If 2x - 1-2x - 4- -find- -i-4x2 - 1-4x2- ii 8x3 - 1-8x3

(i) 2x 1/2x = 4 ⇒(2x 1/2x)^2 = (4)^2 ⇒ (2x)^2 + ( 1/2x)^2 2 × 2x ×1/2x = 16 ⇒ 4x^2 + 1/4x^2 2 = 16 ⇒ 4x^2 + 1/4x^2 = 16 + 2 ⇒ 4x^2 + 1/4x^2 = 18. (ii) 2x ...

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Byju's Answer

Standard VIII

Mathematics

Expansion of (x-y)^3

If 2x - 1/2x ...

Question

If $2 x - \frac{1}{2 x} = 4; f i n d : (i) 4 x^{2} + \frac{1}{4 x^{2}} (i i) 8 x^{3} - \frac{1}{8 x^{3}}$

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Solution

(i) $2 x - \frac{1}{2 x} = 4 \Rightarrow {(2 x - \frac{1}{2 x})}^{2} = (4)^{2} \Rightarrow (2 x)^{2} + {(\frac{1}{2 x})}^{2} - 2 \times 2 x \times \frac{1}{2 x} = 16 \Rightarrow 4 x^{2} + \frac{1}{4 x^{2}} - 2 = 16 \Rightarrow 4 x^{2} + \frac{1}{4 x^{2}} = 16 + 2 \Rightarrow 4 x^{2} + \frac{1}{4 x^{2}} = 18$ .

(ii) $2 x - \frac{1}{2 x} = 4 \Rightarrow {(2 x - \frac{1}{2 x})}^{3} = (4)^{3} \Rightarrow (2 x)^{3} - {(\frac{1}{2 x})}^{3} - 3 \times 2 x \times \frac{1}{2 x} (2 x - \frac{1}{2 x}) = 64 \Rightarrow 8 x^{3} - \frac{1}{8 x^{3}} - 3 (4) = 64 \Rightarrow 8 x^{3} - \frac{1}{8 x^{3}} - 12 = 64 \Rightarrow 8 x^{3} - \frac{1}{8 x^{3}} = 64 + 12 \Rightarrow 8 x^{3} - \frac{1}{8 x^{3}} = 76$ .

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Q. State True or False.

{[\frac{3}{2} x + 1]}^{3}

= \frac{27}{8} x^{3} + \frac{27}{4} x^{2} + \frac{9}{2} x + 1

Q. State True or False.
Expanded form of

{(\frac{3}{2} x + 1)}^{3} =

\frac{27}{8} x^{3} + \frac{27}{4} x^{2} + \frac{9}{2} x + 1

Q. Prove that

{tan}^{- 1} (\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}) - {tan}^{- 1} (\frac{4 x}{1 - 4 x^{2}}) = {tan}^{- 1} 2 x; | 2 x | < \frac{1}{\sqrt{3}}

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If 2x−12x=4; find :(i) 4x2+14x2(ii)8x3−18x3

(i) 2x−12x=4⇒(2x−12x)2=(4)2⇒(2x)2+(12x)2−2×2x×12x=16⇒4x2+14x2−2=16⇒4x2+14x2=16+2⇒4x2+14x2=18. (ii) 2x−12x=4⇒(2x−12x)3=(4)3⇒(2x)3−(12x)3−3×2x×12x(2x−12x)=64⇒8x3−18x3−3(4)=64⇒8x3−18x3−12=64⇒8x3−18x3=64+12⇒8x3−18x3=76.

If $2 x - \frac{1}{2 x} = 4; f i n d : (i) 4 x^{2} + \frac{1}{4 x^{2}} (i i) 8 x^{3} - \frac{1}{8 x^{3}}$