RD Sharma 2020 solution class 9 chapter 4 Algebraic Identities MCQS

MCQS

Page-4.29

Question 1:

Mark the correct alternative in each of the following:

(1) If $x+\frac{1}{x}=5$, then $x^2+\frac{1}{x^2}=$

(a) 25

(b) 10

(c) 23

(d) 27

Answer 1:

In the given problem, we have to find the value of
Given
We shall use the identity
Here put,

Hence the value of is
Hence the correct choice is (c).

Question 2:

If $x+\frac{1}{x}=2$, then $x^3+\frac{1}{x^3} =$

(a) 64

(b) 14

(c) 8

(d) 2

Answer 2:

In the given problem, we have to find the value of
Given
We shall use the identity
Here putting,

Hence the value of is
Hence the correct choice is (d).

Question 3:

If $x+\frac{1}{x}$ = 4, then $x^4+\frac{1}{x^4}=$

(a) 196

(b) 194

(c) 192

(d) 190

Answer 3:

In the given problem, we have to find the value of
Given
We shall use the identity
Here put,

Squaring on both sides we get, 

Hence the value of is
Hence the correct choice is (b).

Question 4:

If $x+\frac{1}{x}=3$, then $x^6+\frac{1}{x^6}$ =

(a) 927

(b) 414

(c) 364

(d) 322

Answer 4:

In the given problem, we have to find the value of
Given
We shall use the identityand
Here put,

Take Cube on both sides we get,

Hence the value of is
Hence the correct choice is (d).

Question 5:

 If $x^2+\frac{1}{x^2} =102$, then $x-\frac{1}{x}$ =

(a) 8

(b) 10

(c) 12

(d) 13

Answer 5:

In the given problem, we have to find the value of
Given
We shall use the identity
Here putting,


Hence the value of is
Hence the correct choice is (b).

Question 6:

If $x^3+\frac{1}{x^3}=110$, then $x+\frac{1}{x}=$

(a) 5

(b) 10

(c) 15

(d) none of these

Answer 6:

In the given problem, we have to find the value of
Given
We shall use the identity

Put we get,

Substitute y = 5 in the above equation we get

The Equation satisfy the condition that
Hence the value of is 5
The correct choice is (a).

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Question 7:

If $x^3 - \frac{1}{x^3}=14$, then $x-\frac{1}{x} =$

(a) 5

(b) 4

(c) 3

(d) 2

Answer 7:

In the given problem, we have to find the value of
Given
We shall use the identity

Put we get,

Substitute y = 2 in above equation we get,

The Equation satisfy the condition that
Hence the value of is 2
Hence the correct choice is (d).

Question 8:

If a + b + c = 9 and ab + bc + ca = 23, then a² + b² + c² =

(a) 35

(b) 58

(c) 127

(d) none of these

Answer 8:

We have to find
Given
Using identity we get,

By transposing +46 to left hand side we get,

Hence the value of is
The correct choice is (a).

Question 9:

(a − b)³ + (b − c)³ + (c − a)³ =

(a) (a + b + c) (a² + b² + c² − ab − bc − ca)

(b) (a − b) (b − c) (c − a)

(c) 3(a − b) ( b− c) (c − a)

(d) none of these

Answer 9:

Given
Using identity
Here

Hence the Value of is
The correct choice is .

Question 10:

If $\frac{a}{b}+\frac{b}{a}=-1$, then a³ − b³ =

(a) 1

(b) −1

(c) $\frac{1}{2}$

(d) 0

Answer 10:

Given
Taking Least common multiple in we get,


Using identity

Hence the value of is
The correct choice is (d).

Question 11:

If a − b = −8 and ab  = −12, then a³ − b³ =

(a) −244

(b) −240

(c) −224

(d) −260

Answer 11:

To find the value of a³ − b³
Given
Using identity
Here we get


Transposing -288 to left hand side we get 

Hence the value of is -224
The correct choice is .

