RBSE Solutions Class 11 Maths Chapter 7 Binomial Theorem Exercise 7.5

Question 1. If y is too small as compared to x, then prove that \( \frac { (x - y)^n }{ (x + y)^n } = 1 - \frac { 2ny }{ x } \), where \( y^2 \) and higher powers are negligible.
Answer: We need to prove the given identity by considering y to be very small compared to x, allowing us to ignore \( y^2 \) and higher powers of y.
Let's start with the left-hand side (L.H.S.):
\( \frac { (x - y)^n }{ (x + y)^n } = \left( \frac { x - y }{ x + y } \right)^n \)
We can factor out x from both numerator and denominator:
\( = \left( \frac { x(1 - y/x) }{ x(1 + y/x) } \right)^n \)
\( = \left( \frac { 1 - y/x }{ 1 + y/x } \right)^n \)
\( = (1 - y/x)^n (1 + y/x)^{-n} \)
Now, using the binomial expansion \((1 + z)^m \approx 1 + mz \) when z is very small, and ignoring terms with \( y^2 \) or higher powers:
\( (1 - y/x)^n \approx 1 + n(-y/x) = 1 - ny/x \)
\( (1 + y/x)^{-n} \approx 1 + (-n)(y/x) = 1 - ny/x \)
Substitute these approximations back into the equation:
\( = (1 - ny/x)(1 - ny/x) \)
\( = 1 - ny/x - ny/x + (-ny/x)(-ny/x) \)
\( = 1 - 2ny/x + n^2 y^2/x^2 \)
Since \( y^2 \) and higher powers are negligible, we ignore the term \( n^2 y^2/x^2 \):
\( \implies = 1 - \frac { 2ny }{ x } \)
This matches the Right-Hand Side (R.H.S.).
Hence Proved.
In simple words: When y is much smaller than x, we can simplify the expression. By using a special formula that ignores very tiny numbers squared or cubed, the left side of the equation becomes exactly the same as the right side.

🎯 Exam Tip: Remember to use the binomial approximation \((1+x)^n \approx 1+nx \) for small x and clearly state which terms are neglected. Pay attention to negative signs in exponents.

Question 2. If x is too small such that \( x^2 \) and higher powers are negligible then find the value of the following expressions:
(i) \( \frac { (9+2x)^{1/2} (3+4x) }{ (1+x)^{1/5} } \)
(ii) \( \frac { \sqrt{1-2x} + (1+3x)^{4/3} }{ 3+x+\sqrt{4-x} } \)
(iii) \( \frac { (1+3x)^{-4} \sqrt{16-3x} }{ (8+x)^{2/3} } \)
Answer: We will simplify each expression by using the binomial approximation \((1+z)^m \approx 1+mz \) and ignoring terms with \( x^2 \) or higher powers, as x is very small.

(i) \( \frac { (9+2x)^{1/2} (3+4x) }{ (1+x)^{1/5} } \)
First, simplify the terms individually:
\( (9+2x)^{1/2} = [9(1 + 2x/9)]^{1/2} = 9^{1/2} (1 + 2x/9)^{1/2} = 3(1 + \frac { 1 }{ 2 } \cdot \frac { 2x }{ 9} ) = 3(1 + x/9) \)
\( (1+x)^{1/5} = (1 + \frac { 1 }{ 5 } \cdot x) = (1 + x/5) \)
Now substitute these into the expression:
\( \frac { 3(1 + x/9) (3+4x) }{ (1 + x/5) } \)
\( = 3(1 + x/9)(3+4x)(1 + x/5)^{-1} \)
Approximate \( (1 + x/5)^{-1} \approx (1 - x/5) \)
\( = 3(1 + x/9)(3+4x)(1 - x/5) \)
Multiply the first two terms:
\( 3(3 + 4x + x/3 + 4x^2/9) \)
Ignore \( x^2 \) term:
\( = 3(3 + 4x + x/3) = 3(3 + 13x/3) = 9 + 13x \)
Now multiply by the last term:
\( = (9 + 13x)(1 - x/5) \)
\( = 9 - 9x/5 + 13x - 13x^2/5 \)
Ignore \( x^2 \) term:
\( = 9 + 13x - 9x/5 \)
\( = 9 + (\frac { 65 - 9 }{ 5 } )x = 9 + \frac { 56 }{ 5 } x \)
So, \( \frac { (9+2x)^{1/2} (3+4x) }{ (1+x)^{1/5} } = 9 + \frac { 56 }{ 5 } x \)
In simple words: We break down the complex fraction into simpler parts. We use a rule that says for small 'x', numbers like \( (1+x)^m \) can be approximated as \( 1+mx \). After applying this rule to each part and ignoring any \( x^2 \) terms, we combine them to get a simplified answer.

