Class 9 Maths: Algebraic Identities | Practice Extra Questions
Q. 1 If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.
Check Solution
Ans: B
Explanation: We are given that $a + b + c = 0$. We want to find the value of $a^3 + b^3 + c^3$.
We can use the identity:
$a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca)$
Since $a + b + c = 0$, we have
$a^3 + b^3 + c^3 – 3abc = (0)(a^2 + b^2 + c^2 – ab – bc – ca)$
$a^3 + b^3 + c^3 – 3abc = 0$
Therefore, $a^3 + b^3 + c^3 = 3abc$.
Correct Option: B
Q. 2 What is the square root of $y^2 + \frac{1}{y^2} + 2$?
Check Solution
Ans: A
Explanation: We can rewrite the expression inside the square root as a perfect square trinomial. Notice that $y^2 + \frac{1}{y^2} + 2$ is similar to the form $a^2 + 2ab + b^2$, which can be factored to $(a+b)^2$.
In our case, we can let $a = y$ and $b = \frac{1}{y}$. Then $a^2 = y^2$, $b^2 = \frac{1}{y^2}$ and $2ab = 2 \cdot y \cdot \frac{1}{y} = 2$.
Therefore, $y^2 + \frac{1}{y^2} + 2 = (y + \frac{1}{y})^2$.
Taking the square root, we get $\sqrt{(y + \frac{1}{y})^2} = |y + \frac{1}{y}|$. Since the options do not include an absolute value, the assumption is made that $y+\frac{1}{y}$ is positive. Therefore, the answer is $y + \frac{1}{y}$.
Correct Option: A
Q. 3 If $49x^2 – b = (7x + 1/2)(7x – 1/2)$, then the value of b is ______ .
Check Solution
Ans: C
Explanation: We are given the equation $49x^2 – b = (7x + 1/2)(7x – 1/2)$.
We can expand the right side of the equation using the difference of squares formula, which states that $(a+c)(a-c) = a^2 – c^2$. In this case, $a = 7x$ and $c = 1/2$.
So, $(7x + 1/2)(7x – 1/2) = (7x)^2 – (1/2)^2 = 49x^2 – 1/4$.
Now we have $49x^2 – b = 49x^2 – 1/4$.
For the two sides of the equation to be equal for all values of x, the coefficients of the corresponding terms must be equal. Therefore, we must have $-b = -1/4$.
Multiplying both sides by -1, we get $b = 1/4$.
Correct Option: C
Q. 4 Find the constant term in the expansion of $(x+3)^3$.
Check Solution
Ans: B
Explanation: The binomial theorem states that $(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$. In our case, $a=x$, $b=3$, and $n=3$. We want the constant term, which means the term without any $x$ variables. This will happen when the power of $x$ is 0.
So we have:
$\binom{3}{k} x^{3-k} 3^k$
For the constant term, we need $3-k = 0$, so $k=3$.
Therefore, the constant term is $\binom{3}{3} x^{3-3} 3^3 = 1 \cdot x^0 \cdot 27 = 27$.
Alternatively, expanding directly:
$(x+3)^3 = (x+3)(x+3)(x+3) = (x^2 + 6x + 9)(x+3) = x^3 + 6x^2 + 9x + 3x^2 + 18x + 27 = x^3 + 9x^2 + 27x + 27$. The constant term is 27.
Correct Answer: 27
Q. 5 Factorize the following:
$9x^2 – 12x + 4$
Check Solution
Ans: D
Solution:
$(3x – 2)^2$
Answer: $(3x – 2)^2$
Q. 6 Factorise the following:
$4x^2 – 8x + 4$
Check Solution
Ans: B
Solution:
$4x^2 – 8x + 4 = 4(x^2 – 2x + 1) = 4(x-1)^2$
Answer: $4(x-1)^2$
Q. 7 The common quantity that must be added to each term of $a^2:b^2$ to make it equal to $a:b$ is:
Check Solution
Ans: A
Explanation: Let $x$ be the common quantity that must be added to each term of $a^2:b^2$ to make it equal to $a:b$.
