If f(a,b)`=(f(b)-f(a))/(b-a)` and
`f(a,b,c)=(f(b,c)-f(a,b))/(c-a)` prove that `f(a,b,c)=|{:(f(a),f(b),f(c)),(1,1,1),(a,b,c):}|-:|{:(1,1,1),(a,b,c),(a^(2),b^(2),c^(2)):}|`.

If f(x)=log_e((1-x)/(1+x)); prove that f(a)+f(b)=f((a+b)/(1+a b))

If f(x)=(x-1)/(x+1) , then prove that: (f(b)-f(a))/(1+f(b)*f(a))=(b-a)/(1+ab)

f(x)=x/(x-1) then (f(a))/(f(a+1)) is equal to a. f(-a) b. f(1//a) c. f(a^2) d. f(-a/(a-1))

Using properties of determinant, If f(x)= a + bx + cx^(2) , prove that |(a,b,c),(b,c,a),(c,a,b)|= -f(1)f(omega)f(omega^(2)), omega is a cube root of unity

If f(x)=ax^(2)+bx+c and f(x+1)=f(x)+x+1 , then the value of (a+b) is __

If g(x)=(f(x))/((x-a)(x-b)(x-c)) ,where f(x) is a polynomial of degree <3 , then intg(x)dx=|[1,a,f(a)log|x-a|],[1,b,f(b)log|x-b|],[1,c,f(c)log|x-c|]|-:|[1,a, a^2],[ 1,b,b^2],[ 1,c,c^2]|+k (dg(x))/(dx)=|[1,a,-f(a)(x-a)^(-2)],[1,b,-f(b)(x-b)^(-2)],[1,c,-f(c)(x-c)^(-2)]|:-|[1,a,a^2],[1,b,b^2],[1,c,c^2]|

Suppose we define the definite integral using the following formula int_a^b f(x)dx=(b-a)/2 (f(x)+f(b)) , for more accurate result for c in (a,b), F(c)=(c-a)/2 (f(a)+f(c))+(b-c)/2(f(b)+f(c)) and when c=(a+b)/2, int_a^b f(x)dx=(b-a)/4(f(a)+f(b)+2f(c))

Suppose we define integral using the following formula int_(a)^(b)f(x)dx = (b-a)/(2) (f(a)+f(b)) , for more accurate result for c in (a, b), F(c) = (c-a)/(2) (f(a)+f(c)) + (b-c)/(2)(f(b) + f(c)) . When c = (a+b)/(2) , then int_(a)^(b) f(x)dx = (b-a)/(4)(f(a) + f(b) + 2f(c)) . lim_(trarra)(int_(a)^(t)f(x) dx -((t-a))/(2)(f(t)+f(a)))/((t-a)^(3))=0 forall a Then the degree of f(x) can at most be

If y=f(x) is a monotonic function in (a,b), then the area bounded by the ordinates at x=a, x=b, y=f(x) and y=f(c)("where "c in (a,b))" is minimum when "c=(a+b)/(2) . "Proof : " A=int_(a)^(c)(f(c)-f(x))dx+int_(c)^(b)(f(c))dx =f(c)(c-a)-int_(a)^(c)(f(x))dx+int_(a)^(b)(f(x))dx-f(c)(b-c) rArr" "A=[2c-(a+b)]f(c)+int_(c)^(b)(f(x))dx-int_(a)^(c)(f(x))dx Differentiating w.r.t. c, we get (dA)/(dc)=[2c-(a+b)]f'(c)+2f(c)+0-f(c)-(f(c)-0) For maxima and minima , (dA)/(dc)=0 rArr" "f'(c)[2c-(a+b)]=0(as f'(c)ne 0) Hence, c=(a+b)/(2) "Also for "clt(a+b)/(2),(dA)/(dc)lt0" and for "cgt(a+b)/(2),(dA)/(dc)gt0 Hence, A is minimum when c=(a+b)/(2) . If the area bounded by f(x)=(x^(3))/(3)-x^(2)+a and the straight lines x=0, x=2, and the x-axis is minimum, then the value of a is

If y=f(x) is a monotonic function in (a,b), then the area bounded by the ordinates at x=a, x=b, y=f(x) and y=f(c)("where "c in (a,b))" is minimum when "c=(a+b)/(2) . "Proof : " A=int_(a)^(c)(f(c)-f(x))dx+int_(c)^(b)(f(c))dx =f(c)(c-a)-int_(a)^(c) (f(x))dx+int_(a)^(b)(f(x))dx-f(c)(b-c) rArr" "A=[2c-(a+b)]f(c)+int_(c)^(b)(f(x))dx-int_(a)^(c)(f(x))dx Differentiating w.r.t. c, we get (dA)/(dc)=[2c-(a+b)]f'(c)+2f(c)+0-f(c)-(f(c)-0) For maxima and minima , (dA)/(dc)=0 rArr" "f'(c)[2c-(a+b)]=0(as f'(c)ne 0) Hence, c=(a+b)/(2) "Also for "clt(a+b)/(2),(dA)/(dc)lt0" and for "cgt(a+b)/(2),(dA)/(dc)gt0 Hence, A is minimum when c=(a+b)/(2) . If the area enclosed by f(x)= sin x + cos x, y=a between two consecutive points of extremum is minimum, then the value of a is