(ii) \( \frac { \sqrt{1-2x} + (1+3x)^{4/3} }{ 3+x+\sqrt{4-x} } \)
Simplify numerator terms:
\( \sqrt{1-2x} = (1-2x)^{1/2} \approx 1 + \frac { 1 }{ 2 } (-2x) = 1 - x \)
\( (1+3x)^{4/3} \approx 1 + \frac { 4 }{ 3 } (3x) = 1 + 4x \)
Numerator becomes: \( (1-x) + (1+4x) = 2 + 3x \)
Simplify denominator terms:
\( \sqrt{4-x} = [4(1 - x/4)]^{1/2} = 4^{1/2} (1 - x/4)^{1/2} = 2(1 + \frac { 1 }{ 2 } (-x/4) ) = 2(1 - x/8) = 2 - x/4 \)
Denominator becomes: \( 3+x+(2-x/4) = 5 + x - x/4 = 5 + 3x/4 \)
Now, the entire expression is:
\( \frac { 2 + 3x }{ 5 + 3x/4 } \)
Factor out 5 from the denominator:
\( = \frac { 2 + 3x }{ 5(1 + 3x/20) } = \frac { 1 }{ 5 } (2+3x)(1 + 3x/20)^{-1} \)
Approximate \( (1 + 3x/20)^{-1} \approx (1 - 3x/20) \)
\( = \frac { 1 }{ 5 } (2+3x)(1 - 3x/20) \)
\( = \frac { 1 }{ 5 } (2 - 6x/20 + 3x - 9x^2/20) \)
Ignore \( x^2 \) term:
\( = \frac { 1 }{ 5 } (2 - 3x/10 + 3x) = \frac { 1 }{ 5 } (2 + \frac { 30x - 3x }{ 10 } ) = \frac { 1 }{ 5 } (2 + 27x/10) \)
\( = \frac { 2 }{ 5 } + \frac { 27 }{ 50 } x \)
So, \( \frac { \sqrt{1-2x} + (1+3x)^{4/3} }{ 3+x+\sqrt{4-x} } = \frac { 2 }{ 5 } + \frac { 27 }{ 50 } x \)
In simple words: We simplify the top and bottom parts of the fraction separately. For terms like square roots or powers, we use the approximation that works for very small 'x'. Then, we combine the simplified top and bottom parts to get the final answer.

