OP Malhotra Class 9 Maths Solutions Chapter 3 Expansions Chapter Test

S Chand Class 9 ICSE Maths Solutions Chapter 3 Expansions Chapter Test

Question 1. The coefficient of x in the product (2 - 3x) (5 - 2x) is
(a) 19
(b) - 19
(c) 15
(d) 6
Answer: (b) - 19
Multiply the given expressions:
\( (2 - 3x) (5 - 2x) = 2(5) + 2(-2x) - 3x(5) - 3x(-2x) \)
\( = 10 - 4x - 15x + 6x^2 \)
\( = 10 - 19x + 6x^2 \)
The term with x is \( -19x \).
So, the coefficient of x is \( -19 \). The coefficient is just the number part that multiplies the variable.
In simple words: When you multiply the two brackets, the numbers that combine with 'x' add up to -19. This -19 is the coefficient of x.

🎯 Exam Tip: Always pay attention to the signs when multiplying terms, especially with negative numbers, to avoid errors in the final coefficient.

Question 2. If \( 3x^4 + kx^2 - 8 = (3x^2 - 2) (x^2 + 4) \) for all x, then the value of k is:
(a) - 2
(b) 12
(c) 10
(d) - 8
Answer: (c) 10
We are given the equation \( 3x^4 + kx^2 - 8 = (3x^2 - 2) (x^2 + 4) \).
First, expand the right side of the equation:
\( (3x^2 - 2) (x^2 + 4) = 3x^2(x^2) + 3x^2(4) - 2(x^2) - 2(4) \)
\( = 3x^4 + 12x^2 - 2x^2 - 8 \)
\( = 3x^4 + 10x^2 - 8 \)
Now, compare this expanded form with the left side of the given equation:
\( 3x^4 + kx^2 - 8 = 3x^4 + 10x^2 - 8 \)
By comparing the coefficients of \( x^2 \) on both sides, we find that \( k \) must be 10. This method works because the equation holds for all values of x.
In simple words: To find 'k', first multiply out the right side of the equation. Then, compare the numbers that are with \( x^2 \) on both sides. These numbers must be the same, so k is 10.

🎯 Exam Tip: When comparing polynomial equations, ensure all terms are expanded and simplified before equating coefficients of like powers of the variable.

Question 3. The coefficient of \( x^2 \) in \( (3x + x^3) (x + \frac{1}{x}) \) is
(a) 3
(b) 1
(c) 4
(d) 2
Answer: (c) 4
We need to find the coefficient of \( x^2 \) in the product \( (3x + x^3) (x + \frac{1}{x}) \).
Let's multiply the terms:
\( (3x + x^3) (x + \frac{1}{x}) = 3x(x) + 3x(\frac{1}{x}) + x^3(x) + x^3(\frac{1}{x}) \)
\( = 3x^2 + 3 + x^4 + x^2 \)
Combine the \( x^2 \) terms:
\( = 3x^2 + x^2 + 3 + x^4 \)
\( = 4x^2 + x^4 + 3 \)
The terms containing \( x^2 \) sum up to \( 4x^2 \). Therefore, the coefficient of \( x^2 \) is 4. This expansion involves basic exponent rules like \( x^a \cdot x^b = x^{a+b} \) and \( \frac{x^a}{x^b} = x^{a-b} \).
In simple words: First, multiply everything in the two brackets. Then, look for all the terms that have \( x^2 \). Add the numbers in front of these \( x^2 \) terms to get the final coefficient.

🎯 Exam Tip: Remember that \( x \cdot \frac{1}{x} \) simplifies to 1. Be careful when distributing terms to make sure all multiplications are done correctly.