Question 12:

If the volume of a cuboid is 3x² − 27, then its possible dimensions are

(a) 3, x², − 27x

(b) 3, x − 3, x + 3

(c) 3, x², 27x

(d) 3, 3, 3

Answer 12:

We have to find the possible dimension of cuboid 
Given: volume of cuboid 


Take 3 as common factor

Using identity
We get,

Here the dimension of cuboid is 3,
The correct alternate is .

Question 13:

75 × 75 + 2 × 75 × 25 + 25 × 25 is equal to

(a) 10000

(b) 6250

(c) 7500

(d) 3750

Answer 13:

We have to find the product of
Using identity

Here



Hence the product of is 10,000
The correct choice is .

Question 14:

(x − y) (x + y) (x² + y²) (x⁴ + y⁴) is equal to

(a) x¹⁶ − y¹⁶

(b) x⁸ − y⁸

(c) x⁸ + y8

(d) x¹⁶ + y¹⁶
 

Answer 14:

Given
Using the identity

Hence is equal to
The correct choice is .

Question 15:

If $x^4+\frac{1}{x^4}=623$, then $x+\frac{1}{x}=$

(a) 27

(b) 25

(c) $3\sqrt{3}$

(d) $-3\sqrt{3}$

Answer 15:

In the given problem, we have to find the value of
Given
We shall use the identity
Here put,

We shall use the identitywe get,

Taking square root on both sides we get,

Hence the value of is
Hence the correct choice is (c).

Question 16:

If $x^4+\frac{1}{x^4}=194,$ then $x^3+\frac{1}{x^3} =$

(a) 76

(b) 52

(c) 64

(d) none of these

Answer 16:

Given
Using identity
Here,


Again using identity
Here

Substituting

Using identity
Here

Hence the value of is
The correct choice is (b).  

Question 17:

If $x-\frac{1}{x}=\frac{15}{4}$, then $x+\frac{1}{x}$=

(a) 4

(b) $\frac{17}{4}$

(c) $\frac{13}{4}$

(d) $\frac{1}{4}$

Answer 17:

In the given problem, we have to find the value of
Given
We shall use the identity
Here putting,


Substitute in we get,


Hence the value of is
Hence the correct choice is (b).

Question 18:

If $3x+\frac{2}{x}=7$, then $\left(9x^2-\frac{4}{x^2}\right)=$

(a) 25

(b) 35

(c) 49

(d) 30

Answer 18:

We have to find the value of
Given
Using identity we get,
Here

Substituting we get,


By transposing left hand side we get,

Again using identity we get,

Substituting we get 

Using identity we get 
Here

Substituting we get,

The value of is
The correct choice is (b) 

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Question 19:

If a² + b² + c² − ab − bc − ca =0, then

(a) a + b + c

(b) b + c = a

(c) c + a = b

(d) a = b = c

Answer 19:

Given
Multiplying both sides by 2 we get,



Therefore the sum of positive quantities is zero if and only if each quantity is zero.


If, then
The correct choice is (d).

Question 20:

If a + b + c = 0, then $\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}=$

(a) 0

(b) 1

(c) −1

(d) 3

Answer 20:

We have to find
Given
Using identity


Hence the value of
The correct choice is (d).

Question 21:

If a^1/3 + b^1/3 + c^1/3 = 0, then

(a) a + b + c = 0

(b) (a + b + c)³ =27abc

(c) a + b + c = 3abc

(d) a³ + b³ + c³ = 0

Answer 21:

Given
Using identity we get
Here

Taking Cube on both sides we get,

Hence the value of is
The correct choice is .

Question 22:

If a + b + c = 9 and ab + bc + ca =23, then a³ + b³ + c³ − 3abc =

(a) 108

(b) 207

(c) 669

(d) 729

Answer 22:

We have to find the value of
Given
Using identity we get,

By transposing +46 to left hand side we get,

Using identity

The value of is
Hence the correct choice is .

