Show that:

Question:

Show that:

(i) $\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1$

(ii) $\left[\left\{\frac{x^{a(a-b)}}{x^{a(a+b)}}\right\} \div\left\{\frac{x^{b(b-a)}}{x^{b(b+a)}}\right\}\right]^{a+b}=1$

(iii) $\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}}\left(x^{\frac{1}{b-c}}\right)^{\frac{1}{b-a}}\left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}=1$

(iv) $\left(\frac{x^{a^{2}+b^{2}}}{x^{a b}}\right)^{a+b}\left(\frac{x^{b^{2}+c^{2}}}{x^{b c}}\right)^{b+c}\left(\frac{x^{c^{2}+a^{2}}}{x^{a c}}\right)^{a+c}=x^{2\left(a^{3}+b^{3}+c^{3}\right)}$

(v) $\left(x^{a-b}\right)^{a+b}\left(x^{b-c}\right)^{b+c}\left(x^{c-a}\right)^{c+a}=1$

(vi) $\left\{\left(x^{a-a^{-1}}\right)^{\frac{1}{a-1}}\right\}^{\frac{a}{a+1}}=x$

(vii) $\left(\frac{a^{x+1}}{a^{y+1}}\right)^{x+y}\left(\frac{a^{y+2}}{a^{z+2}}\right)^{y+z}\left(\frac{a^{z+3}}{a^{x+3}}\right)^{z+x}=1$

(viii) $\left(\frac{3^{a}}{3^{b}}\right)^{a+b}\left(\frac{3^{b}}{3^{c}}\right)^{b+c}\left(\frac{3^{c}}{3^{a}}\right)^{c+a}=1$

 

Solution:

(i)

$\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}$

$=\frac{1}{1+\frac{x^{a}}{x^{b}}}+\frac{1}{1+\frac{x^{b}}{x^{a}}}$

$=\frac{x^{b}}{x^{b}+x^{a}}+\frac{x^{a}}{x^{a}+x^{b}}$

$=\frac{x^{b}+x^{a}}{x^{a}+x^{b}}$

$=1$

(ii)xaa-bxaa+b÷xbb-axbb+aa+b=xaa-bxaa+b×xbb+axbb-aa+b=xa2-abxa2+ab×xb2+abxb2-aba+b">

$\left[\left\{\frac{x^{a(a-b)}}{x^{a(a+b)}}\right\} \div\left\{\frac{x^{b(b-a)}}{x^{b(b+a)}}\right\}\right]^{a+b}$

$=\left[\left\{\frac{x^{a(a-b)}}{x^{a(a+b)}}\right\} \times\left\{\frac{x^{b(b+a)}}{x^{b(b-a)}}\right\}\right]^{a+b}$

$=\left[\left\{\frac{x^{a^{2}-a b}}{x^{a^{2}+a b}}\right\} \times\left\{\frac{x^{b^{2}+a b}}{x^{b^{2}-a b}}\right\}\right]^{a+b}$

$=\left[\left\{x^{a^{2}-a b-a^{2}-a b}\right\} \times\left\{x^{b^{2}+a b-b^{2}+a b}\right\}\right]^{a+b}$

$=\left[x^{-2 a b} \times x^{2 a b}\right]^{a+b}$

$=\left[x^{-2 a b+2 a b}\right]^{a+b}$

$=\left[x^{0}\right]^{a+b}$

$=1$

(iii)

$\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}}\left(x^{\frac{1}{b-c}}\right)^{\frac{1}{b-a}}\left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}$

$=(x)^{\frac{1}{a-b} \times \frac{1}{a-c}}(x)^{\frac{1}{b-c} \times \frac{1}{b-a}}(x)^{\frac{1}{c-a} \times \frac{1}{c-b}}$

$=(x)^{\frac{1}{a-b}} \times \frac{1}{a-c}+\frac{1}{b-c} \times \frac{1}{b-a}+\frac{1}{c-a} \times \frac{1}{c-b}$

$=(x)^{\frac{(b-c)-(a-c)+(a-b)}{(a-b)(a-c)(b-c)}}$

$=x^{0}$

$=1$

(iv)

$\left(\frac{x^{a^{2}+b^{2}}}{x^{a b}}\right)^{a+b}\left(\frac{x^{b^{2}+c^{2}}}{x^{b c}}\right)^{b+c}\left(\frac{x^{c^{2}+a^{2}}}{x^{a c}}\right)^{a+c}$

$=\left(x^{a^{2}+b^{2}-a b}\right)^{a+b}\left(x^{b^{2}+c^{2}-b c}\right)^{b+c}\left(x^{c^{2}+a^{2}-a c}\right)^{a+c}$

$=\left[x^{(a+b)\left(a^{2}+b^{2}-a b\right)}\right]\left[x^{(b+c)\left(b^{2}+c^{2}-b c\right)}\right]\left[x^{(a+c)\left(c^{2}+a^{2}-a c\right)}\right]$

$=\left(x^{a^{3}+b^{3}}\right)\left(x^{b^{3}+c^{3}}\right)\left(x^{a^{3}+c^{3}}\right)$

$=x^{2\left(a^{3}+b^{3}+c^{3}\right)}$

(v)

$\left(x^{a-b}\right)^{a+b}\left(x^{b-c}\right)^{b+c}\left(x^{c-a}\right)^{c+a}$

$=\left[x^{(a-b)(a+b)}\right]\left[x^{(b-c)(b+c)}\right]\left[x^{(c-a)(c+a)}\right]$

$=x^{\left(a^{2}-b^{2}\right)} x^{\left(b^{2}-c^{2}\right)} x^{\left(c^{2}-a^{2}\right)}$

$=x^{a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}}$

$=x^{0}$

$=1$

(vi)

$\left\{\left(x^{a-a^{-1}}\right)^{\frac{1}{a-1}}\right\}^{\frac{a}{a+1}}$

$=\left\{\left(x^{a-\frac{1}{a}}\right)^{\frac{1}{a-1} \times \frac{a}{a+1}}\right\}$

$=\left\{x^{\frac{a^{2}-1}{a}}\right\}^{\frac{a}{a^{2}-1}}$

$=x^{\frac{a^{2}-1}{a} \times \frac{a}{a^{2}-1}}$

$=x^{1}$

$=x$

(vii)

$\left(\frac{a^{x+1}}{a^{y+1}}\right)^{x+y}\left(\frac{a^{y+2}}{a^{z+2}}\right)^{y+z}\left(\frac{a^{z+3}}{a^{x+3}}\right)^{z+x}$

$=\left(a^{x+1-y-1}\right)^{x+y}\left(a^{y+2-z-2}\right)^{y+z}\left(a^{z+3-x-3}\right)^{z+x}$

$=\left(a^{x-y}\right)^{x+y}\left(a^{y-z}\right)^{y+z}\left(a^{z-x}\right)^{z+x}$

$=a^{x^{2}-y^{2}+y^{2}-z^{2}+z^{2}-x^{2}}$

$=a^{0}$

$=1$

(viii)

$\left(\frac{3^{a}}{3^{b}}\right)^{a+b}\left(\frac{3^{b}}{3^{c}}\right)^{b+c}\left(\frac{3^{c}}{3^{a}}\right)^{c+a}$

$=\left(3^{a-b}\right)^{a+b}\left(3^{b-c}\right)^{b+c}\left(3^{c-a}\right)^{c+a}$

$=3^{a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}}$

$=3^{0}$

$=1$

 

 

 

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