Mathematics Paper 2 (Algebra - II) Free MCQ Test with solutions for CTET

Given:

6x2 - x - 2 = 0

Calculation:

6x2 - x - 2 = 0

⇒ 6x2 - 4x + 3x - 2 = 0

⇒ 2x(3x - 2) + 1(3x - 2)

⇒ (2x + 1)(3x - 2) = 0

So, 2x + 1 = 0 and 3x - 2 = 0

⇒ x = $-\frac{1}{2}$ and x = 

∴ The correct answer is option 3.

Formula used:

(a - b)² = a² - 2ab + b²

Calculation:

Form option (3)

[(√3 x)² + (√2)² - 2√6 x] - x²= 2x² + 2x

3x² + 2 - 2√6 x - x² = 2x² + 2x

2x² + 2 - 2√6 x  = 2x² + 2x

2 - 2√6 x  =  2x

Since there is no x² term left. 

∴  is not a quadratic equation.

Formula used

(a + b)² = a² + b² + 2ab

Calculation

(4x + 1)2 − (4x + 3) (4x − 1)

⇒ 16x² + 1 + 2 × 4x × 1 - (16x² - 4x + 12x - 3)

⇒ 1 + 8x × 1 - 8x + 3

⇒ 4

The value is 4.

Given: 

Quadratic equation = x2 - 3x + 4 = 0

Formula used:

Quadratic equation: ax2 + bx + c = 0

The sum of the roots = α + β = 

Product of the roots = α × β =  

Calculation:

The sum of the roots = α + β=  = 3

Product of the roots = α × β = ${\displaystyle \frac{4}{1}}$ = 4

∴ 3 and 4 are the sum and product of the roots of the equation x2 - 3x + 4 = 0.

Concept -

let the polynomial be p(x) then p(x) = 0 gives you the zeros of the polynomial.

Explanation -

We have the polynomial  9x² - 4

Now for the zeros of the polynomial -

 9x² - 4  = 0

⇒  9x² = 4

⇒  x² = 4/9

Hence option (1) is true.

Concept -

If α and β are the roots of the polynomial f(x) = ax2 + bx + c then

sum of roots = -b/a

product of roots = c/a

Explanation -

we have the polynomial 3x2 - 5x - 2

Now the sum of the roots(zeros) is = 

Hence option(3) is true.

Concept Used:

A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is ax² + bx + c = 0, where a, b, and c are constants, and x is an unknown variable.

Calculation:

3x2 - 2x + 3 = 2

⇒ 3x2 - 2x + 1 = 0, This equation is the form of ax² + bx + c = 0.

Option 2 - x2 + 2x + 3, this is the not the form of ax² + bx + c = 0

Option 3 - x + 5 = 2x - 8, I this equation second degree equation not formed.

Option 4 - x3 + 2x2 + 3x + 1 = 0, In this equation degree is 3, in quadratic degree is 2.

∴ This equation can also be written in standard form by subtracting 2 from both sides to get 3x² - 2x + 1 = 0. Option (1) is correct.

Solution:

As -2 is root of ay2 + ay + 3 = 0,

⇒ a × (-2)² + a(-2) + 3 = 0

⇒ 4a - 2a + 3 = 0

⇒ 2a + 3 = 0

⇒ 2a = -3

⇒ a = -3/2

and -2 also root of y2 + y + b = 0

⇒ (-2)² + (-2) + b =0 

⇒ 4 - 2 + b = 0

⇒ 2 + b = 0

⇒ b = -2

⇒ a²b = (-3/2)² × (-2) = 9/4 × (-2) = -9/2

Thus, value of  a2b is -9/2.

Given:

Sum of squares of two numbers = 12,

Product of the numbers = 4

Solution:

Let the numbers be 'a' and 'b'.

ab = 4

Taking the square root of both sides, we get:

a - b = ±2

Therefore, the difference between the numbers is ±2.

Concept:

1. For the quadratic equation ax2 + bx + c = 0

Sum of root (α + β) = -b/a

Product of root = c/a

2. a² + b2 = (a + b)2 - 2ab

3. a⁴ + b⁴ = (a² + b²)2 - 2(ab)²

Calculation:

x2 - 7x + 1 = 0

As α & β be roots of the quadratic equation

α + β = -(-7)/1

α + β  = 7

αβ = 1

By using the above identity

α² + β² = (α + β)² - 2αβ = 7² - 2

α2 + β2 = 47

Now we can use the identity:

α⁴ + β⁴ = (α2 + β2)2 - 2α2β2 

Substituting in the value of α² + β² and αβ = 1, we get:

α⁴4 + β⁴ = (47)² - 2 = 2207

∴ The value of α4 + β4 is 2207.