Question 23:

$\frac{(a^2-b^2{)}^3+(b^2-c^2{)}^3+(c^2-a^2{)}^3}{(a-b{)}^3 + (b-c{)}^3 + (c-a{)}^3}=$

(a) 3(a + b) ( b+ c) (c + a)

(b) 3(a − b) (b − c) (c − a)

(c) (a − b) (b − c) (c − a)

(d) none of these

Answer 23:

We have to find the value of
Using Identity we get,
 

Hence the value of is
The correct choice is .

Question 24:

The product (a + b) (a − b) (a² − ab + b²) (a² + ab + b²) is equal to

(a) a⁶ + b⁶

(b) a⁶ − b⁶

(c) a³ − b³

(d) a³ + b³

Answer 24:

We have to find the product of
Using identity 


We can rearrange as 

Again using the identity
Here

Hence the product of is
The correct choice is .

Question 25:

The product (x²−1) (x⁴ + x² + 1) is equal to

(a) x⁸ − 1

(b) x⁸ + 1

(c) x⁶ − 1

(d) x⁶ + 1

Answer 25:

We have to find the product of
Using identity
Here

Hence the product value of is
The correct alternate is .

Question 26:

If $\frac{a}{b}+\frac{b}{a}= 1$, then a³ + b³ =

(a) 1

(b) −1

(c) $\frac{1}{2}$

(d) 0

Answer 26:

Given


Using identity we get,

Hence the value of is .
The correct choice is (d).

Question 27:

If 49a² − b = $\left(7a+\frac{1}{2}\right) \left(7a-\frac{1}{2}\right)$, then the value of b is

(a) 0

(b) $\frac{1}{4}$

(c) $\frac{1}{\sqrt{2}}$

(d) $\frac{1}{2}$

Answer 27:

We have to find the value of b
Given
Using identity
We get

Equating ‘b’ on both sides we get 

Hence the value of b is
The correct choice is .

Question 28:

One of the factors of (25x² – 1) + (1 + 5x)² is
(a) 5 + x
(b) 5 – x
(c) 5x – 1
(d) 10x

Answer 28:

$$\begin{gathered} \left(25x^2-1\right)+{\left(1+5x\right)}^2 \\ =\left({\left(5x\right)}^2-{\left(1\right)}^2\right)+{\left(1+5x\right)}^2 \\ =\left(5x-1\right)\left(5x+1\right)+\left(1+5x\right)\left(1+5x\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^2-b^2=\left(a+b\right)\left(a-b\right)\right) \\ =\left(5x+1\right)\left(5x-1+1+5x\right) \\ =\left(5x+1\right)\left(10x\right) \\ \mathrm{Therefore}, \left(25x^2-1\right)+{\left(1+5x\right)}^2 \mathrm{has} \mathrm{two} \mathrm{factors} \left(5x+1\right) \mathrm{and} \left(10x\right). \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{d}\right). \end{gathered}$$

Question 29:

^{If $9x^2-b=\left(3x+\frac{1}{2}\right) \left(3x-\frac{1}{2}\right)$,} then the value of b is
(a) 0

(b) $\frac{1}{\sqrt{2}}$

(c) $\frac{1}{4}$

(d) $\frac{1}{2}$
 

Answer 29:

$$\begin{gathered} \mathrm{Given}: \\ 9x^2-b=\left(3x+\frac{1}{2}\right)\left(3x-\frac{1}{2}\right) \\ 9x^2-b=\left(3x+\frac{1}{2}\right)\left(3x-\frac{1}{2}\right) \\ \Rightarrow 9x^2-b={\left(3x\right)}^2-{\left(\frac{1}{2}\right)}^2 \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^2-b^2=\left(a+b\right)\left(a-b\right)\right) \\ \Rightarrow 9x^2-b=9x^2-\frac{1}{4} \\ \Rightarrow -b=-\frac{1}{4} \\ \Rightarrow b=\frac{1}{4} \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{c}\right). \end{gathered}$$

