(x)((a)/(2b)+(2b)/(a))^(2)-((2b)/(a)-(a)/(2b))^(2)

If f(x)=((a+x)/(b+x))^(a+b+2x) then prove that f'(0)=[2log((a)/(b))+(b^(2)-a^(2))/(ab)]((a)/(b))^(a+b)

If f(x)=((a+x)/(b+x))^(a+b+2x) , prove that, f'(0) =[2"log"(a)/(b) +(b^(2)-a^(2))/(ab)]((a)/(b))^(a+b) .

Evaluate : (iii) ((a)/(2b ) + (2b )/( a) ) ( ( a)/( 2b ) - (2b)/( a))

Area of the quadrilateral formed with the foci of the hyperbola (x^(2))/(a^(2))-(y^(2))/(b^(2))=1 and (x^(2))/(a^(2))-(y^(2))/(b^(2))=-1( a) 4(a^(2)+b^(2))( b) 2(a^(2)+b^(2))( c) (a^(2)+b^(2))(d)(1)/(2)(a^(2)+b^(2))

Solve (by any method) : (a-b)/x+(a+b)/y=(2(a^(2)+b^(2)))/((a^(2)-b^(2))) (a+b)/x+(a-b)/y=2

If x=a sec theta and y=b tan theta, then b^(2)x^(2)-a^(2)y^(2)=( a )ab(b)a^(2)-b^(2)(c)a^(2)+b^(2) (d) a^(2)b^(2)

If af(x+1)+bf(1/(x+1))=x ,x!=-1,a!=b ,t h e nf(2) is equal to (a) (2a+b)/(2(a^2-b^2)) (b) a/(a^2-b^2) (c) (a+2b)/(a^2-b^2) (d) none of these

The value of c for which the equation a x^2+2b x+c=0 has equal roots is (a) (b^2)/a (b) (b^2)/(4a) (c) (a^2)/b (d) (a^2)/(4b)

The value of c for which the equation a x^2+2b x+c=0 has equal roots is (b^2)/a (b) (b^2)/(4a) (c) (a^2)/b (d) (a^2)/(4b)