a * b = \frac{a^2+b^2}{ab} , a > 0 , b > 0 ⇒ \frac{(1*2)*(1*1)}{1*3} =

TS ICET 24 May 2019 Shift 1 Paper

Question : 74 of 200

Marks: +1, -0

a * b =

{a^2+b^2}/{ab}

, a > 0 , b > 0 ⇒

{(1*2)*(1*1)}/{1*3}

Solution:

a^{*} b=\;{a^{2}+b^{2}}/{a b}, a>0 b>0

(1^{*} 2) ⇒ \;{1^{2}+2^{2}}/{1 × 2}=\;{5}/{2}

(1^{*} 1) ⇒ \;{1^{2}+1^{2}}/{1 × 1}=\;{2}/{1}

(1 * 2) *(1 * 1)=\;{({5}/{2})^{2}+({2}/{1})^{2}}/{{5}/{2} × 2}=\;{\;{25}/{4}+4}/{5}=\;{41}/{20}

(1+3) ⇒ \;{1^{2}+3^{2}}/{1 × 3}=\;{1+9}/{3}=\;{10}/{3}

\;{(1 * 2) *(1 * 1)}/{(1 * 3)}=\;{\;{41}/{20}}/{{10}/{3}}=\;{41 × 3}/{20 × 10}=\;{123}/{200}

Go to Question:

Prev Question

Next Question