(iii) \( \frac { (1+3x)^{-4} \sqrt{16-3x} }{ (8+x)^{2/3} } \)
Simplify numerator terms:
\( (1+3x)^{-4} \approx 1 + (-4)(3x) = 1 - 12x \)
\( \sqrt{16-3x} = [16(1 - 3x/16)]^{1/2} = 16^{1/2} (1 - 3x/16)^{1/2} = 4(1 + \frac { 1 }{ 2 } (-3x/16) ) = 4(1 - 3x/32) = 4 - 3x/8 \)
Numerator becomes: \( (1 - 12x)(4 - 3x/8) \)
\( = 4 - 3x/8 - 48x + 36x^2/8 \)
Ignore \( x^2 \) term:
\( = 4 - x(\frac { 3 }{ 8 } + 48) = 4 - x(\frac { 3 + 384 }{ 8 } ) = 4 - \frac { 387 }{ 8 } x \)
Simplify denominator term:
\( (8+x)^{2/3} = [8(1 + x/8)]^{2/3} = 8^{2/3} (1 + x/8)^{2/3} \)
\( = (\sqrt[3]{8})^2 (1 + \frac { 2 }{ 3 } \cdot x/8) = 2^2 (1 + x/12) = 4(1 + x/12) = 4 + x/3 \)
Now, the entire expression is:
\( \frac { 4 - 387x/8 }{ 4 + x/3 } \)
Factor out 4 from both numerator and denominator:
\( = \frac { 4(1 - 387x/32) }{ 4(1 + x/12) } = (1 - 387x/32)(1 + x/12)^{-1} \)
Approximate \( (1 + x/12)^{-1} \approx (1 - x/12) \)
\( = (1 - 387x/32)(1 - x/12) \)
\( = 1 - x/12 - 387x/32 + 387x^2/(32 \cdot 12) \)
Ignore \( x^2 \) term:
\( = 1 - x(\frac { 1 }{ 12 } + \frac { 387 }{ 32 } ) \)
\( = 1 - x(\frac { 8 + 3 \cdot 387 }{ 96 } ) = 1 - x(\frac { 8 + 1161 }{ 96 } ) = 1 - \frac { 1169 }{ 96 } x \)
So, \( \frac { (1+3x)^{-4} \sqrt{16-3x} }{ (8+x)^{2/3} } = 1 - \frac { 1169 }{ 96 } x \)
In simple words: We use the same simplification method as before. We approximate each part of the numerator and denominator using the binomial theorem, ignoring any \( x^2 \) terms since x is tiny. Then, we combine these simplified parts to arrive at the final, simple expression.

🎯 Exam Tip: Remember to factor out constants from inside binomial terms like \( (9+2x)^{1/2} \) to get it into the \( (1+z)^m \) form before applying the approximation. This is a common step and helps in correct simplification.

Question 3. Find the value of:
(i) \( \sqrt{30} \) upto 4 places of decimal
(ii) \( (1.03)^{1/3} \) upto 4 places of decimal
(iii) \( \frac { 1 }{ (8.16)^{1/3} } \) upto 4 places of decimal
(iv) Cube root of 126 upto 5 places of decimal
Answer: We will use binomial approximation to find the values.

(i) Value of \( \sqrt{30} \) upto 4 places of decimal
We can write \( \sqrt{30} \) as a term near a perfect square:
\( \sqrt{30} = \sqrt{36 - 6} = \sqrt{36(1 - 6/36)} = 6\sqrt{1 - 1/6} \)
\( = 6 (1 - 1/6)^{1/2} \)
Using the binomial expansion \( (1+x)^n = 1 + nx + \frac { n(n-1) }{ 2! } x^2 + \frac { n(n-1)(n-2) }{ 3! } x^3 + ... \)
Here, \( n = 1/2 \) and \( x = -1/6 \).
\( = 6 [1 + \frac { 1 }{ 2 } (-1/6) + \frac { (1/2)(1/2 - 1) }{ 2! } (-1/6)^2 + \frac { (1/2)(1/2 - 1)(1/2 - 2) }{ 3! } (-1/6)^3 + ...] \)
\( = 6 [1 - \frac { 1 }{ 12 } + \frac { (1/2)(-1/2) }{ 2 } (1/36) + \frac { (1/2)(-1/2)(-3/2) }{ 6 } (-1/216) + ...] \)
\( = 6 [1 - \frac { 1 }{ 12 } - \frac { 1 }{ 8 } \cdot \frac { 1 }{ 36 } + \frac { 1 }{ 16 } \cdot \frac { 1 }{ 216 } + ...] \)
\( = 6 [1 - \frac { 1 }{ 12 } - \frac { 1 }{ 288 } + \frac { 1 }{ 3456 } + ...] \)
Calculate the decimal values:
\( = 6 [1 - 0.083333 - 0.003472 + 0.000289 + ...] \)
\( = 6 [0.913484] \)
\( = 5.480904 \)
Upto 4 decimal places, \( \sqrt{30} \approx 5.4809 \)
In simple words: We want to find the square root of 30. We rewrite it as \( \sqrt{36-6} \) to make it fit a special math formula. We then use this formula to calculate the value, adding more terms for better accuracy until we have four decimal places.

