Algebra Important Questions: Definition and Solved Examples

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Short Answer Question [2 Marks]

Ques - Find the Angle between the Given Two Vectors 4i + 5j – k and 2i – j + k. 

Ans: Let, a = 4i + 5j – k and b = 2i – j + k

So, the dot product is,

a.b = (4i + 5j – k).(2i – j + k)

= (4)(2) + (5)(-1) + (-1)(1)

= 8 - 5 - 1

= 2

Ques- Simplify the expression: 12m² – 9m + 5m – 4m² – 7m + 10. 

Ans: Given that: 

12m² – 4m² +5m – 9m – 7m +10

= (12 – 4)m² + (5 – 9 – 7)m+10

=8m² + (-4-7)m + 10

= 8m² + 11m + 10

Ques- Find out the value of n at which the given two vectors A = (- 2n, – 3, – 2) and B = (8, 3, 2) are found to be the inverses of one another. 

Ans: The vector will be the inverses of each other when:

- 2n = – 8

n = 8/2

n = 4

Hence, when n = 4, the two vectors A and B will be the negatives of each other, as determined by comparing the components of the two vectors.

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Short Answer Questions [3 Marks]

Ques- If the quadratic equation is x² – 3x – m (m +3 ) = 0 and the value of m is constant. Find the root of the equation. 

Ans: x² – 3x – m(m+3) = 0

a = 1, b= -3, c = -m(m+3) 

D = (b² – 4ac)

So,

D= (-3)² -4 (1) = 9 + 4m (m+3)

= 4m² +12m +9

= (2m+3)²

(α,ß) = [-3±√(2m+3)²]/2x1x – m(m+3)

= (α,ß) = (m+3,-m)

Ques- The quadratic equations x² – ax + b = 0 and x² – px + q = zero have a commonplace root and the second equation has equal roots, displaying that b + q = ap/2. 

Ans: With the aid of thinking about α and β to be the roots of equation (i) and α to be the common root, we can clear up the problem via the use of the sum fabricated from the roots system.

The given quadratic equations are

x² – ax + b = zero ………. (i)

x² – px + q = 0 ………..(ii)

From equation (i), α + β = a, α = b

From equation (ii), 2α = p, α² = q

b + q = αβ + α² = α (α + β) = ap/2

Ques- Compute the angle between two vectors given that they are unit vectors p and q where p x q = 1 / 3i + 1 / 4j.

Ans: |a| = |b| = 1

Where,

| a x b | = √ ( (1 / 3)² + ( 1 / 4)²) = 1 / 5

Put the given values in the above formula, we get

| a x b | = |a| |b| sin θ

1 / 5 = (1) (1) sin θ

θ = sin^-1 (1/ 5)

θ = 30º

Ques- Solve for 4x³ + 24x² – 64x = 0. 

Ans: 4x(x²+ 6x – 16) = 0

∴ 4x = 0 or x²+ 6x – 16 = 0 (4x = 0, x = 0)

∴ x² + 6x – 16 = 0

∴ x² + 8x − 2x – 16 = 0 

∴ (x − 2)(x + 8) = 0

∴ x = 2 or x = −8 

∴ x = −8, 0 or 2

Ques- Solve for x⁵– 41x³+ 400x = 0. 

Ans: x(x⁴ - 41x² + 400) = 0

∴ x = 0 or x⁴ – 41x²+ 400 = 0

Let x²= a

∴ a² – 41a + 400 = 0

∴ a² – 25a − 16a + 400 = 0

∴ (a – 16) (a – 25) = 0

∴ a = 16 or a = 25

∴ x²= 16 or x2= 25

∴ x = ±4 or x = ±5

∴ x = −4 or -5 or 0 or 5 or 4

Ques- Find the unit vector of 2i + j + 4k. 

Ans: Let’s assume the given vector to be R.

So, R = 2i + j +4k

|R | = √22 + 12 + 42 = √4 + 1 + 16 = √21

Unit vector of R shall be denoted as R.

R = [ 1/ |R |] R

R = [ 1/ √21] 2i + j + 4k

Hence the unit vector of R is 121 (2i + j +4k).

Ques- Determine the value of n for which A = (-5, -1, 3n) and B = (-5, -1, -9) are the inverses of each other. 

Ans: When the magnitudes of two vectors are equal and their orientations are opposite, we know they are equal. This is how we figure out what the value of the unknown n:

A = - B

⇒ (-5, -1, -3n) = - (-5, -1, -9)

We can reach the following result by putting the appropriate components equal to each other:

-5 = 5, 

-3n = -9 and 

-1 = 1.

When we simplify the equation above, we get n = 3

As a result, when n = 3, the two vectors A and B are inverses of one another.

Ques- Find the zeros of a biquadratic equation x⁴ – 3x² + 2 = 0. 

Ans: Given f (x) = x⁴ – 3x² + 2

On substituting x2 = z in the given equation we get,

f(x) = z² – 3z + 2 = 0

z² – 2z – z + 2 = 0

z(z – 2) -1(z – 2) = 0

∴ z = 1 and z = 2

Hence, x = ±√1 and x = ±√2 [Since, z = x²].

Ques- Find the positive value of λ for which the coefficient of x² in the expression x²[√x + (λ/x²)]10 is 720. 