Then, we can write the equation as:
$\frac{a^2 + x}{b^2 + x} = \frac{a}{b}$
Cross-multiplying, we get:
$b(a^2 + x) = a(b^2 + x)$
$a^2b + bx = ab^2 + ax$
$bx – ax = ab^2 – a^2b$
$x(b – a) = ab(b – a)$
If $a \ne b$, then we can divide both sides by $(b – a)$:
$x = ab$
If $a = b$, then the original ratio becomes $a^2:a^2$ which is $1:1$ and the target ratio is $a:a$ which is also $1:1$. The question doesn’t specify if $a\ne b$.
If $a=b$ then both ratios are equal regardless of $x$. Thus, we can conclude that the answer is $ab$.
Correct Option: A
Q. 8 What is the value of $101^2 – 99^2$?
Check Solution
Ans: B
Solution:
Using the difference of squares factorization, $a^2 – b^2 = (a+b)(a-b)$.
$101^2 – 99^2 = (101+99)(101-99) = (200)(2) = 400$.
Answer: 400
Q. 9 Factorise $x^2 – 4 – 4b – b^2$.
Check Solution
Ans: B
Solution:
$x^2 – 4 – 4b – b^2 = x^2 – (4 + 4b + b^2) = x^2 – (2+b)^2 = (x – (2+b))(x + (2+b)) = (x-2-b)(x+2+b)$
Answer: $(x-2-b)(x+2+b)$
Q. 10 Without actually calculating the cubes, evaluate the expression $(30)^3 + (-18)^3 + (-12)^3$.
Check Solution
Ans: C
Explanation: We can use the identity: If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.
In this case, let $a = 30$, $b = -18$, and $c = -12$. Then, $a + b + c = 30 + (-18) + (-12) = 30 – 18 – 12 = 0$.
Therefore, $(30)^3 + (-18)^3 + (-12)^3 = 3(30)(-18)(-12) = 3(30)(216) = 90(216) = 19440$.
Correct Answer: 19440
Q. 11 What is the value of
$\frac{a^3 + b^3 + c^3 – 3abc}{ab + bc + ca – a^2 – b^2 – c^2},$
when $a = -5$, $b = -6$, and $c = 10$?
Check Solution
Ans: D
Explanation:
We can factor the numerator using the identity:
$a^3 + b^3 + c^3 – 3abc = (a+b+c)(a^2 + b^2 + c^2 – ab – bc – ca).$
The denominator can be rewritten as:
$ab + bc + ca – a^2 – b^2 – c^2 = -(a^2 + b^2 + c^2 – ab – bc – ca).$
Therefore, the given expression simplifies to:
$\frac{a^3 + b^3 + c^3 – 3abc}{ab + bc + ca – a^2 – b^2 – c^2} = \frac{(a+b+c)(a^2 + b^2 + c^2 – ab – bc – ca)}{-(a^2 + b^2 + c^2 – ab – bc – ca)}.$
If $a^2 + b^2 + c^2 – ab – bc – ca \neq 0$, we can cancel the terms in the numerator and denominator, which yields:
$\frac{(a+b+c)(a^2 + b^2 + c^2 – ab – bc – ca)}{-(a^2 + b^2 + c^2 – ab – bc – ca)} = -(a+b+c).$
Substituting $a = -5$, $b = -6$, and $c = 10$, we get
$-(a+b+c) = -(-5 – 6 + 10) = -(-11 + 10) = -(-1) = 1.$
Now we check if $a^2 + b^2 + c^2 – ab – bc – ca \neq 0$. We have:
$a^2 + b^2 + c^2 = (-5)^2 + (-6)^2 + 10^2 = 25 + 36 + 100 = 161$
$ab + bc + ca = (-5)(-6) + (-6)(10) + (10)(-5) = 30 – 60 – 50 = -80$
$a^2 + b^2 + c^2 – ab – bc – ca = 161 – (-80) = 161 + 80 = 241 \neq 0$.