(ii) Value of \( (1.03)^{1/3} \) upto 4 places of decimal
We can write \( (1.03)^{1/3} \) as \( (1 + 0.03)^{1/3} \).
Using the binomial expansion \( (1+x)^n = 1 + nx + \frac { n(n-1) }{ 2! } x^2 + \frac { n(n-1)(n-2) }{ 3! } x^3 + ... \)
Here, \( n = 1/3 \) and \( x = 0.03 \).
\( = 1 + \frac { 1 }{ 3 } (0.03) + \frac { (1/3)(1/3 - 1) }{ 2! } (0.03)^2 + \frac { (1/3)(1/3 - 1)(1/3 - 2) }{ 3! } (0.03)^3 + ... \)
\( = 1 + 0.01 + \frac { (1/3)(-2/3) }{ 2 } (0.0009) + \frac { (1/3)(-2/3)(-5/3) }{ 6 } (0.000027) + ... \)
\( = 1 + 0.01 + (-\frac { 1 }{ 9 } ) (0.0009) + (\frac { 10 }{ 81 } ) (0.000027) + ... \)
\( = 1 + 0.01 - 0.0001 + 0.0000033 + ... \)
\( = 1.0099033 \)
Upto 4 decimal places, \( (1.03)^{1/3} \approx 1.0099 \)
In simple words: We want to find the cube root of 1.03. We write it as \( (1 + 0.03)^{1/3} \) and use the binomial formula for \( (1+x)^n \). We calculate the first few terms of this formula and add them up to get the answer, rounded to four decimal places.

(iii) Value of \( \frac { 1 }{ (8.16)^{1/3} } \) upto 4 places of decimal
\( \frac { 1 }{ (8.16)^{1/3} } = (8.16)^{-1/3} \)
We can write \( (8.16)^{-1/3} \) as \( (8 + 0.16)^{-1/3} \).
Factor out 8:
\( = [8(1 + 0.16/8)]^{-1/3} = 8^{-1/3} (1 + 0.02)^{-1/3} \)
\( = \frac { 1 }{ \sqrt[3]{8} } (1 + 0.02)^{-1/3} = \frac { 1 }{ 2 } (1 + 0.02)^{-1/3} \)
Using the binomial expansion \( (1+x)^n = 1 + nx + \frac { n(n-1) }{ 2! } x^2 + ... \)
Here, \( n = -1/3 \) and \( x = 0.02 \).
\( = \frac { 1 }{ 2 } [1 + (-\frac { 1 }{ 3 } )(0.02) + \frac { (-1/3)(-1/3 - 1) }{ 2! } (0.02)^2 + ...] \)
\( = \frac { 1 }{ 2 } [1 - \frac { 0.02 }{ 3 } + \frac { (-1/3)(-4/3) }{ 2 } (0.0004) + ...] \)
\( = \frac { 1 }{ 2 } [1 - 0.006666 + \frac { 4/9 }{ 2 } (0.0004) + ...] \)
\( = \frac { 1 }{ 2 } [1 - 0.006666 + \frac { 2 }{ 9 } (0.0004) + ...] \)
\( = \frac { 1 }{ 2 } [1 - 0.006666 + 0.000088 + ...] \)
\( = \frac { 1 }{ 2 } [0.993422] \)
\( = 0.496711 \)
Upto 4 decimal places, \( \frac { 1 }{ (8.16)^{1/3} } \approx 0.4967 \)
In simple words: To find the value of one divided by the cube root of 8.16, we first rewrite it as \( (8.16)^{-1/3} \). We then take out the number 8 to get \( (1+x)^n \) form and use the binomial formula. After calculating the terms and summing them up, we divide by 2 and round to four decimal places.

