 3x + 5)(x^2 - 3x + 5)$

For making perfect square to $x^4 + x^2 + 25$
We add $+10x^2$ and $-10x^2$ to it.
$= x^4 + x^2 + 25$
$= x^4 + x^2 + 25 + 10x^2 - 10x^2$
$= [x^4 + 10x^2 + 25] - 9x^2$
$= (x^2 + 5)^2 + (3x)^2$
$= [(x^2 + 5) + 3x][(x^2 + 5) - 3x]$
$= (x^2 + 3x + 5)(x^2 - 3x + 5)$

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MCQ 81 Mark

The factors of $x^3 - x^2y - xy^2{ }+ y^3 $are:

A
$(x + y)(x^2 - xy + y^2)$
B
$(x + y)(x^2 + xy + y^2)$
✓
$(x + y)^2(x - y)$
D
$(x - y)^2(x + y)$

Answer

Correct option: C.

$(x + y)^2(x - y)$

$x^3 - x^2y - xy^2 + y^3 = x^3 + y^3 - xy(x + y)$
Now by identity $x^3 + y^3$
$= (x + y)(x^2 + y^2 - xy),$ we have
$x^3 - x^2y - xy^2 + y^3 $
$= (x + y)(x^2 + y^2 - xy) - xy(x + y)$
$= (x + y)(x^2 + y^2 - xy - xy)$
$= (x + y)(x^2 + y^2 - 2xy)$
$= (x + y)(x - y)^2$

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MCQ 91 Mark

$(x + y)^3 - (x - y)^3$ can be factorized as:

A
$2y(3x^2 + y^2)$
✓
$2x(3x^2 + y^2)$
C
$2y(3y^2 + x^2)$
D
$2x(x^2+ 3y^2)$

Answer

Correct option: B.

$2x(3x^2 + y^2)$

We know the identity
$a^3 - b^3 = (a - b)(a^2 + b^2 + ab)$
Let $x + y = a$ and $x - y = b$
Then,
$a^3 - b^3$
$= (x + y)^3 - (x - y)^3$
$= [(x + y) - (x - y)][(x + y)^2 + (x - y)^2 + (x + y)(x - y)]$
$= 2y[x^2 + y^2 + 2xy + x^2 + y^2 - 2xy + x^2 - y^2]$
$= 2y(3x^2 + y^2)$

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MCQ 101 Mark

If $x^3 - 3x^2 + 3x - 7 = (x + 1)(ax^2 + bx + c),$ then $a + b + c =$

A
$4$
B
$12$
✓
$-10$
D
$3$

Answer

Correct option: C.

$-10$

The given equation is
$x^3 - 3x^2 + 3x - 7 = (x + 1)(ax^2 + bx + c)$
This can be written as
$x^3 - 3x^2 + 3x - 7 = (x + 1)(ax^2 + bx + c)$
$= x^3 - 3x^2 + 3x - 7 = ax^3 + bx^2 + cx + ax^2 + bx + c$
$= x^3 - 3x^2 + 3x - 7 = ax^3 + (a + b)x^2 + (b + c)x + c$
Comparing the cofficients on both sides of the equation.
We get,
$a = 1 ...(1)$
$a + b = 3 ...(2)$
$b + c = 3 ...(3)$
$c = -7 ...(4)$
Putting the value of a form $(1)$ in $(2)$
We get,
$1 + b = 3,$
$b = -3 - 1$
$b= -4$
So the value of $a, b$ and $c$ is $1, -4$ and $-7$ respectively.
Therefore,
$a + b + c = 1 - 4 - 7 = -10$

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MCQ 111 Mark

The factors of $x^3 - 1 + y^3 + 3xy$ are:

✓
$(x - 1 + y)(x^2 + 1 + y^2 + x + y - xy)$
B
$(x + y + 1)(x^2 + y^2 + 1 - xy - x - y)$
C
$(x - 1 + y)(x^2 - 1 - y^2 + x + y + xy)$
D
$3(x + y - 1)(x^2 + y^2 - 1)$

Answer

Correct option: A.

