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CBSE Questions for Class 9 Maths Polynomials Quiz 8 - MCQExams.com
CBSE
Class 9 Maths
Polynomials
Quiz 8
If $$g (x) = \displaystyle 3x^{2}-2x-5 $$, what is the value of $$g(-1)$$?
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0%
$$-4$$
0%
$$-10$$
0%
$$-6$$
0%
$$0$$
Explanation
Given, $$g(x)=x^{2}-2x-5$$
Then $$g(-1)=3(-1)^{2}-2(-1)-5$$
$$=3+2-5$$
$$=0$$
For what value of $$k$$, $$-2$$ is a zero of the polynomial
$$3x^2 + 4x + 2k$$?
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0%
$$5$$
0%
$$3$$
0%
$$-8$$
0%
$$-2$$
Explanation
If $$-2$$ is a zero of $$f(x)=3x^2+4x+2k$$
Thus,
$$f(-2)=0$$
$$\therefore 3(-2)^2+4(-2)+2k=0$$
$$=>12-8+2k=0$$
$$=>2k=-4$$
$$=>k=-2$$
The value of the polynomial $$x^{3} -27$$ when $$x = 3$$ is
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0%
$$26$$
0%
$$25$$
0%
$$0$$
0%
$$1$$
Explanation
Let $$f(x)=x^3-27.$$
Therefore, at $$x=3$$,
$$f(3) = 3^3-27$$
$$ = 27-27$$
$$= 0$$
Hence, option $$C$$ is correct.
$$(a -b)^2 -(a + b)^2$$ is equal to:
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0%
$$2ab$$
0%
$$4ab$$
0%
$$6ab$$
0%
$$-4ab$$
Explanation
Given, $$(a -b)^2-(a + b)^2 $$.
We know, $$(x+y)^2=x^2+2xy+y^2$$
and
$$(x-y)^2=x^2-2xy+y^2$$.
Then,
$$(a -b)^2-(a + b)^2 $$
$$= (a^2 + b^2 -2ab) -(a^2 + b^2 + 2ab) $$
$$= a^2 + b^2 -2ab -a^2 - b^2 - 2ab) $$
$$= -4ab$$.
Therefore, option $$D$$ is correct.
Say true or false:
The zeros of the polynomial $$x^{2} - 14x + 49$$ is equal to $$7$$.
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0%
True
0%
False
Explanation
The given polynomial
$$x^2-14x+49=0$$
can be factorized as follows:
$$x^{ 2 }-14x+49=0\\ \Rightarrow x^{ 2 }-7x-7x+49=0\\ \Rightarrow x(x-7)-7(x-7)=0\\ \Rightarrow (x-7)(x-7)=0\\ \Rightarrow (x-7)=0,\quad (x-7)=0\\ \Rightarrow x=7,\quad x=7$$
Hence, the zeroes of
$$x^2-14x+49$$
is
$$7$$
.
Say true or false:
The zero of $$x-5$$ is $$5$$.
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0%
True
0%
False
Explanation
To find the zeroes of any polynomial, we need to equate the polynomial expression to zero.
Here, the polynomial is $$x-5$$ that is $$x-5=0$$
$$\implies x=5$$.
Hence, the zero of $$x-5$$ is $$5$$.
The value of the polynomial $$7x^{4} + 6x^{2} + x$$ when $$x = -2$$ is
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0%
$$131$$
0%
$$132$$
0%
$$134$$
0%
$$136$$
Explanation
given, $$f(x) = 7x^4+6x^2+x$$
$$f(-2) = 7(-2)^4+6(-2)^2+(-2)$$
$$f(-2) = 7(16)+6(4)-2$$
$$f(-2) = 112+22 = 134$$
The value of the polynomial $$4x^{2} + 3x - 7$$ at $$x=1$$ is:
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0%
$$1$$
0%
$$2$$
0%
$$0$$
0%
None of the above
Explanation
$$f(x) = 4x^2+3x-7$$
So at $$x=1$$,
$$f(1) = 4(1)^2+3(1)-7 $$
$$f(1) = 4+3-7 = 0$$
State the following statement is True or False
The zero of the polynomial $$x + 2 + 4$$ is $$-2$$.