(iv) Cube root of 126 upto 5 places of decimal
We want to find \( \sqrt[3]{126} \). We can write it as a term near a perfect cube:
\( \sqrt[3]{126} = \sqrt[3]{125 + 1} = \sqrt[3]{125(1 + 1/125)} \)
\( = \sqrt[3]{125} (1 + 1/125)^{1/3} = 5 (1 + 0.008)^{1/3} \)
Using the binomial expansion \( (1+x)^n = 1 + nx + \frac { n(n-1) }{ 2! } x^2 + \frac { n(n-1)(n-2) }{ 3! } x^3 + ... \)
Here, \( n = 1/3 \) and \( x = 0.008 \).
\( = 5 [1 + \frac { 1 }{ 3 } (0.008) + \frac { (1/3)(1/3 - 1) }{ 2! } (0.008)^2 + \frac { (1/3)(1/3 - 1)(1/3 - 2) }{ 3! } (0.008)^3 + ...] \)
\( = 5 [1 + 0.00266666 + \frac { (1/3)(-2/3) }{ 2 } (0.000064) + \frac { (1/3)(-2/3)(-5/3) }{ 6 } (0.000000512) + ...] \)
\( = 5 [1 + 0.00266666 - \frac { 1 }{ 9 } (0.000064) + \frac { 10 }{ 81 } (0.000000512) + ...] \)
\( = 5 [1 + 0.00266666 - 0.00000711 + 0.00000006 + ...] \)
\( = 5 [1.00265961] \)
\( = 5.01329805 \)
Upto 5 decimal places, \( \sqrt[3]{126} \approx 5.01330 \)
In simple words: To find the cube root of 126, we first express it as \( \sqrt[3]{125+1} \). This allows us to use the binomial formula for \( (1+x)^n \). We calculate the terms, add them up, and then multiply by 5 to get the final answer, rounded to five decimal places.

🎯 Exam Tip: When finding roots of numbers, always try to express the number as \( (a^k \pm \text{small value}) \) so you can factor out \( a^k \) and use the \( (1 \pm x)^n \) binomial expansion. Choose the perfect square or cube closest to the given number for x to be small.

Question 4. If x is approximately equal to 1, then prove that:
(i) \( \frac { mx^m - nx^n }{ m-n } = x^{m+n} \)
(ii) \( \frac { ax^b - bx^a }{ x^b - x^a } = \frac { 1 }{ 1-x } \)
Answer: We will prove these identities by substituting \( x = 1 - h \) where h is very small, and neglecting higher powers of h.

(i) To prove \( \frac { mx^m - nx^n }{ m-n } = x^{m+n} \)
Let \( x = 1 - h \), where h is very small. Then \( h \to 0 \).
Using the binomial approximation \( (1-h)^k \approx 1 - kh \):
\( x^m = (1-h)^m \approx 1 - mh \)
\( x^n = (1-h)^n \approx 1 - nh \)
Substitute these into the L.H.S.:
L.H.S. \( = \frac { m(1-mh) - n(1-nh) }{ m-n } \)
\( = \frac { m - m^2 h - n + n^2 h }{ m-n } \)
\( = \frac { (m-n) - (m^2 - n^2)h }{ m-n } \)
\( = \frac { (m-n) - (m-n)(m+n)h }{ m-n } \)
\( = 1 - (m+n)h \)
Now, let's look at the R.H.S.:
R.H.S. \( = x^{m+n} = (1-h)^{m+n} \)
Using the binomial approximation:
\( \approx 1 - (m+n)h \)
Since L.H.S. \( = \) R.H.S., the identity is proved.
In simple words: Since x is almost 1, we can write it as \( 1-h \) where h is a tiny number. We then use a simple formula to expand expressions like \( x^m \) and \( x^n \). By plugging these back into the original equation and removing all tiny \( h^2 \) terms, both sides of the equation become the same.

