Answer
424.8k+ views
Hint: To prove this, we simplify both the fractions separately from \[\dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}}\] by using laws of indices given below. Since, we have the same base i.e. x, so we can apply the following laws.
Laws of indices: $(i)$ ${a^m} \cdot {a^n} = {a^{m + n}}$
$(ii)$ $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$
$(iii)$ ${a^{ - n}} = \dfrac{1}{{{a^n}}}$
Complete step-by-step answer:
Simplify \[\dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}}\] by using laws of indices.
Use the law of indices ${a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}$ and rewrite the terms.
\[\therefore \dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}} = \dfrac{1}{{1 + \dfrac{{{x^a}}}{{{x^b}}}}} + \dfrac{1}{{1 + \dfrac{{{x^b}}}{{{x^a}}}}}\]
Find the least common denominator and take LCM in the denominator of both fractions.
$ \Rightarrow \dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}} = \dfrac{1}{{\dfrac{{{x^b} + {x^a}}}{{{x^b}}}}} + \dfrac{1}{{\dfrac{{{x^a} + {x^b}}}{{{x^a}}}}}$
Remember, that dividing by fraction is the same as flipping the fraction in the denominator and then multiplying it with the numerator.
So flip the fraction in the denominator and multiply it to the numerator.
$ \Rightarrow \dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}} = \dfrac{{{x^b}}}{{{x^a} + {x^b}}} + \dfrac{{{x^a}}}{{{x^a} + {x^b}}}$
Take LCM of both the fractions.
\[ \Rightarrow \dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}} = \dfrac{{{x^b} + {x^a}}}{{{x^a} + {x^b}}}\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow \dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}} = 1\]
Hence proved.
Additional Information: * When the base is the same then only we can apply the law of indices, be it multiplication of numbers with the same base or division of numbers with the same base.
* Always keep in mind that any number with power 0 will be equal to 1.
\[ \Rightarrow {a^0} = 1\]
* If powers are in fraction form, then we can write them as
\[ \Rightarrow {a^{\dfrac{m}{n}}} = {({a^m})^{\dfrac{1}{n}}}\] which means we take nth root of the term inside the bracket, so we can write
\[ \Rightarrow {a^{\dfrac{m}{n}}} = \sqrt[n]{{{a^m}}}\]
* Any number with the power m means that the number is multiplied to itself m number of times.
So, we can write \[{x^m} = \underbrace {x \times x \times x........ \times x}_m\]
Note: Students make mistake of assuming the values \[{x^{b - a}},{x^{a - b}}\]as some variables and try solving the fraction which is wrong way to proceed with the question. So, keep in mind we have to use formulas which make solutions easy and not more complex.
Laws of indices: $(i)$ ${a^m} \cdot {a^n} = {a^{m + n}}$
$(ii)$ $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$
$(iii)$ ${a^{ - n}} = \dfrac{1}{{{a^n}}}$
Complete step-by-step answer:
Simplify \[\dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}}\] by using laws of indices.
Use the law of indices ${a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}$ and rewrite the terms.
\[\therefore \dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}} = \dfrac{1}{{1 + \dfrac{{{x^a}}}{{{x^b}}}}} + \dfrac{1}{{1 + \dfrac{{{x^b}}}{{{x^a}}}}}\]
Find the least common denominator and take LCM in the denominator of both fractions.
$ \Rightarrow \dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}} = \dfrac{1}{{\dfrac{{{x^b} + {x^a}}}{{{x^b}}}}} + \dfrac{1}{{\dfrac{{{x^a} + {x^b}}}{{{x^a}}}}}$
Remember, that dividing by fraction is the same as flipping the fraction in the denominator and then multiplying it with the numerator.
So flip the fraction in the denominator and multiply it to the numerator.
$ \Rightarrow \dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}} = \dfrac{{{x^b}}}{{{x^a} + {x^b}}} + \dfrac{{{x^a}}}{{{x^a} + {x^b}}}$
Take LCM of both the fractions.
\[ \Rightarrow \dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}} = \dfrac{{{x^b} + {x^a}}}{{{x^a} + {x^b}}}\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow \dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}} = 1\]
Hence proved.
Additional Information: * When the base is the same then only we can apply the law of indices, be it multiplication of numbers with the same base or division of numbers with the same base.
* Always keep in mind that any number with power 0 will be equal to 1.
\[ \Rightarrow {a^0} = 1\]
* If powers are in fraction form, then we can write them as
\[ \Rightarrow {a^{\dfrac{m}{n}}} = {({a^m})^{\dfrac{1}{n}}}\] which means we take nth root of the term inside the bracket, so we can write
\[ \Rightarrow {a^{\dfrac{m}{n}}} = \sqrt[n]{{{a^m}}}\]
* Any number with the power m means that the number is multiplied to itself m number of times.
So, we can write \[{x^m} = \underbrace {x \times x \times x........ \times x}_m\]
Note: Students make mistake of assuming the values \[{x^{b - a}},{x^{a - b}}\]as some variables and try solving the fraction which is wrong way to proceed with the question. So, keep in mind we have to use formulas which make solutions easy and not more complex.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)