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0%
True
0%
False
Explanation
Rewrite the polynomial $$x+2+4$$ as $$x+6$$ and equate it to $$0$$ as follows:
$$x+6=0\\ \Rightarrow x=-6$$
Hence, the only zero of the polynomial
$$x+2+4$$
is $$-6$$.
Using standard identity, find the value of $$102^2$$.
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0%
$$10402$$
0%
$$10408$$
0%
$$10204$$
0%
$$10404$$
Explanation
Given, $$102^2$$
$$=(100+2)^2$$.
We know, $$(a+b)^2=a^2+2ab+b^2$$.
Then,
$$102^2$$
$$=(100+2)^2$$
$$=100^2+2^2+2(2)(100) $$
$$=10000+4+400 $$
$$= 10404$$.
Therefore, option $$D$$ is correct.
Using standard identity, find $$(2x-y)^2$$.
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0%
$$4x^2+y^2-4xy$$
0%
$$4x^2+y^2+4xy$$
0%
$$2x^2+y^2-4xy$$
0%
$$4x^2+y^2-2xy$$
Explanation
Given,
$$(2x-y)^2$$
We know, $$(a-b)^2=a^2-2ab+b^2$$.
Then,
$$(2x-y)^2$$
$$=(2x)^2+y^2-2\times 2x\times y$$
$$= 4x^2+y^2-4xy$$.
Therefore, option $$A$$ is correct.
Using standard identity, find the value of $$(a+2b)^2$$.
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0%
$$a^2+2b^2+4ab$$
0%
$$a^2+4b^2+2ab$$
0%
$$a^2+4b^2+4ab$$
0%
$$2a^2+4b^2+4ab$$
Explanation
Given,
$$(a+2b)^2$$.
We know, $$(x+y)^2=x^2+2xy+y^2$$.
Then,
$$(a+2b)^2$$
$$=a^2+(2b)^2 +2\times a\times 2b$$
$$=a^2+4b^2+4ab$$.
Therefore, option $$C$$ is correct.
The value of the polynomial $$2x^{5} - 5x^{3} - 10x + 9$$ when $$x = -1$$
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0%
$$21$$
0%
$$22$$
0%
$$23$$
0%
None of the above
Explanation
$$f(x) = 2x^5-5x^3-10x+9$$
$$f(-1) = 2(-1)^5-5(-1)^3-10(-1)+9$$
$$f(-1) = 2(-1)-5(-1)+10+9$$
$$f(-1) = -2+5+19$$
$$f(-1) = 22$$
Using standard identity, find the value of $$99^2$$.
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0%
9901
0%
9801
0%
1001
0%
9701
Explanation
Given,
$$=99^2$$
$$=(100-1)^2$$
We know, $$(a-b)^2=a^2-2ab+b^2$$
Then,
$$99^2$$
$$=(100-1)^2$$
$$=100^2+1^2 - 2\times 100\times 1$$
$$= 10000+1- 200$$
$$=10001-200$$
$$=9801$$
Therefore, option $$B$$ is correct.
Which of the following is correct?
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0%
$$(x-y)^2=x^2+2xy-y^2$$
0%
$$(x-y)^2=x^2-2xy+y^2$$
0%
$$(x-y)^2=x^2-y^2$$
0%
$$(x+y)^2=x^2+2xy-y^2$$
Explanation
We know, $$(a-b)^2=a^2-2ab+b^2$$.
Then,
$$(x-y)^2$$
$$=x^2-2(x)(y)+y^2$$ $$= x^2-2xy+y^2$$.
To verify this, let us consider:
$$(x-y)^2$$
$$=(x-y)$$
$$(x-y)$$
$$=x(x-y)-y(x-y)$$
$$=x^2-xy-yx+y^2$$
$$=x^2-xy-xy+y^2$$
$$=x^2-2xy+y^2$$.
Hence,
$$(x-y)^2$$
$$= x^2-2xy+y^2$$
Therefore, option $$B$$ is correct.
Find the value of $$49^2$$ using standard identity.