Ans: ⇒ x² [¹⁰Cr . (√x)^10-r . (λ/x²)^r] = x² [¹⁰Cr . λ^r . x^(10-r)/2 . x^-2r]

= x² [¹⁰Cr . λ^r . x^(10-5r)/2]

Therefore, r = 2

Hence, ¹⁰C₂ . λ² = 720

⇒ λ² = 16

⇒ λ = ±4.

Ques- Find the values of k for which the quadratic expression (x – a) (x – 10) + 1 = 0 has integral roots. 

Ans: The given equation can be rewritten as, x² – (10 + k)x + 1 + 10k = 0.

D = b² – 4ac = 100 + k² + 20k – 40k = k² – 20k + 96 = (k – 10)² – 4

The quadratic equation will have integral roots, if the value of discriminant > 0, D is a perfect square, a = 1 and b and c are integers.

i.e. (k – 10)² – D = 4

Since discriminant is a perfect square. Hence, the difference between two perfect squares in R.H.S will be 4 only when D = 0 and (k – 10)² = 4.

⇒ k – 10 = ± 2. Therefore, the values of k = 8 and 12.

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Long Answer Questions [4 Marks]

Ques- If f: R → R is defined by f(x) = x² − 3x + 2, find F(f(x)).

Ans: Given:

f(x) = x² − 3x + 2

Therefore, to find F(f(x))

F(f(x)) = f(x)² − 3f(x) + 2.

= (x²–3x+2)² – 3(x²–3x+2) + 2

Now, by applying the formula

(a – b + c)² = a² + b² + c² - 2ab + 2ac - 2ab,

We get,

= (x²)² + (3x)² + 22 – 2x² (3x) + 2x²(2) – 2x²(3x) – 3(x² – 3x + 2) + 2

Now, on substituting the values

= x⁴ + 9x² + 4 – 6x³ – 12x + 4x² – 3x² + 9x – 6 + 2

= x⁴ – 6x³ + 9x² + 4x² – 3x² – 12x + 9x – 6 + 2 + 4

therefore, we get,

F(f(x)) = x⁴ – 6x³ + 10x² – 3x

Ques- If the coefficient of x in the quadratic equation x² + bx + c =0 was taken as 17 in place of 13, its roots were found to be -2 and -15. Find the roots of the original quadratic equation.

Ans: Since there is no change in the coefficient of x² and c, therefore, the product of zeros will remain the same for both equations.

Therefore, the product of zeros (c) = -2 × -15 = 30,

Since, the original value of b is 13.

∴ Sum of zeros = -b/a = -13.

Hence, the original quadratic equation is:

x² – (Sum of Zeros)x + (Product of Zeros) = 0

x² + 13x + 30 = 0

∴ (x + 10) (x + 3) = 0

Therefore, the roots of the original quadratic equations are -3 and -10.

Ques- Determine where, if anywhere, the tangent line to f(x) = x³ − 5x² + x is parallel to the line y= 4x + 23. 

Ans: The first thing that we’ll need of course is the slope of the tangent line. So, all we need to do is take the derivative of the function.

f′(x)=3x² − 10x + 1

Two lines that are parallel will have the same slope and so all we need to do is determine where the slope of the tangent line will be 4, the slope of the given line. In other words, we’ll need to solve,

f′(x) = 4 → 3x² − 10x + 1 = 4 → 3x² − 10x − 3 = 0

This quadratic doesn’t factor and so a quick use of the quadratic formula will solve this for us.

x = (10±√136)/6 = (10±2√34)/6 = (5±√34)/3

So, the tangent line will be parallel to y=4x+23y=4x+23 at,

x = (5−√34)/3 = −0.276984

x = (5+√34)3 = 3.61032

Long Answer Questions [5 Marks]

Ques- From the function f: A →B, which is defined as f(x) = (x - 2)/(x -3). Find out if f is one-one and onto? Here, Let A = R {3} and B = R – {1}. Also, justify your answer. 

Ans: he Function is as follows:

f(x) = (x-2)/(x-3)

Now, check for one-one function.

Here,

f(x1) = (x1 – 2)/(x1 – 3)

f(x2) = (x2 - 2)/(x2 - 3)

Therefore, on putting f(x1) = f(x2)

(x1 - 2)/(x1 - 3) = (x2 - 2)/(x2 - 3)

(x1 - 2)(x2 – 3) = (x1 – 3)(x2 - 2)

x1(x2 – 3) - 2(x2 - 3) = x1(x2 – 2) – 3(x2 – 2)

x1x2 - 3x1 - 2x2 + 6 = x1x2 – 2x1 - 3x2 + 6

-3x1 – 2x2 = -2x1 - 3x2

3x2 - 2x2 = – 2x1 + 3x1

x1 = x2

Hence,

If, f (x1) = f (x2),

Then, x1 = x2

Thus, the function f is one - one function.

Now, Checking for onto function:

f (x) = (x-2)/(x-3)

Let f(x) = y such that y B or y ∈ R – {1}

So, y = (x -2) / (x- 3)

y(x -3) = x - 2

xy - 3y = x-2

xy-x = 3y-2

x(y -1) = 3y- 2

x = (3y -2)/(y-1)

For y=1, x isn’t defined

But it is given that. y ∈ R – {1}

Therefore,

x = (3y- 2)/(y- 1) ∈ R - {3}

Hence, f is onto.

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