A curve `g(x)=intx^(27)(1+x+x^2)^6(6x^2+5x+4)dx` is passing through origin. Then `g(1)=(3^7)/7` (b) `g(1)=(2^7)/7` `g(-1)=1/7` (d) `g(-1)=(3^7)/(14)`

A point on the line (x-1)/1=(y-2)/2=(z+1)/3 at a distance sqrt(6) from the origin is (A) ((-5)/7, (-10)/7,13/7) (B) (5/7,10/7,(-13)/7) (C) (1,2,-1) (D) (-1,-2,1)

f(x)=x^(3)-6x^(2)-19x+84,g(x)=x-7

((6x^(4)+8x^(3)+27x^(2)+7)/(3x^(2)+4x+1))

(3x+1)/(6)+(2x-3)/(7)=(x+3)/(8)+(3x-1)/(14)

((2)/(3)x+4)((3)/(2)x+6)-((1)/(7)x-1)((1)/(7)x+1)

Let f(x)=x^(2)-x+5,x>(1)/(2) and g(x) is its inverse function,then g'(7) equals

If (1-x^(7))/(x(1+x^(7)))dx=a ln|x|+b ln|x^(7)+1|+c then a=1,b=(2)/(7) (b) a=-1,b=(2)/(7)a=1,b=-(2)/(7)(d)a=-1,b=-(2)/(7)

" 2" f(x)=4x^(4)-3x^(3)-2x^(2)+x-7,g(x)=x-1