" 3."(1)/(x sqrt(ax-x^(2)))[" संकेत ":x=(a)/(t)" रखिप्य "

(1)/(sqrt(ax-x^(2)))

Integrate the functions (1)/(x sqrt(ax-x))[ Hints : Put x=(alpha)/(t)

Differentiate w.r.t. time. (i) y=t^(2) " " (ii) x=t^(3//2)" " (iii) y=(1)/(sqrt(t)) (iv) x=4t^(3) " " (v) y=2sqrt(t) " " (vi) y=2t^(2)+t-1 (vii) y=3sqrt(t)+(2)/(sqrt(t)) (viii) y=t^(3)sin t " " (ix) x=te^(t) (x) x= sqrt(t)(1-t)

If y=log((x^2+x+1)/(x^2-x+1))+2/(sqrt(3))t a n^(-1)((sqrt(3)x)/(1-x^2)),"f i n d"(dy)/(dx)

(i) int (dx)/(x sqrt(x^2-1)) (ii) int (dx)/(x sqrt(ax -x^2))

If int (x sqrt(x) -1)/(sqrt(x) -1) dx = ax^(2) + bx^(3//2) + cx + d then ascending order of a, b, c, is

IfI=int(dx)/((2ax+x^(2))^((3)/(2))), then I is equal to (a) -(x+a)/(sqrt(2ax+x^(2)))+c(b)-(1)/(a)(x+a)/(sqrt(2ax+x^(2)))+c(c)-(1)/(a^(2))(x+a)/(sqrt(2ax+x^(2)))+c(d)-(1)/(a^(3))(x+a)/(sqrt(2ax+x^(3)))+c

The equation of tangent to the curve y=int_(x^(2))^(x^(3))(dt)/(sqrt(1+t^(2))) at x=1 is sqrt(3)x+1=y (b) sqrt(2)y+1=x sqrt(3)x+y=1(d)sqrt(2)y=x

Which of the following expressions are polynomials ? In case of a polynomial , write its degree. (i) x^(5)-2x^(3)+x+sqrt(3) (ii) y^(3)+sqrt(3)y (iii) t^(2)-(2)/(5)t+sqrt(5) (iv) x^(100)-1 (v) (1)/(sqrt(2))x^(2)-sqrt(2)x+2 (vi) x^(-2)+2x^(-1)+3 (vii) 1 (viii) (-3)/(5) (ix) (x^(2))/(2)-(2)/(x^(2)) (x) root(3)(2)x^(2)-8 (xi) (1)/(2x^(2)) (xii) (1)/(sqrt(5))x^(1//2)+1 (xiii) (3)/(5)x^(2)-(7)/(3)x+9 (xiv) x^(4)-x^(3//2)+x-3 (xv) 2x^(3)+3x^(2)+sqrt(x)-1