RD Sharma Solutions for Class 10 Chapter 2 Polynomials Exercise 2.2

RD Sharma Solutions Chapter 2 Polynomials Exercise 2.2 Class 10 Maths

Chapter Name	RD Sharma Chapter 2 Polynomials
Book Name	RD Sharma Mathematics for Class 10
Other Exercises	Exercise 2.1 Exercise 2.3
Related Study	NCERT Solutions for Class 10 Maths

Exercise 2.2 Solutions

1.  Verify that the numbers given alongside of the cubic polynomials below are their zeros. Also, verify the relationship between the zeros and coefficients in each case: 

(i) f(x) = 2x³ + x²  - 5x + 2; 1/2, 1 , -2

(ii) g(x) = x³ - 4x² + 5x - 2; 2, 1, 1 

Solution

(i) f(x) = 2x³ + x² - 5x + 2 
f(1/2) = 2(1/2)³ + (1/2)² - 5(1/2) + 2 
= 2/8 + 1/4 - 5/2 + 2 = -4/2 + 2 = 0 
f(1) = 2(1)³ + (1)² - 5(1) + 2 
= 2 + 1 - 5+ 2 = 0 
f(-2) = q(-2)³ + (-2)² - 5(-2) + 2
= -16 + 4 + 10 + 2
= -16 + 16 = 0 
= α + ß +  γ = -b/a
⇒ 1/2 + 1 - 2 = -1/2 
⇒ 1/2 - 1 = -1/2 
⇒ 1/2 = -1/2 
αß.ßγ +  γα = c/α 
⇒ 1/2 ×1 + 1 × - 2 + -2 ×1/2 = -5/2 
⇒ 1/2 - 2 - 1 = - 5/2 
⇒ -5/2 = -5/2 

(ii) g(x) = x³ - 4x² + 5x - 2 

g(2) = (2)³ - 4(2)² + 5(2) - 2 = 8 - 16 + 10 - 2 = 18 - 18 = 0 
g(1) = [1]³ - 4(1)² + 5[1] - 2 = 1 - 4 + 5 - 2 = 6 - 6 = 0 
α + ß + γ = -b/a(2) + 1 + 1 = -(-4) = 4 = 4 
αß + ßγ + γα = c/a
⇒ 2×1 + 1×1 + 1×2 = 5 
⇒ 2 + 1 + 2 = 5 
⇒ 5 = 5 
αßγ = -(-2)
⇒ 2×1×1 = 2 
⇒ 2= 2 

2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and product of its zeros as 3, -1 and -3 respectively. 

Solution

Any cubic polynomial is of the form ax³ + bx² + cx +d
= x³ - Sum of zeroes(x²) [product of zeroes] + sum of the products of its zeroes x - product of zeroes 

= x³ - 2x² + (3 - x)+3
= k[x³ - 3x² - x - 3]
k is any non - zero real numbers 

3. If the zeros of the polynomial f(x) = 2x³ - 15x² + 37x - 30 are in A.P., find them. 

Solution

Let α = a -d, ß= a and γ = a + d be the zeroes of polynomial. 
f(x) = 2x³ - 15x² + 37x - 30 
α + ß + γ = -(-15/2) = 15/2 
αßγ = -(-30/2) = 15 
a - d + a + a +d = 15/2 and a(a - d)(a+ a) = 15 
3a = 15/2, a = 5/2 
a(a² - d²) = 15 
⇒ a² - a²  = (15×2)/5 
⇒ (5/2)² - d² = 6

⇒ (25-6)/4 = d² 
d² = 1/4

⇒ d = 1/2 
∴ α = 5/2 - 1/2 = 4/2 = 2 
ß = 5/2 = 5/2 
γ = 5/2 + 1/2 = 3 

4. Find the condition that the zeros of the polynomial f(x) = x³ + 3px² +3qx + r may be in A.P.

Solution

f(x) = x³ + 3px² + 3qx +q 
Let a - d, a, a+d be the zeroes of the polynomial 

The sum of zeroes  -b/a 
a + a - d + a +d = b/a 
3a = -3p 
a = -p 
Since a is the zero of the polynomial f(x) therefore f(a) = 0 

⇒ [a]³ + 3pa² + 3qa + r = 0  
∴ f(a) = 0

⇒ [a]³ + 3pa² + 3qa + r = 0 
⇒ p³ + 3p(-p)² + 3q(-p) + r = 0 
⇒ -p³ + 3p²  - pq + r = 0 
⇒ 2p³ - pq + r = 0 

5. If the zeroes of the polynomial f(x) = ax³ + 3bx² + 3cx + d are in A.P. , prove that 2b³ - 3abc + a²d = 0 

Solution

Let a-d, a, a+ d be the zeroes of the polynomial f(x) 
The sum of zeroes

⇒ a -d + a + a +d = -3b/a 
⇒ +3a = -3b/a

⇒ a  = -3b/(a×3)

⇒ a = -b/a
f(a)  = 0

⇒ a(a)² + 3b(a)² + 3c(a) + d = 0 
= a(-b/a)³ + 3b²/ a² - (3bc/a) +d = 0 
⇒ 2b³/a² - 3bc/a  + d = 0 
⇒ (2b³ - 3abc + a²d)/a²  = 0 
⇒ 2b³ - 3abc + a²d = 0 

6. If the zeroes of the polynomial f(x) = x³ - 12x² + 39x + k are in A.P., find the value of k. 

Solution

f(x) = x³ - 12x² + 39x - k 
Let a -d, a, a + d  be the zeroes of the polynomial f(x) 
The sum of the zeroes = 12 
3a = 12 
⇒ a = 4 
f(a) = -a(x)³ - l²(4)² + 39(4) + k = 0 
⇒ 64 - 192 + 156 + k = 0 
⇒ -28 = k 
⇒ k = -28