Question 4. If a = 3 + b, prove that \( a^3 - b^3 - 9ab = 27 \).
Answer:
We are given \( a = 3 + b \).
We can rearrange this to \( a - b = 3 \).
Now, cube both sides of this equation:
\( (a - b)^3 = (3)^3 \)
Using the algebraic identity \( (x - y)^3 = x^3 - y^3 - 3xy(x - y) \):
\( a^3 - b^3 - 3ab(a - b) = 27 \)
We know that \( (a - b) = 3 \), so substitute this value into the equation:
\( a^3 - b^3 - 3ab(3) = 27 \)
\( a^3 - b^3 - 9ab = 27 \)
This proves the given statement. This identity is very useful for problems involving cubes of binomials.
In simple words: Start with \( a = 3 + b \), which means \( a - b = 3 \). If you cube both sides of this new equation and use a special formula for \( (a-b)^3 \), you can show that the final expression matches 27.

🎯 Exam Tip: Knowing common algebraic identities like \( (a-b)^3 \) is crucial for solving such proofs. Remember to substitute known values back into the expanded identity.

Question 5. Simplify: \( \frac{\left(a^2-b^2\right)^3+\left(b^2-c^2\right)^3+\left(c^2-a^2\right)^3}{(a-b)^3+(b-c)^3+(c-a)^3} \)
Answer:
We use the algebraic identity: If \( x + y + z = 0 \), then \( x^3 + y^3 + z^3 = 3xyz \).

For the numerator:
Let \( x = a^2 - b^2 \), \( y = b^2 - c^2 \), and \( z = c^2 - a^2 \).
Now, check their sum:
\( x + y + z = (a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2) = 0 \)
Since the sum is 0, the numerator \( (a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3 \) simplifies to \( 3(a^2-b^2)(b^2-c^2)(c^2-a^2) \).

For the denominator:
Let \( p = a - b \), \( q = b - c \), and \( r = c - a \).
Now, check their sum:
\( p + q + r = (a - b) + (b - c) + (c - a) = 0 \)
Since the sum is 0, the denominator \( (a-b)^3+(b-c)^3+(c-a)^3 \) simplifies to \( 3(a-b)(b-c)(c-a) \).

Now substitute these back into the original expression:
\( \frac{3(a^2-b^2)(b^2-c^2)(c^2-a^2)}{3(a-b)(b-c)(c-a)} \)
We can cancel out the 3 from the numerator and denominator.
\( = \frac{(a^2-b^2)(b^2-c^2)(c^2-a^2)}{(a-b)(b-c)(c-a)} \)
Use the difference of squares identity: \( x^2 - y^2 = (x - y)(x + y) \).
\( = \frac{(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)}{(a-b)(b-c)(c-a)} \)
Cancel the common terms \( (a-b), (b-c), (c-a) \) from numerator and denominator.
\( = (a+b)(b+c)(c+a) \)
This algebraic trick is very powerful for simplifying expressions that look complicated at first.
In simple words: If you have three things that add up to zero, and you need to add their cubes, the answer is three times their product. Apply this rule to both the top and bottom parts of the fraction. Then, use the difference of squares formula to simplify even more by canceling out common terms.

🎯 Exam Tip: Always look for conditions where \( x+y+z=0 \) when dealing with sums of cubes, as the identity \( x^3+y^3+z^3=3xyz \) can significantly simplify the problem.

Question 6. If \( a + \frac{1}{(a+2)} = 0 \), then the value of \( (a + 2)^3 + \frac{1}{(a+2)^3} \) is
(a) 6
(b) 4
(c) 3
(d) 2
Answer: (d) 2
Given the equation \( a + \frac{1}{(a+2)} = 0 \).
Multiply the entire equation by \( (a+2) \) to clear the denominator:
\( a(a+2) + \frac{1}{(a+2)}(a+2) = 0 \cdot (a+2) \)
\( a^2 + 2a + 1 = 0 \)
This is a perfect square trinomial, which can be factored as:
\( (a+1)^2 = 0 \)

\( \implies \) \( a+1 = 0 \)

\( \implies \) \( a = -1 \)

Now we need to find the value of \( (a + 2)^3 + \frac{1}{(a+2)^3} \).
Substitute \( a = -1 \) into this expression:
\( (-1 + 2)^3 + \frac{1}{(-1+2)^3} \)
\( = (1)^3 + \frac{1}{(1)^3} \)
\( = 1 + 1 \)
\( = 2 \)
This type of problem often leads to a simple numerical answer after the initial algebraic manipulation.
In simple words: First, solve the given equation to find the value of 'a'. You will find that 'a' is -1. Then, put this value of 'a' into the expression you need to solve. It will simplify to \( 1^3 + \frac{1}{1^3} \), which gives 2.

