c is any real number and `c ne 0`. Prove that `abs(f(c) - f(-c)) = 2`, where f(x) = `absx / x`

Let the functions f and g on the set of real numbers RR be defined by, f(x)= cos x and g(x) =x^(3) . Prove that, (f o g) ne (g o f).

If f(x) is a twice differentiable function such that f(a)=0, f(b)=2, f(c)=-1,f(d)=2, f(e)=0 where a < b < c < de, then the minimum number of zeroes of g(x) = f'(x)^2+f''(x)f(x) in the interval [a, e] is

If f(x)=(|x|)/(x) and c be a non-zero real number, show that |f(c)-f(-c)|=2.

Let f(x)=ax^(2)+bx+c , a ne 0 , a , b , c in I . Suppose that f(1)=0 , 50 lt f(7) lt 60 and 70 lt f(8) lt 80 . The least value of f(x) 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 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

Let f(x)=Ax^2+Bx+C where A, B, C are three real constants, if f(x) is integer for integral values of x, then prove that each of 2A, (A+B) and C is an integer. Conversely, if each of 2A, (A+B) , C is an integer then f(x) will be integer for integral values of x.

Let f(x)=a x^2+bx +c a ,b ,c in R . If f(x) takes real values for real values of x and non-real values for non-real values of x , then (a) a=0 (b) b=0 (c) c=0 (d) nothing can be said about a ,b ,c .

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) . The value of the parameter a for which the area of the figure bounded by the abscissa axis, the graph of the function y=x^(3)+3x^(2)+x+a , and the straight lines, which are parallel to the axis of ordinates and cut the abscissa axis at the point of extremum of the function, which is the least, is

Let f(x+y) = f(x) . f(y) for all x,y where f(0) ne 0. If f(5) =2 and f'(0) =3 , then f'(5) is equal to