Question 30:

The coefficient of x in (x + 3)³ is
(a) 1
(b) 9
(c) 18
(d) 27

Answer 30:

$$\begin{gathered} {\left(x+3\right)}^3 \\ ={\left(x\right)}^3+{\left(3\right)}^3+3\left(x\right)\left(3\right)\left(x+3\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: {\left(a+b\right)}^3=a^3+b^3+3ab\left(a+b\right)\right) \\ =x^3+27+9x\left(x+3\right) \\ =x^3+27+9x^2+27x \\ =x^3+9x^2+27x+27 \\ \mathrm{Thus}, \mathrm{the} \mathrm{coefficient} \mathrm{of} x \mathrm{is} 27. \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{d}\right). \end{gathered}$$

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Question 31:

The value of 249² – 248² is
(a) 1
(b) 477
(c) 487
(d) 497

Answer 31:

$$\begin{gathered} {\left(249\right)}^2-{\left(248\right)}^2 \\ =\left(249+248\right)\left(249-248\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^2-b^2=\left(a+b\right)\left(a-b\right)\right) \\ =\left(497\right)\left(1\right) \\ =497 \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{d}\right). \end{gathered}$$

Question 32:

Which of the following is a factor of (x + y)³ – (x³ + y³)?
(a) x² + 2xy + y²
(b) x² – xy + y²
(c) xy²
(d) 3xy

Answer 32:

$$\begin{gathered} {\left(x+y\right)}^3-\left(x^3+y^3\right) \\ ={\left(x\right)}^3+{\left(y\right)}^3+3\left(x\right)\left(y\right)\left(x+y\right)-\left(x^3+y^3\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: {\left(a+b\right)}^3=a^3+b^3+3ab\left(a+b\right)\right) \\ =x^3+y^3+3xy\left(x+y\right)-x^3-y^3 \\ =3xy\left(x+y\right) \\ \mathrm{Thus}, {\left(x+y\right)}^3-\left(x^3+y^3\right) \mathrm{has} \mathrm{two} \mathrm{factors} \left(3xy\right) \mathrm{and} \left(x+y\right). \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{d}\right). \end{gathered}$$

Question 33:

If $\frac{x}{y}+\frac{y}{x}=-1\left(x,y\ne 0\right)$, the value of x³ – y³ is
(a) 1
(b) –1
(c) 0
(d) $\frac{1}{2}$

Answer 33:

$$\begin{gathered} \mathrm{Given}: \\ \frac{x}{y}+\frac{y}{x}=-1 \\ \Rightarrow \frac{x^2+y^2}{xy}=-1 \\ \Rightarrow x^2+y^2=-xy \\ \Rightarrow x^2+y^2+xy=0 ...\left(1\right) \\ \mathrm{Now}, \\ \left(x^3-y^3\right) \\ =\left(x-y\right)\left(x^2+y^2+xy\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^3-b^3=\left(a-b\right)\left(a^2+b^2+ab\right)\right) \\ =\left(x-y\right)\times 0 \left(\mathrm{From} \left(1\right)\right) \\ =0 \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{c}\right). \end{gathered}$$

Question 34:

If $\frac{x}{y}+\frac{y}{x}=1\left(x, y\ne 0\right)$, the value of x³ + y³ is
(a) 1
(b) –1
(c) 0
(d) $-\frac{1}{2}$

Answer 34:

$$\begin{gathered} \mathrm{Given}: \\ \frac{x}{y}+\frac{y}{x}=1 \\ \Rightarrow \frac{x^2+y^2}{xy}=1 \\ \Rightarrow x^2+y^2=xy \\ \Rightarrow x^2+y^2-xy=0 ...\left(1\right) \\ \mathrm{Now}, \\ \left(x^3+y^3\right) \\ =\left(x+y\right)\left(x^2+y^2-xy\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\right) \\ =\left(x+y\right)\times 0 \left(\mathrm{From} \left(1\right)\right) \\ =0 \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{c}\right). \end{gathered}$$