🎯 Exam Tip: When a problem involves an expression like \( x + \frac{1}{x} \) or similar, look for ways to simplify the initial condition to find the base value, then substitute. Factoring perfect squares is a key skill here.

Question 7. If \( a + \frac { 1 }{ a } + 2 = 0 \), then the value of \( a^{37}-\frac{1}{a^{100}} \) is
(a) 0
(b) – 2
(c) 1
(d) 2
Answer: (b) - 2
Given the equation \( a + \frac{1}{a} + 2 = 0 \).
Multiply the entire equation by \( a \) to clear the denominator:
\( a(a) + \frac{1}{a}(a) + 2(a) = 0 \cdot a \)
\( a^2 + 1 + 2a = 0 \)
Rearrange the terms:
\( a^2 + 2a + 1 = 0 \)
This is a perfect square trinomial, which can be factored as:
\( (a+1)^2 = 0 \)

\( \implies \) \( a+1 = 0 \)

\( \implies \) \( a = -1 \)

Now we need to find the value of \( a^{37} - \frac{1}{a^{100}} \).
Substitute \( a = -1 \) into this expression:
\( (-1)^{37} - \frac{1}{(-1)^{100}} \)
Remember that an odd power of -1 is -1, and an even power of -1 is 1.
So, \( (-1)^{37} = -1 \) and \( (-1)^{100} = 1 \).
\( = -1 - \frac{1}{1} \)
\( = -1 - 1 \)
\( = -2 \)
Understanding how negative numbers behave with exponents is a fundamental concept here.
In simple words: First, solve the initial equation to find 'a'. It will turn out to be -1. Then, put this -1 into the powers of 'a' in the second part of the question. Remember that -1 raised to an odd power is -1, and -1 raised to an even power is +1. Calculate these values and subtract them to get the final answer.

🎯 Exam Tip: Always carefully evaluate powers of negative numbers. An odd exponent keeps the negative sign, while an even exponent makes the result positive.

Question 8. If \( (a - 1)^2 + (b + 2)^2 + (c + 1)^2 = 0 \) then the value of \( 2a - 3b + 7c \) is
(a) 12
(b) 3
(c) - 11
(d) 1
Answer: (d) 1
We are given the equation \( (a - 1)^2 + (b + 2)^2 + (c + 1)^2 = 0 \).
For the sum of squares of real numbers to be zero, each individual square term must be zero. This is because squares of real numbers are always non-negative (0 or positive).

So, we must have:
\( (a - 1)^2 = 0 \implies a - 1 = 0 \implies a = 1 \)
\( (b + 2)^2 = 0 \implies b + 2 = 0 \implies b = -2 \)
\( (c + 1)^2 = 0 \implies c + 1 = 0 \implies c = -1 \)

Now that we have the values for a, b, and c, we can find the value of the expression \( 2a - 3b + 7c \).
Substitute the values:
\( 2(1) - 3(-2) + 7(-1) \)
\( = 2 - (-6) + (-7) \)
\( = 2 + 6 - 7 \)
\( = 8 - 7 \)
\( = 1 \)
This principle is useful in many algebraic contexts, showing a unique solution for variables in such equations.
In simple words: When you add up squares of numbers and the total is zero, it means each square must be zero by itself. So, find 'a', 'b', and 'c' by making each part in the brackets equal to zero. Then, put these numbers into the expression \( 2a - 3b + 7c \) to get the final answer.

🎯 Exam Tip: Remember the fundamental property that the sum of squares of real numbers is zero if and only if each term is zero. This simplifies complex-looking equations significantly.

