`2=x+1/{1+1/{3+1/4}} Then x value=(a)18/17 (b)21/17 (c)13/17 (d)12/17`

If x-1/x=(15)/4, then x+1/x\ =? (a) 4 (b) (17)/4 (c) (13)/4\ (d) 1/4

The additive inverse of 17 is (a) -17 (b) 17 (c) 1/(17) (d) -1/(17)

If the mode of the data: 16 ,\ 15 ,\ 17 ,\ 16 ,\ 15 ,\ x ,\ 19 ,\ 17 ,\ 14 is 15 , then x= (a) 15 (b) 16 (c) 17 (d) 19

P(x) be a polynomial of degree 3 satisfying P(-1) =10 , P(1) =-6 and p(x) has maxima at x = -1 and p(x) has minima at x=1 then The value of P(2) is (a) -15 (b) -16 (c) -17 (d) -22 (c) -17 (d) -22

If the median of the data: 6, 7, x-2,\ x , 17, 20, written in ascending order, is 16. Then x= (a) 15 (b) 16 (c) 17 (d) 18

The value of 1/{3+1/{2-1/{7/9}}}+17/22 is :

If vec(DA)=veca,vec(AB)=vecb and vec(CB)=kveca where kgt0 and X,Y are the midpoint of DB and AC respectively such that |veca|=17 and |vec(XY)|=4, then k is equal to (A) 9/17 (B) 8/17 (C) 25/17 (D) 4/17

A root of the equation 17x^2 + 17x tan [2 tan^-1(1/5) - pi/4] - 10 = 0 is (i) 10/17 (ii) -1 (iii) -7/17 (iv) 1

If x : y=1:1,\ t h e n\ (3x+4y)/(5x+6y)= (a) 7/(11) (b) (17)/(11) (c) (17)/(23) (d) 4/5

If x : y=1:1,\ t h e n\ (3x+4y)/(5x+6y)= (a) 7/(11) (b) (17)/(11) (c) (17)/(23) (d) 4/5