$(x - 1 + y)(x^2 + 1 + y^2 + x + y - xy)$

By using identity
$a^3 + b^3 + c^3 - 3abc $
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
We can write,
$x^3 - 1 + y^3 + 3xy$
$= (x^3) + (-1)^3 + (y^3) - 3(-1)(x)(y)$
$= [x + (-1) + y][x^2 + (-1)^2 + y^2 - x(-1) - y(-1) - xy]$
$= (x - 1 + y)(x^2 + 1 + y^2+ x + y - xy)$

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MCQ 121 Mark

The factors of $8a^3 + b^3 - 6ab + 1$ are:

A
$(2a + b - 1)(4a^2 + b^2 + 1 - 3ab - 2a)$
B
$(2a - b + 1)(4a^2 + b^2 - 4ab + 1 - 2a + b)$
✓
$(2a + b + 1)(4a^2 + b^2 + 1 -2ab - b - 2a)$
D
$(2a - 1 + b)(4a^2 + 1 - 4a - b - 2ab)$

Answer

Correct option: C.

$(2a + b + 1)(4a^2 + b^2 + 1 -2ab - b - 2a)$

We know the identity
$a^3 + b^3 + c^3 - 3abc$
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
So by using identity, we can write given expression as
$(2a)^3 + (b)^3 + (1)^3 - 3(2a)(b)(1)$
$= (2a + b + 1)[(2a)^2 + b^2 + 1^2 -2a \times b - b \times 1 - 2a \times 1]$
$= (2a + b + 1)(4a^2 + b^2 + 1 -2ab - b - 2a)$

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Question 131 Mark

The value of $\frac{(2.3)^3-0.027}{(2.3)^2+0.69+0.09},$ is:

2
3
2.327
2.273

Answer

2

Solution:
$\frac{(2.3)^3-0.027}{(2.3)^2+0.69+0.09}$
$=\frac{(2.3)^3-(0.3)^3}{(2.3)^2+(0.3)^3+(2.3)(0.3)}$
$=\frac{(2.3 - 0.3)\{(2.3)^2+(0.3)^2+(2.3)(0.3)\}}{((2.3)^2+(0.3)^2+(2.3)(0.3))}$
$=2.3 - 0.3$
$=2$
Hence, correct option is (a).

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MCQ 141 Mark

If $(x + y)^3 - (x - y)^3 - 6y(x^2 - y^2) = ky^2,$ then $k =$

A
$1$
B
$2$
C
$4$
✓
$8$

Answer

Correct option: D.

$8$

Let $x + y = A$ and $x - y = B$
Now,$ (A - B)^3 = A^3 - B^3 - 3AB(A - B)$
$\Rightarrow [(x + y) - (x - y)]^3$
$ = (x + y)^3 - (x - y)^3 - 3(x + y)(x - y)[(x + y) - (x - y)]$
$= (x + y)^3 - (x - y)^3 - 3(x^2 - y^2)(2y)$
$= (x + y)^3 - (x - y)^3 - 6y(x^2 - y^2)$
But, $(x + y)^3 - (x - y)^3 - 6y(x^2 - y^2) = ky^3$
$\Rightarrow [(x + y) - (x - y)]^3 = (2y)^3 = k8y^3$
$\Rightarrow (2y)^3 = ky^3$
$\Rightarrow 8y^3 = ky^3$
$\Rightarrow k = 8$

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MCQ 151 Mark

The value of $\frac{(0.013)^3+(0.007)^3}{(0.013)^2-0.013\times0.007+(0.007)^2},$ is:

A
$0.006$
✓
$0.02$
C
$0.0091$
D
$0.00185$

Answer

Correct option: B.

$0.02$

By using identity $a^3 + b^3 = (a + b)(a^2 + b^2 - ab),$ we have
$\frac{(0.013)^3+(0.007)^3}{(0.013)^2-0.013\times0.007+(0.007)^2}$
$=\frac{\{(0.013)+(0.007)\}(0.013)^2-(0.013)(0.007)+(0.007)^2}{(0.013)^2-(0.013)(0.007)+(0.007)^2}$
$=0.013+0.007$
$=0.020$
$=0.02$

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M.C.Q - MATHS STD 9 Questions - Vidyadip