RD Sharma Class 9 Solutions Chapter 2 Exercise 2.1 (Updated for 2024)

RD Sharma Class 9 Solutions Chapter 2 Exercise 2.1: RD Sharma Solutions Class 9 Mathematics Chapter 2 Exercise 2.1 Exponents of Real Numbers are provided here. Class 9 Chapter 2 solutions are solved by subject experts in accordance with the CBSE syllabus for the 9th standard. Get ready to score excellent percentages in your upcoming Class 9 Maths exams with the RD Sharma Solutions Class 9 Maths. The solutions are easy to understand and all your doubts will be cleared. To know more about the RD Sharma Class 9 Solutions Chapter 2  Exercise 2.1, read the whole blog. 

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RD Sharma Class 9 Solutions Chapter 2 Exercise 2.1

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RD Sharma Class 9 Chapter 2 Exponents of Real Numbers Ex 2.1

Question 1: Simplify the following

(i) 3(a⁴ b³)¹⁰ x 5 (a² b²)³

(ii) (2x ^-2 y³)³

Solution:

Using laws: (a^m)ⁿ = a^mn , a⁰ = 1, a^-m = 1/a and a^m x aⁿ = a^m+n]

(i) 3(a⁴ b³)¹⁰ x 5 (a² b²)³

On simplifying the given equation, we get;

= 3(a⁴⁰ b³⁰) x 5 (a⁶ b⁶)

= 15 (a⁴⁶ b³⁶)

[using laws: (a^m)ⁿ = a^mn and a^m x aⁿ = a^m+n]

(ii) (2x ^-2 y³)³

On simplifying the given equation, we get;

= (2³ x ^{-2 × 3} y^3×3)

= 8 x ^-6 y⁹

(iii)

Question 2: If a = 3 and b =-2, find the values of:

(i) a^a+ b^b

(ii) a^b + b^a

(iii) (a+b)^ab

Solution:

(i) a^a+ b^b

Now putting the values of ‘a’ and ‘b’, we get;

= 3³ + (−2)⁻²

= 3³ + (−1/2)²

= 27 + 1/4

= 109/4

(ii) a^b + b^a

Now putting the values of ‘a’ and ‘b’, we get;

= 3⁻² + (−2)³

= (1/3)² + (−2)³

= 1/9 – 8

= −71/9

(iii) (a+b)^ab

Now putting the values of ‘a’ and ‘b’, we get;

= (3 + (−2))³⁽⁻²⁾

= (3–2))⁻⁶

= 1⁻⁶

= 1

Question 3: Prove that

Solution:

(i) L.H.S. =

= R.H.S.

(ii) We have to prove here;

L.H.S. =

=R.H.S.

(iii) L.H.S. =

Question 4: Prove that

Solution:

(i) L.H.S

= R.H.S.

(ii) L.H.S

= R.H.S.

Question 5: Prove that

Solution:

(i) L.H.S.

= R.H.S.

(ii)

L.H.S.

= R.H.S.

Question 6: If abc = 1, show that

Solution:

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