RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2 (Updated for 2024)

RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2: An algebraic expression made up of two words is called a binomial expression. For students who want to learn the correct steps to solve such problems, RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2 is available in PDF format, and students can easily download the PDF from the link provided below.

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1. Find the 11^th term from the beginning and the 11^th term from the end in the expansion of (2x – 1/x²)²⁵.

Solution:

Given:

(2x – 1/x²)²⁵

The given expression contains 26 terms.

So, the 11^th term from the end is the (26 − 11 + 1) ^th term from the beginning.

In other words, the 11^th term from the end is the 16^th term from the beginning.

Then,

T₁₆ = T₁₅₊₁ = ²⁵C₁₅ (2x)^25-15 (-1/x²)¹⁵

= ²⁵C₁₅ (2¹⁰) (x)¹⁰ (-1/x³⁰)

= – ²⁵C₁₅ (2¹⁰ / x²⁰)

Now, we shall find the 11^th term from the beginning.

T₁₁ = T₁₀₊₁ = ²⁵C₁₀ (2x)^25-10 (-1/x²)¹⁰

= ²⁵C₁₀ (2¹⁵) (x)¹⁵ (1/x²⁰)

= ²⁵C₁₀ (2¹⁵ / x⁵)

2. Find the 7^th term in the expansion of (3x² – 1/x³)¹⁰.

Solution:

Given:

(3x² – 1/x³)¹⁰

Let us consider the 7^th term as T₇

So,

T₇ = T₆₊₁

= ¹⁰C₆ (3x²)^10-6 (-1/x³)⁶

= ¹⁰C₆ (3)⁴ (x)⁸ (1/x¹⁸)

= [10×9×8×7×81] / [4×3×2×x¹⁰]

= 17010 / x¹⁰

∴ The 7^th term of the expression (3x² – 1/x³)¹⁰ is 17010 / x¹⁰.

3. Find the 5^th term in the expansion of (3x – 1/x²)¹⁰.

Solution:

Given:

(3x – 1/x²)¹⁰

The 5^th term from the end is the (11 – 5 + 1)th, is., 7^th term from the beginning.

So,

T₇ = T₆₊₁

= ¹⁰C₆ (3x)^10-6 (-1/x²)⁶

= ¹⁰C₆ (3)⁴ (x)⁴ (1/x¹²)

= [10×9×8×7×81] / [4×3×2×x⁸]

= 17010 / x⁸

∴ The 5^th term of the expression (3x – 1/x²)¹⁰ is 17010 / x⁸.

4. Find the 8^th term in the expansion of (x^3/2 y^1/2 – x^1/2 y^3/2)¹⁰.

Solution:

Given:

(x^3/2 y^1/2 – x^1/2 y^3/2)¹⁰

Let us consider the 8^th term as T₈

So,

T₈ = T₇₊₁

= ¹⁰C₇ (x^3/2 y^1/2)^10-7 (-x^1/2 y^3/2)⁷

= -[10×9×8]/[3×2] x^9/2 y^3/2 (x^7/2 y^21/2)

= -120 x⁸y¹²

∴ The 8^th term of the expression (x^3/2 y^1/2 – x^1/2 y^3/2)¹⁰ is -120 x⁸y¹².

5. Find the 7^th term in the expansion of (4x/5 + 5/2x) ⁸.

Solution:

Given:

(4x/5 + 5/2x) ⁸

Let us consider the 7^th term as T₇

So,

T₇ = T₆₊₁

∴ The 7^th term of the expression (4x/5 + 5/2x) ⁸ is 4375/x⁴.

6. Find the 4^th term from the beginning and 4^th term from the end in the expansion of (x + 2/x) ⁹.

Solution:

Given:

(x + 2/x) ⁹

Let T_r+1 be the 4th term from the end.

Then, T_r+1 is (10 − 4 + 1)th, i.e., 7th, the term from the beginning.

7. Find the 4^th term from the end in the expansion of (4x/5 – 5/2x) ⁹.

Solution:

Given:

(4x/5 – 5/2x) ⁹

Let T_r+1 be the 4th term from the end of the given expression.

Then, T_r+1 is (10 − 4 + 1)th term, i.e., the 7th term, from the beginning.

T₇ = T₆₊₁

∴ The 4^th term from the end is 10500/x³.

8. Find the 7th term from the end in the expansion of (2x² – 3/2x) ⁸.

Solution:

Given:

(2x² – 3/2x) ⁸

Let T_r+1 be the 4th term from the end of the given expression.

Then, T_r+1 is (9 − 7 + 1)th term, i.e., 3rd term, from the beginning.

T₃ = T₂₊₁

∴ The 7^th term from the end is 4032 x¹⁰.