(ii) To prove \( \frac { ax^b - bx^a }{ x^b - x^a } = \frac { 1 }{ 1-x } \)
Let \( x = 1 - h \), where h is very small. Then \( h \to 0 \).
Using the binomial approximation \( (1-h)^k \approx 1 - kh \):
\( x^b = (1-h)^b \approx 1 - bh \)
\( x^a = (1-h)^a \approx 1 - ah \)
Substitute these into the L.H.S.:
L.H.S. \( = \frac { a(1-bh) - b(1-ah) }{ (1-bh) - (1-ah) } \)
\( = \frac { a - abh - b + abh }{ 1 - bh - 1 + ah } \)
\( = \frac { a - b }{ ah - bh } \)
\( = \frac { a - b }{ h(a - b) } \)
\( = \frac { 1 }{ h } \)
Now, let's look at the R.H.S.:
R.H.S. \( = \frac { 1 }{ 1-x } = \frac { 1 }{ 1-(1-h) } = \frac { 1 }{ 1-1+h } = \frac { 1 }{ h } \)
Since L.H.S. \( = \) R.H.S., the identity is proved.
In simple words: Again, we replace x with \( 1-h \) because x is very close to 1. We then use the approximation for powers of \( (1-h) \) on both sides of the equation. After simplifying everything, and cancelling common terms, we find that the left side becomes identical to the right side, proving the statement.

🎯 Exam Tip: When x approaches 1, it's often useful to substitute \( x = 1-h \) (or \( x = 1+h \)) and use the binomial approximation for small h. Remember to simplify both sides of the identity to show they are equal.

Question 5. If \( p = q + h \) where h is too small whose higher order terms can be neglected, prove that \( \frac { q+2p }{ p+2q } = \frac { q+2(q+h) }{ (q+h)+2q } = 1 + \frac { h }{ 3q } \).
Answer: We need to prove the given identity by substituting \( p = q+h \) and simplifying the expression, neglecting higher powers of h.

Let's start with the Left-Hand Side (L.H.S.) and substitute \( p = q+h \):
L.H.S. \( = \frac { q+2p }{ p+2q } \)
\( = \frac { q+2(q+h) }{ (q+h)+2q } \)
\( = \frac { q+2q+2h }{ q+h+2q } \)
\( = \frac { 3q+2h }{ 3q+h } \)
To simplify this expression, we can divide both the numerator and the denominator by 3q:
\( = \frac { \frac{3q+2h}{3q} }{ \frac{3q+h}{3q} } \)
\( = \frac { 1 + \frac{2h}{3q} }{ 1 + \frac{h}{3q} } \)
Now, we can rewrite this as:
\( = (1 + \frac { 2h }{ 3q } ) (1 + \frac { h }{ 3q } )^{-1} \)
Since h is very small, \( \frac { h }{ 3q } \) is also very small. We can use the binomial approximation \( (1+x)^n \approx 1 + nx \) for small x.
So, \( (1 + \frac { h }{ 3q } )^{-1} \approx 1 + (-1) \frac { h }{ 3q } = 1 - \frac { h }{ 3q } \)
Substitute this approximation back into the expression:
\( = (1 + \frac { 2h }{ 3q } ) (1 - \frac { h }{ 3q } ) \)
Now, multiply these two terms:
\( = 1 - \frac { h }{ 3q } + \frac { 2h }{ 3q } - \frac { 2h^2 }{ 9q^2 } \)
We are told to neglect higher order terms, which means we can ignore the \( \frac { 2h^2 }{ 9q^2 } \) term because \( h^2 \) is extremely small.
\( \implies = 1 - \frac { h }{ 3q } + \frac { 2h }{ 3q } \)
\( = 1 + \frac { h }{ 3q } \)
This matches the Right-Hand Side (R.H.S.).
Hence Proved.
In simple words: We start by replacing 'p' with 'q+h' in the given expression. Then we simplify the fraction. Because 'h' is very small, we use a simple rule to approximate one part of the fraction. After multiplying and ignoring any super tiny terms with 'h' squared, the whole expression becomes \( 1 + h/(3q) \), which is what we needed to prove.

🎯 Exam Tip: When simplifying expressions involving small quantities, remember to use the binomial approximation \((1+x)^n \approx 1+nx \) and identify all terms that can be neglected (typically those with \( x^2 \) or higher powers).