**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/x^{2})^{25}.**

**Solution:**

Given:

(2x – 1/x^{2})^{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,

T_{16} = T_{15+1} = ^{25}C_{15} (2x)^{25-15} (-1/x^{2})^{15}

= ^{25}C_{15} (2^{10}) (x)^{10} (-1/x^{30})

= – ^{25}C_{15} (2^{10} / x^{20})

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

T_{11} = T_{10+1} = ^{25}C_{10} (2x)^{25-10} (-1/x^{2})^{10}

= ^{25}C_{10} (2^{15}) (x)^{15} (1/x^{20})

= ^{25}C_{10} (2^{15} / x^{5})

**2. Find the 7 ^{th} term in the expansion of (3x^{2} – 1/x^{3})^{10}.**

**Solution:**

Given:

(3x^{2} – 1/x^{3})^{10}

Let us consider the 7^{th} term as T_{7}

So,

T_{7} = T_{6+1}

= ^{10}C_{6} (3x^{2})^{10-6} (-1/x^{3})^{6}

= ^{10}C_{6} (3)^{4} (x)^{8} (1/x^{18})

= [10×9×8×7×81] / [4×3×2×x^{10}]

= 17010 / x^{10}

∴ The 7^{th} term of the expression (3x^{2} – 1/x^{3})^{10} is 17010 / x^{10}.

**3.** **Find the 5 ^{th} term in the expansion of (3x – 1/x^{2})^{10}.**

**Solution:**

Given:

(3x – 1/x^{2})^{10}

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

So,

T_{7} = T_{6+1}

= ^{10}C_{6} (3x)^{10-6} (-1/x^{2})^{6}

= ^{10}C_{6} (3)^{4} (x)^{4} (1/x^{12})

= [10×9×8×7×81] / [4×3×2×x^{8}]

= 17010 / x^{8}

∴ The 5^{th} term of the expression (3x – 1/x^{2})^{10} is 17010 / x^{8}.

**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 T_{8}

So,

T_{8} = T_{7+1}

= ^{10}C_{7} (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 x^{8}y^{12}

∴ The 8^{th} term of the expression (x^{3/2} y^{1/2} – x^{1/2} y^{3/2})^{10} is -120 x^{8}y^{12}.

**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 T_{7}

So,

T_{7} = T_{6+1}

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

**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.

T_{7} = T_{6+1}

∴ The 4^{th} term from the end is 10500/x^{3}.

**8. Find the 7th term from the end in the expansion of (2x ^{2} – 3/2x)^{ 8}.**

**Solution:**

Given:

(2x^{2} – 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.

T_{3} = T_{2+1}

∴ The 7^{th} term from the end is 4032 x^{10}.

**9. Find the coefficient of:**

**(i) x ^{10} in the expansion of (2x^{2} – 1/x)^{20}**

**(ii) x ^{7} in the expansion of (x – 1/x^{2})^{40}**

**(iii) x ^{-15} in the expansion of (3x^{2} – a/3x^{3})^{10}**

**(iv) x ^{9} in the expansion of (x^{2} – 1/3x)^{9}**

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

**(vi) x in the expansion of (1 – 2x ^{3} + 3x^{5}) (1 + 1/x)^{8}**

**(vii) a ^{5}b^{7} in the expansion of (a – 2b)^{12}**

**(viii) x in the expansion of (1 – 3x + 7x ^{2}) (1 – x)^{16}**

**Solution:**

**(i) **x^{10} in the expansion of (2x^{2} – 1/x)^{20}

Given:

(2x^{2} – 1/x)^{20}

If x^{10 }occurs in the (r + 1)th term in the given expression.

Then, we have:

T_{r+1 }= ^{n}C_{r} x^{n-r} a^{r}

**(ii) **x^{7} in the expansion of (x – 1/x^{2})^{40}

Given:

(x – 1/x^{2})^{40}

If x^{7 }occurs at the (r + 1) th term in the given expression.

Then, we have:

T_{r+1 }= ^{n}C_{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 (3x^{2} – a/3x^{3})^{10}

Given:

(3x^{2} – a/3x^{3})^{10}

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

Then, we have:

T_{r+1 }= ^{n}C_{r} x^{n-r} a^{r}

**(iv) **x^{9} in the expansion of (x^{2} – 1/3x)^{9}

Given:

(x^{2} – 1/3x)^{9}

If x^{9} occurs at the (r + 1)th term in the above expression.

Then, we have:

T_{r+1 }= ^{n}C_{r} x^{n-r} a^{r}

For this term to contain x^{9}, 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 }= ^{n}C_{r} x^{n-r} a^{r}

**(vi) **x in the expansion of (1 – 2x^{3} + 3x^{5}) (1 + 1/x)^{8}

Given:

(1 – 2x^{3} + 3x^{5}) (1 + 1/x)^{8}

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

Then, we have:

(1 – 2x^{3} + 3x^{5}) (1 + 1/x)^{8}** = **(1 – 2x^{3} + 3x^{5}) (^{8}C_{0} + ^{8}C_{1} (1/x) + ^{8}C_{2} (1/x)^{2} + ^{8}C_{3} (1/x)^{3} + ^{8}C_{4} (1/x)^{4} + ^{8}C_{5} (1/x)^{5} + ^{8}C_{6} (1/x)^{6} + ^{8}C_{7} (1/x)^{7} + ^{8}C_{8} (1/x)^{8})

So, ‘x’ occurs in the above expression at -2x^{3}.^{8}C_{2} (1/x^{2}) + 3x^{5}.^{8}C_{4} (1/x^{4})

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

= -56 + 210

= 154

**(vii) **a^{5}b^{7} in the expansion of (a – 2b)^{12}

Given:

(a – 2b)^{12}

If a^{5}b^{7} occurs at the (r + 1)th term in the given expression.

Then, we have:

T_{r+1 }= ^{n}C_{r} x^{n-r} a^{r}

**(viii) **x in the expansion of (1 – 3x + 7x^{2}) (1 – x)^{16}

Given:

(1 – 3x + 7x^{2}) (1 – x)^{16}

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

Then, we have:

(1 – 3x + 7x^{2}) (1 – x)^{16} = (1 – 3x + 7x^{2}) (^{16}C_{0} + ^{16}C_{1} (-x) + ^{16}C_{2} (-x)^{2} + ^{16}C_{3} (-x)^{3} + ^{16}C_{4} (-x)^{4} + ^{16}C_{5} (-x)^{5} + ^{16}C_{6} (-x)^{6} + ^{16}C_{7} (-x)^{7} + ^{16}C_{8} (-x)^{8} + ^{16}C_{9} (-x)^{9} + ^{16}C_{10} (-x)^{10} + ^{16}C_{11} (-x)^{11} + ^{16}C_{12} (-x)^{12} + ^{16}C_{13} (-x)^{13} + ^{16}C_{14} (-x)^{14} + ^{16}C_{15} (-x)^{15} + ^{16}C_{16} (-x)^{16})

So, ‘x’ occurs in the above expression at ^{16}C_{1} (-x) – 3x^{16}C_{0}

∴ 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} = ^{n}C_{r} x^{n-r} a^{r}

We have included all the information regarding CBSE RD Sharma Solutions Class 11 Maths Chapter 18 Exercise 18.2. If you have any queries feel free to ask in the comment section.

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