Question

(a) Using the factor theorem, show that (x – 2) is a factor of x3 + x2 – 4x – 4. [3]
Hence, factorise the polynomial completely.
(b) Prove that :
(cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1 [3]
(c) In an Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively.
Find the : [4]
(i) first term
(ii) common difference
(iii) sum of the first 20 terms.​

Answer 1

Question :

Using the factor theorem, show that (x – 2) is a factor of x³ + x² – 4x – 4. Factorise it.

Solution :

Here,

• g(x) = (x - 2)
• f(x) = x³ + x² - 4x - 4

By using the factor theorem, we get :

⠀⠀=> x³ + x² - 4x - 4 = 0

⠀⠀=> (2)³ + (2)² - 4(2) - 4 = 0

⠀⠀=> 8 + 4 - 8 - 4 = 0

⠀⠀=> 0 = 0

Hence, it is proved that (x - 2) is a factor of (x³ + x² - 4x - 4).

Now dividing the f(x) by g(x), we get :

⠀⠀⠀⠀⠀⠀x - 2)x³ + x² - 4x - 4(x² + 3x + 2

⠀⠀⠀⠀⠀⠀⠀⠀⠀x³ - 2x

⠀⠀⠀⠀⠀⠀⠀⠀⠀× 3x² - 4x

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀3x² - 6x

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀×⠀⠀2x - 4

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀2x - 4

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀×

Hence, when the f(x) is divided by g(x) the quotient is (x³ + x² - 4x - 4)

Now by factorising it, we get :

⠀⠀⠀=> x² + 3x + 2 = 0

⠀⠀⠀=> x² + 2x + x + 2 = 0

⠀⠀⠀=> x(x + 2) + 1(x + 2) = 0

⠀⠀⠀=> (x + 2)(x + 1) = 0

Hence the factors of the equation are (x + 2) and (x + 1).

So the factors of the f(x) are :

⠀⠀=> x³ + x² - 4x - 4 = 0

⠀⠀=> (x - 2)(x² + 3x + 2) = 0

⠀⠀=> (x - 2)(x + 2)(x + 1) = 0

Hence the factorization of (x³ + x² - 4x - 4) is (x - 2)(x + 2)(x + 1).

$\rule{400}{2}$

Question :

Prove that :-

$\sf{[cosec(\theta) - sin(\theta)][sec(\theta) - cos(\theta)][tan(\theta) + cot(\theta)] = 1}$

Solution :

$:\implies \sf{[cosec(\theta) - sin(\theta)][sec(\theta) - cos(\theta)][tan(\theta) + cot(\theta)] = 1}$

We know that,

• cosec(θ) = 1/sin(θ)
• sec(θ) = 1/cos(θ)
• tan(θ) = sin(θ)/cos(θ)
• cot(θ) = cos(θ)/sin(θ).

Now by substituting them in the equation, we get

$:\implies \sf{\bigg[\dfrac{1}{sin(\theta)} - sin(\theta)\bigg]\bigg[\dfrac{1}{cos(\theta)} - cos(\theta)\bigg]\bigg[\dfrac{sin(\theta)}{cos(\theta)} + \dfrac{cos(\theta)}{sin(\theta)}\bigg] = 1} \\ \\ \\ :\implies \sf{\bigg[\dfrac{1 - sin^{2}(\theta)}{sin(\theta)}\bigg]\bigg[\dfrac{1 - cos^{2}(\theta)}{cos(\theta)}\bigg]\bigg[\dfrac{sin^{2} (\theta) + cos^{2}(\theta)}{cos(\theta)sin(\theta)}\bigg] = 1}$

We know that,

• [1 - sin²(θ)] = cos²(θ)
• [1 - cos²(θ)] = sin²(θ)
• [sin²(θ) + cos²(θ)] = 1

Now by substituting them in the equation, we get :

$:\implies \sf{\dfrac{cos^{2}(\theta)}{sin(\theta)} \times \dfrac{sin^{2}(\theta)}{cos(\theta)} \times \dfrac{1}{cos(\theta)sin(\theta)} = 1} \\ \\ \\ :\implies \sf{\dfrac{cos^{2}(\theta)sin^{2}(\theta)}{cos^{2}(\theta)sin^{2}(\theta)} = 1} \\ \\ \\ :\implies \sf{1 = 1} \\ \\ \\ \boxed{\therefore \sf{[cosec(\theta) - sin(\theta)][sec(\theta) - cos(\theta)][tan(\theta) + cot(\theta)] = 1}}$

Proved!!

$\rule{400}{2}$

Question :

In an Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively.

Find the : 

• First term of the AP.
• Common difference of the AP.
• Sum of first 20 terms of the AP.

Explanation :

Given :

• 4th term of the AP = 8
• 6th term of the AP = 14

To find :

• First term of the AP, a = ?
• Common difference of the AP, d = ?
• Sum of first 20 terms of the AP, s₂₀ = ?

Knowledge required :

• Formula for nth term of an AP, tn = a + (n - 1)d

[Where : tn = nth term of the AP, a = First term of the AP, n = no. of terms of the AP, d = Common Difference of the AP]

• Formula for sum of n terms of an AP, sn = (n/2)[2a + (n - 1)d]

[Where : sn = sum of n terms of the AP, a = First term of the AP, n = no. of terms of the AP, d = Common Difference of the AP]

Solution :

To find the nth term for the 4th term of the AP :

By using the formula for nth term of an AP and substituting the values in it, we get :

⠀⠀=> tn = a + (n - 1)d

⠀⠀=> 8 = a + (4 - 1)d

⠀⠀=> 8 = a + 3d

⠀⠀⠀∴ a + 3d = 8⠀⠀⠀⠀⠀⠀⠀.....Eq.(i)

To find the nth term for the 6th term of the AP :

By using the formula for nth term of an AP and substituting the values in it, we get :

⠀⠀=> tn = a + (n - 1)d

⠀⠀=> 14 = a + (6 - 1)d

⠀⠀=> 14 = a + 5d

⠀⠀⠀∴ a + 5d = 14⠀⠀⠀⠀⠀⠀⠀.....Eq.(ii)

Now by subtracting Eq.(i) and Eq.(ii), we get :

⠀⠀=> (a + 3d) - (a + 5d) = 8 - 14

⠀⠀=> a + 3d - a - 5d = -6

⠀⠀=> -2d = -6

⠀⠀=> d = -6/-2

⠀⠀=> d = 3

⠀⠀⠀∴ d = 3

Hence the common difference of the AP is 3.

Now by substituting the value of d in the Eq.(i), we get :

⠀⠀=> a + 3d = 8

⠀⠀=> a + 3(3) = 8

⠀⠀=> a + 9 = 8

⠀⠀=> a = 8 - 9

⠀⠀=> a = (-1)

⠀⠀⠀∴ a = (-1)

Hence the first term of the AP is (-1).

To find the sum of 20 term of the AP :

By using the formula for sum of n terms of the AP and substituting the values in it, we get :

⠀⠀=> sn = (n/2)[2a + (n - 1)d]

⠀⠀=> s₂₀ = (20/2) × [2(-1) + (20 - 1)3]

⠀⠀=> s₂₀ = 10 × [-2 + (19)3]

⠀⠀=> s₂₀ = 10 × [-2 + 57]

⠀⠀=> s₂₀ = 10 × 55

⠀⠀=> s₂₀ = 550

⠀⠀⠀∴ s₂₀ = 550

Hence the sum of 20 terms of the AP is 550.⠀⠀