Question 35:

If x² + y² + xy = 1 and x + y = 2, then xy =

(a) –3

(b) 3

(c) $-\frac{3}{2}$

(d) 0

Answer 35:

$$\begin{gathered} \mathrm{Given}: \\ {x}^2+y^2+xy=1 ...\left(1\right) \\ x+y=2 ...\left(2\right) \\ \mathrm{Now}, \\ x+y=2 \\ \mathrm{Squaring} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ \Rightarrow {\left(x+y\right)}^2=2^2 \\ \Rightarrow x^2+y^2+2xy=4 \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: {\left(a+b\right)}^2=a^2+b^2+2ab\right) \\ \Rightarrow x^2+y^2+xy+xy=4 \\ \Rightarrow 1+xy=4 \left(\mathrm{From} \left(1\right)\right) \\ \Rightarrow xy=4-1 \\ \Rightarrow xy=3 \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{b}\right). \end{gathered}$$

Question 36:

If a, b, c are natural numbers such that a² + b² + c² = 29 and ab + bc + ca = 26, and a + b + c = ______.
(a) 9
(b) 6
(c) 7
(d) 10

Answer 36:

$$\begin{gathered} \mathrm{Given}: \\ {a}^2+b^2+c^2=29 ...\left(1\right) \\ ab+bc+ca=26 ...\left(2\right) \\ \mathrm{Now}, \\ {\left(a+b+c\right)}^2 \\ =a^2+b^2+c^2+2\left(ab+bc+ca\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: {\left(a+b+c\right)}^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\right) \\ =29+2\left(26\right) \left(\mathrm{From} \left(1\right) \mathrm{and} \left(2\right)\right) \\ =29+52 \\ =81 \\ \mathrm{Since}, {\left(a+b+c\right)}^2=81 \mathrm{and} a, b, c \mathrm{are} \mathrm{natural} \mathrm{numbers} \\ \mathrm{Therefore}, a+b+c=9. \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{a}\right). \end{gathered}$$

Question 37:

If $2x+\frac{y}{3}=12 \mathrm{and} xy = 30, \mathrm{then} 8x^3+\frac{y^3}{27}=\_\_\_\_\_\_\_$
(a) 1008
(b) 168
(c) 106
(d) none of these

Answer 37:

$$\begin{gathered} \mathrm{Given}: \\ 2x+\frac{y}{3}=12 ...\left(1\right) \\ xy=30 ...\left(2\right) \\ \mathrm{Now}, \\ 2x+\frac{y}{3}=12 \\ \mathrm{Taking} \mathrm{cube} \mathrm{on} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ \Rightarrow {\left(2x+\frac{y}{3}\right)}^3={12}^3 \\ \Rightarrow {\left(2x\right)}^3+{\left(\frac{y}{3}\right)}^3+3\left(2x\right)\left(\frac{y}{3}\right)\left(2x+\frac{y}{3}\right)=1728 \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: {\left(a+b\right)}^3=a^3+b^3+3ab\left(a+b\right)\right) \\ \Rightarrow 8x^3+\frac{y^3}{27}+2xy\left(2x+\frac{y}{3}\right)=1728 \\ \Rightarrow 8x^3+\frac{y^3}{27}+2\times 30\times 12=1728 \left(\mathrm{From} \left(1\right) \mathrm{and} \left(2\right)\right) \\ \Rightarrow 8x^3+\frac{y^3}{27}+720=1728 \\ \Rightarrow 8x^3+\frac{y^3}{27}=1728-720 \\ \Rightarrow 8x^3+\frac{y^3}{27}=1008 \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{a}\right). \end{gathered}$$