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0%
$$2391$$
0%
$$2401$$
0%
$$2411$$
0%
$$2421$$
Explanation
As we know that,
$$(a-b)^2=a^2-2ab+b^2$$
Substitute $$50$$ for $$a$$ and $$1$$ for $$b$$ in above formula,
$$\begin{aligned}{}{(50 - 1)^2}& = {50^2} - 2 \times 50 \times 1 + {1^2}\\{49^2}& = 2500 - 100 + 1\\ &= 2401\end{aligned}$$
Therefore, $$B$$ is correct.
Find the value of $$99\times 101$$ using standard identity.
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0%
$$9999$$
0%
$$9989$$
0%
$$9979$$
0%
$$1009$$
Explanation
Given, $$99\times 101=(100-1)(100+1)$$.
We know, $$(a+b)(a-b) = a^2-b^2$$.
Then,
$$99\times 101=(100-1)(100+1)$$
$$=(100)^2-1^2$$
$$=10000-1$$
$$=9999$$.
Therefore, option $$A$$ is correct.
Square of $$3a-4b$$ is:
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0%
$$9a^2+16b^2-24ab$$
0%
$$6a^2-8b^2$$
0%
$$9a^2-16b^2$$
0%
$$9a^2+16b^2+24ab$$
Explanation
Given, square of $$3a-4b$$
i.e.
$$(3a-4b)^2$$.
We know, $$(a-b)^2=a^2-2ab+b^2$$.
Then,$$(3a-4b)^2$$
$$= (3a)^2-2(3a)(4b)+(4 b)^2$$
$$=9a^2-24ab+16b^2$$.
Therefore, option $$A$$ is correct.
Find the value of $$47\times 53$$.
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0%
$$2451$$
0%
$$2471$$
0%
$$2491$$
0%
$$2501$$
Explanation
Given,
$$47\times 53 $$
$$= (50-3)(50+3)$$.
We know, $$(a+b)(a-b)=a^2-b^2$$.
Then,
$$47\times 53 = (50-3)(50+3)$$
$$=(50)^2-3^2=2500-9=2491$$.
Therefore, option $$C$$ is correct.
Find the value of $$87^2-13^2$$.
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0%
$$7300$$
0%
$$7350$$
0%
$$7400$$
0%
$$7450$$
Explanation
Given,
$$87^2-13^2$$.
We know, $$a^2-b^2=(a+b)(a-b)$$.
Then,
$$=87^2-13^2$$
$$=(87+13)(87-13)$$
$$=100\times 74=7400$$.
Therefore, option $$C$$ is correct.
Find the value of $$52^2$$ using standard identity.
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0%
$$2604$$
0%
$$2704$$
0%
$$2804$$
0%
$$2904$$
Explanation
Given,
$$52^2$$
$$=(50+2)^2$$.
We know, $$(a+b)^2=a^2+b^2+2ab$$.
Then,
$$52^2=(50+2)^2$$
$$=50^2+2^2+2\times 50\times 2$$
$$=2500+4+200$$
$$=2704$$.
Therefore, option $$B$$ is correct.
Find the value of $$199\times 201$$.
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0%
$$39989$$
0%
$$40001$$
0%
$$39999$$
0%
$$400011$$
Explanation
Given,
$$199\times 201$$
$$ =(200-1)(200+1) $$.
We know, $$a^2-b^2=(a-b)(a+b)$$.
Then,
$$199\times 201$$
$$ =(200-1)(200+1) $$
$$= 200^2-1^2$$
$$= 40000-1$$
$$=39999$$.
Therefore, option $$C$$ is correct.
Find the value of $$995^2-5^2$$.
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0%
$$99000$$
0%
$$9900$$
0%
$$99005$$
0%
$$990000$$
Explanation
Given,
$$995^2-5^2$$.
We know, $$a^2-b^2=(a+b)(a-b)$$.
Then,
$$995^2-5^2=(995+5)(995-5)$$
$$=1000\times 990$$
$$=990000$$.
Therefore, option $$D$$ is correct.
Find the value of $$(a+b)^2-(a-b)^2$$.
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0%
$$ab$$
0%
$$2ab$$
0%
$$3ab$$
0%
$$4ab$$
Explanation
Given, $$(a+b)^2-(a-b)^2$$.
We know,
$$(a+b)^2$$
$$=a^2+b^2+2ab$$
and
$$(a-b)^2$$
$$a^2+b^2-2ab$$.