Question 9. If \( ax + by = 3 \), \( bx - ay = 4 \) and \( x^2 + y^2 = 1 \), then the value of \( a^2 + b^2 \) is
(a) – 1
(b) – 25
(c) 1
(d) 25
Answer: (d) 25
We are given three equations:
1. \( ax + by = 3 \)
2. \( bx - ay = 4 \)
3. \( x^2 + y^2 = 1 \)

Square both equation (1) and equation (2):
From (1): \( (ax + by)^2 = 3^2 \)
\( a^2x^2 + b^2y^2 + 2abxy = 9 \)

From (2): \( (bx - ay)^2 = 4^2 \)
\( b^2x^2 + a^2y^2 - 2abxy = 16 \)

Now, add these two squared equations:
\( (a^2x^2 + b^2y^2 + 2abxy) + (b^2x^2 + a^2y^2 - 2abxy) = 9 + 16 \)
Notice that the \( 2abxy \) and \( -2abxy \) terms cancel out:
\( a^2x^2 + b^2y^2 + b^2x^2 + a^2y^2 = 25 \)

Group the terms by \( a^2 \) and \( b^2 \):
\( (a^2x^2 + a^2y^2) + (b^2x^2 + b^2y^2) = 25 \)
Factor out \( a^2 \) from the first group and \( b^2 \) from the second group:
\( a^2(x^2 + y^2) + b^2(x^2 + y^2) = 25 \)
Now, factor out the common term \( (x^2 + y^2) \):
\( (a^2 + b^2)(x^2 + y^2) = 25 \)

From equation (3), we know that \( x^2 + y^2 = 1 \). Substitute this value:
\( (a^2 + b^2)(1) = 25 \)

\( \implies \) \( a^2 + b^2 = 25 \)
This elegant solution shows how combining equations and using algebraic identities can simplify seemingly complex problems.
In simple words: Square the first two equations and then add them together. You will notice some terms cancel out. Then, use the third given equation (\( x^2 + y^2 = 1 \)) to easily find the value of \( a^2 + b^2 \).

🎯 Exam Tip: When given equations with sums of squares or products, consider squaring and adding/subtracting them to eliminate terms or reveal desired expressions.

Question 10. If \( p + q = 10 \) and \( pq = 5 \), then the numerical value of \( \frac{p}{q}+\frac{q}{p} \) will be:
(a) 22
(b) 18
(c) 16<
(d) 20
Answer: (b) 18
We are given:
1. \( p + q = 10 \)
2. \( pq = 5 \)

We need to find the value of \( \frac{p}{q} + \frac{q}{p} \).
First, combine the fractions by finding a common denominator:
\( \frac{p}{q} + \frac{q}{p} = \frac{p \cdot p + q \cdot q}{pq} = \frac{p^2 + q^2}{pq} \)

Now, we need to find the value of \( p^2 + q^2 \). We know the algebraic identity:
\( (p + q)^2 = p^2 + q^2 + 2pq \)
Rearrange this identity to solve for \( p^2 + q^2 \):
\( p^2 + q^2 = (p + q)^2 - 2pq \)

Substitute the given values \( p + q = 10 \) and \( pq = 5 \) into this rearranged formula:
\( p^2 + q^2 = (10)^2 - 2(5) \)
\( p^2 + q^2 = 100 - 10 \)
\( p^2 + q^2 = 90 \)

Finally, substitute the values of \( p^2 + q^2 \) and \( pq \) into the combined fraction expression:
\( \frac{p^2 + q^2}{pq} = \frac{90}{5} \)
\( = 18 \)
This problem shows how fundamental algebraic identities are used to solve problems efficiently.
In simple words: To solve this, first combine the two fractions into one by finding a common bottom part. Then, use the formula for \( (p+q)^2 \) to find \( p^2 + q^2 \). Once you have that, divide it by \( pq \) to get your final answer.

🎯 Exam Tip: Always look for ways to express \( p^2+q^2 \) using \( (p+q) \) and \( pq \), as these are often given values in such problems. The identity \( p^2+q^2 = (p+q)^2 - 2pq \) is a key shortcut.