Question

For x 20, the least value of the expression-

JEE/Engineering Exams

Maths

Solution

4.0 (1 ratings)

( f(x)=frac{1+x^{2}}{1+x} ) ( begin{aligned} f^{prime}(x)=& frac{2 x(1+x)-left(1+x^{2}right)(0+1)}{(1+x)^{2}}=0 f^{prime}(x)=& 2 x+2 x^{2}-1 div x^{2} Rightarrow f^{prime}(x)=x^{2}+2 x-1=0 f^{prime}(x)=frac{-2 pm sqrt{4-4 x(x-1)}}{2} &=-frac{2 pm 2 sqrt{2}}{2} end{aligned} ) ( x f^{prime}(x)=-frac{2 pm 2 sqrt{2}}{2}=frac{-1 pm sqrt{2}}{2} ) So ( x ) thould be ( x geqslant 0 Rightarrow x=-1+sqrt{2} ) Let check for ( x=0 quad x quad x=-1+sqrt{2} ) ( left.Rightarrow f(x)right|_{x=0}=frac{1}{1}=1 ) ( left.f(x)right|_{x=1+sqrt{2}}=frac{1+(-1+sqrt{2})^{2}}{1-1+sqrt{2}}=frac{1+1+2-2 sqrt{2}}{sqrt{2}} ) ( f(x)=frac{4-2 sqrt{2}}{sqrt{2}}=frac{2[2-sqrt{2}]}{sqrt{2}}=2 sqrt{2}-2 ) ( f(x) mid=(sqrt{2}-1) 2 ) ( Rightarrow ) Least value of ( x=1 mid operatorname{atx}=0 )