Thus the cancellation is valid.
Correct Answer: 1
Q. 12 Which of the following is a factor of $(a+b)^3 – (a^3 + b^3)$?
Check Solution
Ans: A
Solution:
$(a+b)^3 – (a^3 + b^3) = a^3 + 3a^2b + 3ab^2 + b^3 – a^3 – b^3 = 3a^2b + 3ab^2 = 3ab(a+b)$.
Factors are 3, a, b, (a+b).
Answer: 3
Q. 13 Factorize: $x^2 + y^2 + 2xy + 4xz + 4yz + 4z^2$
Check Solution
Ans: A
Solution:
$(x^2 + 2xy + y^2) + 4z(x+y) + 4z^2 = (x+y)^2 + 4z(x+y) + 4z^2 = (x+y+2z)^2$
Answer: $(x+y+2z)^2$
Q. 14 If $a + b + c = 0$, then $a^3 + b^3 + c^3$ is equal to ______ .
Check Solution
Ans: D
If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.
Explanation:
We know that $a^3 + b^3 + c^3 – 3abc = (a+b+c)(a^2 + b^2 + c^2 – ab – bc – ca)$.
Since $a+b+c = 0$, we have $a^3 + b^3 + c^3 – 3abc = 0$.
Therefore, $a^3 + b^3 + c^3 = 3abc$.
Q. 15 If $x = \frac{a-b}{a+b}$, $y = \frac{b-c}{b+c}$, $z = \frac{c-a}{c+a}$, and $a=2, b=3, c=5$, find the value of $\frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)}$.
Check Solution
Ans: B
Solution:
$x = \frac{2-3}{2+3} = -\frac{1}{5}$
$y = \frac{3-5}{3+5} = -\frac{2}{8} = -\frac{1}{4}$
$z = \frac{5-2}{5+2} = \frac{3}{7}$
$1+x = \frac{4}{5}$
$1+y = \frac{3}{4}$
$1+z = \frac{10}{7}$
$1-x = \frac{6}{5}$
$1-y = \frac{5}{4}$
$1-z = \frac{4}{7}$
$\frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)} = \frac{\frac{4}{5} \cdot \frac{3}{4} \cdot \frac{10}{7}}{\frac{6}{5} \cdot \frac{5}{4} \cdot \frac{4}{7}} = \frac{\frac{120}{140}}{\frac{120}{140}} = 1$
Answer: 1
Q. 16 One of the factors of $(25x^2 – 1) + (1 + 5x)^2$ is ______ .
Check Solution
Ans: D
Explanation: First, expand the expression:
$(25x^2 – 1) + (1 + 5x)^2 = 25x^2 – 1 + (1 + 10x + 25x^2) = 25x^2 – 1 + 1 + 10x + 25x^2 = 50x^2 + 10x$
Now, factor the expression:
$50x^2 + 10x = 10x(5x + 1)$
We can rewrite $5x+1$ as $5 – (-1+5x)$, so that doesn’t help us find a direct match. The factors of the original expression are 10x and 5x+1. Among the given options, only 10x is a factor. Let us examine other options. We can rewrite the expression as $10x(5x+1)$. Thus, comparing our findings with the options.
A. 5 + x – Not a factor.
B. $5 – x$ – Not a factor.
C. $5x – 1$ – Not a factor.
D. 10x – is a factor.
Correct Option: D
Q. 17 Factorise: $27x^3 + y^3 + z^3 – 9xyz$
Check Solution
Ans: B
Explanation:
We can use the identity: $a^3 + b^3 + c^3 – 3abc = (a+b+c)(a^2 + b^2 + c^2 – ab – bc – ca)$.
In our case, $a = 3x$, $b = y$, and $c = z$.
