RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2 (Updated for 2023)

RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2: Download the Free PDF of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2 from the link given in this blog.  In this exercise, students will learn how to find the value of a polynomial. This study material helps students to master a concept by solving various questions listed in the exercise. Be it your Maths exam or class assignment, we have got you covered. All the solutions in the RD Sharma Solutions Class 9 Maths are designed by subject matter experts and are very reliable. To know more, read the whole blog.

Download RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2 PDF

RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2

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Question 1: If f(x) = 2x³ – 13x² + 17x + 12, find

(i) f (2)

(ii) f (-3)

(iii) f(0)

Solution:

f(x) = 2x³ – 13x² + 17x + 12

(i) f(2) = 2(2)³ – 13(2) ² + 17(2) + 12

= 2 x 8 – 13 x 4 + 17 x 2 + 12

= 16 – 52 + 34 + 12

= 62 – 52

= 10

(ii) f(-3) = 2(-3)³ – 13(-3) ² + 17 x (-3) + 12

= 2 x (-27) – 13 x 9 + 17 x (-3) + 12

= -54 – 117 -51 + 12

= -222 + 12

= -210

(iii) f(0) = 2 x (0)³ – 13(0) ² + 17 x 0 + 12

= 0-0 + 0+ 12

= 12

Question 2: Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases:

(i) f(x) = 3x + 1, x = −1/3

(ii) f(x) = x² – 1, x = 1,−1

(iii) g(x) = 3x² – 2 , x = 2/√3 , −2/√3

(iv) p(x) = x³ – 6x² + 11x – 6 , x = 1, 2, 3

(v) f(x) = 5x – π, x = 4/5

(vi) f(x) = x² , x = 0

(vii) f(x) = lx + m, x = −m/l

(viii) f(x) = 2x + 1, x = 1/2

Solution:

(i) f(x) = 3x + 1, x = −1/3

f(x) = 3x + 1

Substitute x = −1/3 in f(x)

f( −1/3) = 3(−1/3) + 1

= -1 + 1

= 0

Since, the result is 0, so x = −1/3 is the root of 3x + 1

(ii) f(x) = x² – 1, x = 1,−1

f(x) = x² – 1

Given that x = (1 , -1)

Substitute x = 1 in f(x)

f(1) = 1² – 1

= 1 – 1

= 0

Now, substitute x = (-1) in f(x)

f(-1) = (−1)² – 1

= 1 – 1

= 0

Since , the results when x = 1 and x = -1 are 0, so (1 , -1) are the roots of the polynomial f(x) = x² – 1

(iii) g(x) = 3x² – 2 , x = 2/√3 , −2/√3

g(x) = 3x² – 2

Substitute x = 2/√3 in g(x)

g(2/√3) = 3(2/√3)² – 2

= 3(4/3) – 2

= 4 – 2

= 2 ≠ 0

Now, Substitute x = −2/√3 in g(x)

g(2/√3) = 3(-2/√3)² – 2

= 3(4/3) – 2

= 4 – 2

= 2 ≠ 0

Since, the results when x = 2/√3 and x = −2/√3) are not 0. Therefore (2/√3 , −2/√3 ) are not zeros of 3x²–2.

(iv) p(x) = x³ – 6x² + 11x – 6 , x = 1, 2, 3

p(1) = 1³ – 6(1)² + 11x 1 – 6 = 1 – 6 + 11 – 6 = 0

p(2) = 2³ – 6(2)² + 11×2 – 6 = 8 – 24 + 22 – 6 = 0

p(3) = 3³ – 6(3)² + 11×3 – 6 = 27 – 54 + 33 – 6 = 0

Therefore, x = 1, 2, 3 are zeros of p(x).

(v) f(x) = 5x – π, x = 4/5

f(4/5) = 5 x 4/5 – π = 4 – π ≠ 0

Therefore, x = 4/5 is not a zero of f(x).

(vi) f(x) = x² , x = 0

f(0) = 0² = 0

Therefore, x = 0 is a zero of f(x).

(vii) f(x) = lx + m, x = −m/l

f(−m/l) = l x −m/l + m = -m + m = 0

Therefore, x = −m/l is a zero of f(x).

(viii) f(x) = 2x + 1, x = ½

f(1/2) = 2x 1/2 + 1 = 1 + 1 = 2 ≠ 0

Therefore, x = ½ is not a zero of f(x).

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