At what point of curve `y=2/3x^3+1/2x^2,` the tangent makes equal angle with the axis? `(a)(1/2,5/(24))a n d(-1,-1/6)` `(b)(1/2,4/9)a n d(-1,0)` `(c)(1/3,1/7)a n d(-3,1/2)` `(d)(1/3,4/(47))a n d(-1,-1/3)`

Find the equation of the line passing through the points (1,2,3)a n d(-1,0,4)dot

If the normal to the curve y=f(x) at the point (3,4) makes an angle (3pi)/4 with the positive x-axis, then f'(3)= (a) -1 (b) -3/4 (c) 4/3 (d) 1

The angle between the tangents to the curves y=x^2a n dx=y^2a t(1,1) is cos^(-1)(4/5) (b) sin^(-1)(3/5) tan^(-1)(3/4) (d) tan^(-1)(1/3)

The lines ax + by + c = 0, where 3a + 2b + 4c= 0, are concurrent at the point (a) (1/2 ,3/4) (b) (1,3) (c) (3,1) (d) (3/4 ,1/2)

The abscissas of point Pa n dQ on the curve y=e^x+e^(-x) such that tangents at Pa n dQ make 60^0 with the x-axis are. (a)1n((sqrt(3)+sqrt(7))/7)a n d1n((sqrt(3)+sqrt(5))/2) (b)1n((sqrt(3)+sqrt(7))/2) (c)1n((sqrt(7)-sqrt(3))/2) (d)+-1n((sqrt(3)+sqrt(7))/2)

The mean deviation of the series a , a+d , a+2d ,.....,a+2nd from its mean is (a) ((n+1)d)/(2n+1) (b) (n d)/(2n+1) (c) (n(n+1)d)/(2n+1) (d) ((2n+1)d)/(n(n+1))

If y=3^(x-1)+3^(-x-1) , then the least value of y is (a) 2 (b) 6 (c) 2//3 (d) 3//2

If 3tan^(-1)(1/(2+sqrt(3)))-tan^(-1)(1/x)=tan^(-1)(1/3), then x is equal to 1 (b) 2 (c) 3 (d) sqrt(2)

The derivative of y=(1-x)(2-x)....(n-x) at x=1 is (a) 0 (b) (-1)(n-1)! (c) n !-1 (d) (-1)^(n-1)(n-1)!

The domain of the function f(x)=(sin^(-1)(3-x))/(I n(|x|-2)i s (a) [2,4] (b) (2,3)uu(3,4] (c) (0,1)uu(1,oo) (d) (-oo,-3)uu(2,oo)