Then $27x^3 + y^3 + z^3 – 9xyz = (3x)^3 + y^3 + z^3 – 3(3x)(y)(z)$.
Applying the identity, we get $(3x + y + z)((3x)^2 + y^2 + z^2 – (3x)y – yz – z(3x))$.
Simplifying, we have $(3x + y + z)(9x^2 + y^2 + z^2 – 3xy – yz – 3xz)$.
Correct Answer: $(3x+y+z)(9x^2+y^2+z^2-3xy-yz-3xz)$
Q. 18 Factorise: $8x^3 + 27y^3 + z^3 – 18xyz$
Check Solution
Ans: A
Solution:
$8x^3 + 27y^3 + z^3 – 18xyz = (2x)^3 + (3y)^3 + z^3 – 3(2x)(3y)z$
Using the identity $a^3 + b^3 + c^3 – 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$= (2x+3y+z)((2x)^2 + (3y)^2 + z^2 – (2x)(3y) – (3y)z – z(2x))$
$= (2x+3y+z)(4x^2 + 9y^2 + z^2 – 6xy – 3yz – 2xz)$
Answer: $(2x+3y+z)(4x^2+9y^2+z^2-6xy-3yz-2xz)$
Q. 19 If $(x+1)(x+2)(x+3)(x+k) + 1$ is a perfect square, then the value of $k$ is?
Check Solution
Ans: A
Explanation: Let $P(x) = (x+1)(x+2)(x+3)(x+k) + 1$. We can rearrange the terms as:
$P(x) = [(x+1)(x+3)][(x+2)(x+k)] + 1$
$P(x) = (x^2 + 4x + 3)(x^2 + (k+2)x + 2k) + 1$.
Let $x^2 + 4x = y$. Then $x^2 + 4x + 3 = y+3$ and $x^2 + (k+2)x + 2k = x^2 + 4x + kx – 2x + 2k = y + (k-2)x + 2k$.
The given expression can be written as $(x+1)(x+3)(x+2)(x+k) + 1$. We can also pair the factors as $(x+1)(x+k)(x+2)(x+3) + 1 = (x^2 + (k+1)x + k)(x^2+5x+6) + 1$.
For $P(x)$ to be a perfect square, we can try to get the form $(ax^2+bx+c)^2$.
Consider another grouping: $[(x+1)(x+k)][(x+2)(x+3)]+1 = (x^2+(k+1)x+k)(x^2+5x+6)+1$.
If we rearrange the original expression as:
$[(x+1)(x+3)][(x+2)(x+k)] + 1 = (x^2+4x+3)(x^2+(k+2)x+2k) + 1$
Let us check if we can write $(x+1)(x+2)(x+3)(x+k)+1 = (ax^2+bx+c)^2$. Consider the first grouping:
$(x^2+4x+3)(x^2+(k+2)x+2k) + 1 = (x^2+4x+3)(x^2+(k+2)x+2k) + 1$
$(x^2+4x+3)(x^2+(k+2)x+2k)+1$
If $k=4$, we have
$(x^2+4x+3)(x^2+6x+8)+1 = x^4+10x^3+35x^2+50x+24+1=x^4+10x^3+35x^2+50x+25$
We have $(x^2+5x+5)^2=x^4+25x^2+25+10x^3+50x+10x^2 = x^4+10x^3+35x^2+50x+25$
So, if $k=4$, $(x+1)(x+2)(x+3)(x+4)+1 = (x^2+5x+5)^2$.
Correct Option: A
Q. 20 One of the factors of $(4x^2 – 9) + (3 + 4x)^2$ is ______ .
Check Solution
Ans: C
Solution:
$(4x^2 – 9) + (3 + 4x)^2 = 4x^2 – 9 + 9 + 24x + 16x^2 = 20x^2 + 24x = 4x(5x + 6)$.
The factors are $4x$ and $5x + 6$.
Answer: 5x + 6
Practice Extra Questions for Class 9 Maths