9. Find the coefficient of:

(i)  x¹⁰ in the expansion of (2x² – 1/x)²⁰

(ii) x⁷ in the expansion of (x – 1/x²)⁴⁰

(iii) x^-15 in the expansion of (3x² – a/3x³)¹⁰

(iv) x⁹ in the expansion of (x² – 1/3x)⁹

(v) x^m in the expansion of (x + 1/x)ⁿ

(vi) x in the expansion of (1 – 2x³ + 3x⁵) (1 + 1/x)⁸

(vii) a⁵b⁷ in the expansion of (a – 2b)¹²

(viii) x in the expansion of (1 – 3x + 7x²) (1 – x)¹⁶

Solution:

(i)  x¹⁰ in the expansion of (2x² – 1/x)²⁰

Given:

(2x² – 1/x)²⁰

If  x¹⁰ occurs in the (r + 1)th term in the given expression.

Then, we have:

T_r+1 = ⁿC_r x^n-r a^r

(ii) x⁷ in the expansion of (x – 1/x²)⁴⁰

Given:

(x – 1/x²)⁴⁰

If x⁷ occurs at the (r + 1) th term in the given expression.

Then, we have:

T_r+1 = ⁿC_r x^n-r a^r

40 − 3r =7

3r = 40 – 7

3r = 33

r = 33/3

= 11

(iii) x^-15 in the expansion of (3x² – a/3x³)¹⁰

Given:

(3x² – a/3x³)¹⁰

If x⁻¹⁵ occurs at the (r + 1)th term in the given expression.

Then, we have:

T_r+1 = ⁿC_r x^n-r a^r

(iv) x⁹ in the expansion of (x² – 1/3x)⁹

Given:

(x² – 1/3x)⁹

If x⁹ occurs at the (r + 1)th term in the above expression.

Then, we have:

T_r+1 = ⁿC_r x^n-r a^r

For this term to contain x⁹, we must have

18 − 3r = 9

3r = 18 – 9

3r = 9

r = 9/3

= 3

(v) x^m in the expansion of (x + 1/x)ⁿ

Given:

(x + 1/x)ⁿ

If x^m occurs at the (r + 1)th term in the given expression.

Then, we have:

T_r+1 = ⁿC_r x^n-r a^r

(vi) x in the expansion of (1 – 2x³ + 3x⁵) (1 + 1/x)⁸

Given:

(1 – 2x³ + 3x⁵) (1 + 1/x)⁸

If x occurs at the (r + 1)th term in the given expression.

Then, we have:

(1 – 2x³ + 3x⁵) (1 + 1/x)⁸ = (1 – 2x³ + 3x⁵) (⁸C₀ + ⁸C₁ (1/x) + ⁸C₂ (1/x)² + ⁸C₃ (1/x)³ + ⁸C₄ (1/x)⁴ + ⁸C₅ (1/x)⁵ + ⁸C₆ (1/x)⁶ + ⁸C₇ (1/x)⁷ + ⁸C₈ (1/x)⁸)

So, ‘x’ occurs in the above expression at -2x³.⁸C₂ (1/x²) + 3x⁵.⁸C₄ (1/x⁴)

∴ Coefficient of x = -2 (8!/(2!6!)) + 3 (8!/(4! 4!))

= -56 + 210

= 154

(vii) a⁵b⁷ in the expansion of (a – 2b)¹²

Given:

(a – 2b)¹²

If a⁵b⁷ occurs at the (r + 1)th term in the given expression.

Then, we have:

T_r+1 = ⁿC_r x^n-r a^r

(viii) x in the expansion of (1 – 3x + 7x²) (1 – x)¹⁶

Given:

(1 – 3x + 7x²) (1 – x)¹⁶

If x occurs at the (r + 1)th term in the given expression.

Then, we have:

(1 – 3x + 7x²) (1 – x)¹⁶ = (1 – 3x + 7x²) (¹⁶C₀ + ¹⁶C₁ (-x) + ¹⁶C₂ (-x)² + ¹⁶C₃ (-x)³ + ¹⁶C₄ (-x)⁴ + ¹⁶C₅ (-x)⁵ + ¹⁶C₆ (-x)⁶ + ¹⁶C₇ (-x)⁷ + ¹⁶C₈ (-x)⁸ + ¹⁶C₉ (-x)⁹ + ¹⁶C₁₀ (-x)¹⁰ + ¹⁶C₁₁ (-x)¹¹ + ¹⁶C₁₂ (-x)¹² + ¹⁶C₁₃ (-x)¹³ + ¹⁶C₁₄ (-x)¹⁴ + ¹⁶C₁₅ (-x)¹⁵ + ¹⁶C₁₆ (-x)¹⁶)

So, ‘x’ occurs in the above expression at ¹⁶C₁ (-x) – 3x¹⁶C₀

∴ Coefficient of x = -(16!/(1! 15!)) – 3(16!/(0! 16!))

= -16 – 3

= -19

10. Which term in the expansion of contains x and y to one and the same power?

Solution:

Let us consider T_r+1 th term in the given expansion contains x and y to one and the same power.

Then we have,

T_r+1 = ⁿC_r x^n-r a^r

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FAQ: RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2