Then,
$$(a+b)^2-(a-b)^2$$
$$=a^2+b^2+2ab-(a^2+b^2-2ab)$$
$$=a^2+b^2+2ab-a^2-b^2+2ab$$
$$= 4ab$$.
Therefore, option $$D$$ is correct.
The zero of the polynomial $$2x+1$$ is:
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0%
$$x=-\dfrac{1}{2}$$
0%
$$x=-1$$
0%
$$x=-2$$
0%
None of the above
Explanation
Let $$p(x)=2x+1$$
Consider $$p(x)=0$$
$$\Rightarrow 2x+1=0$$
$$\Rightarrow x=\dfrac{-1}{2}$$
Therefore, $$x=\dfrac{-1}{2}$$ is the zero of the polynomial $$2x+1=0$$
A zero of a polynomial:
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0%
needs to be $$0$$.
0%
can be any number including $$0$$.
0%
needs to be an integer.
0%
None of the above
Explanation
Consider the polynomial, $$p(x) = ax^2 + bx +c$$.
To find the zero of a polynomial, we write $$p(x) = 0$$.
Hence, a real number
$$c$$ is said to be a zero of the polynomial
$$p(x)$$, if
$$p(c)=0$$.
E.g. : Suppose, $$p(x)=x^2$$.
When $$x=0$$, we get $$p(x)=(0)^2=0$$.
Clearly, a zero of the polynomial can be any number including $$0$$.
Therefore, option $$B$$ is correct.
The zero of the polynomial $$3x+5$$ is :
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0%
$$\dfrac{-5}{3}$$
0%
$$\dfrac{5}{3}$$
0%
$$\dfrac{1}{2}$$
0%
$$\dfrac{-1}{2}$$
Explanation
We find the zeros of the given polynomial using:
$$3x+5=0$$
$$3x=-5$$
$$x=\dfrac{-5}{3}$$.
$$\left( 3-\sqrt { 7 } \right) \left( 3+\sqrt { 7 } \right) =$$?
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0%
$$4$$
0%
$$2$$
0%
$$6$$
0%
$$8$$
Explanation
Given,
$$({3}-{\sqrt { 7 } })$$
$$({3}+{\sqrt { 7 } })$$.
We know, $$(x-y)(x+y)$$ $$={x}^{2}-{y}^{2}$$.
Thus,
$$({3}-{\sqrt { 7 } })$$
$$({3}+{\sqrt { 7 } })$$
$$={3}^{2}-{(\sqrt { 7 } )}^{2}$$
$$=9-7=2$$.
Therefore, option $$B$$ is correct.
The value of $$(a+b)^2-2(a-b)^2+(a-b)(a+b)$$ is:
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0%
$$4ab-b^2$$
0%
$$2ab-b^2$$
0%
$$3ab-b^2$$
0%
$$6ab-2b^2$$
Explanation
Given, $$(a+b)^2-2(a-b)^2+(a-b)(a+b)$$.
We know,
$$(a+b)^{2}=a^2+2ab+b^2$$,
$$(a-b)^{2}=a^2-2ab+b^2$$
and $$(a+b)(a-b)=a^2-b^2$$.
Then,
$$(a+b)^2-2(a-b)^2+(a-b)(a+b)$$
$$=(a^2+b^2+2ab)-2(a^2+b^2-2ab)+(a^2-b^2)$$
$$=a^2+b^2+2ab-2a^2-2b^2+ 4ab+a^2-b^2$$
$$=2a^2+b^2+6ab-2a^2-3b^2$$
$$=6ab-2b^2$$.
Therefore, option $$D$$ is correct.
Find the value of $$1.05\times 0.95$$ using standard identity.
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0%
$$0.9985$$
0%
$$0.9975$$
0%
$$0.9875$$
0%
$$0.9995$$
Explanation
Given, $$1.05\times 0.95$$
$$=(1+0.05)(1-0.05)$$.
We know,
$$(a+b)(a-b)=a^2-b^2$$.
Then,
$$1.05\times 0.95$$
$$=(1+0.05)(1-0.05)$$
$$=1^2-(0.05)^2$$
$$=1-0.0025$$
$$=0.9975$$.
Therefore, option $$B$$ is correct.
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