Find the product of `7/2s^2t\ a n d\ s+tdot` Verify the result for `s=1/2a n d\ t=5.`

Find the product of (y+2/7y^2)\ a n d\ (7y-y^2) and verify the result for y=3.

Let f(x)=x^3+3x+2a n dg(x) be the inverse of it. Then the area bounded by g(x) , the x-axis, and the ordinate at x=-2a n dx=6 is (a) 9/4s qdotu n i t s (b) 4/3s qdotu n i t s (c) 5/4s qdotu n i t s (d) 7/3s qdotu n i t s

Charge through a cross-section of a conductor is given by Q = 5t^(2) - 2t coulomb . Find the average current through the conductor in the interval t_(1) = 2s to t_(2) = 4s (1) 14s (2) 28 A (3) 56A (4) 7A

If S be the sum P the product and R be the sum of the reciprocals of n terms of a GP then p^2 is equal to S//R b. R//S c. (R//S)^n d. (S//R)^n

The area of the region whose boundaries are defined by the curves y= 2cosx , y= 3tanx and the y-axis is (a) 1+31n(2/(sqrt(3)))s qdotu n i t s (b) 1+3/2 1n3-31n2s qdotu n i t s (c) 1+3/2 1n3-1n2s qdotu n i t s (d) 1n3-1n2s qdotu n i t s

In a harmonic progression t_(1), t_(2), t_(3),……………., it is given that t_(5)=20 and t_(6)=50 . If S_(n) denotes the sum of first n terms of this, then the value of n for which S_(n) is maximum is

If sum of n termsof a sequende is S_n then its nth term t_n=S_n-S_(n-1) . This relation is vale for all ngt-1 provided S_0= 0. But if S_!=0 , then the relation is valid ony for nge2 and in hat cast t_1 can be obtained by the relation t_1=S_1. Also if nth term of a sequence t_1=S_n-S_(n-1) then sum of n term of the sequence can be obtained by putting n=1,2,3,.n and adding them. Thus sum_(n=1)^n t_n=S_n-S_0. if S_0=0, then sum_(n=1)^n t_n=S_n. On the basis of above information answer thefollowing questions:If nth term of a sequence is n/(1+n^2+n^4) then the sum of its first n terms is (A) (n^2+n)/(1+n+n^2) (B) (n^2-n)/(1+n+n^2) (C) (n^2+n)/(1-n+n^2) (D) (n^2+n)/(2(1+n+n^2)

If the extremities of the base of an isosceles triangle are the points (2a ,0) and (0, a), and the equation of one of the side is x=2a , then the area of the triangle is (a) 5a^2s qdotu n i t s (b) (5a^2)/2s qdotu n i t s (25 a^2)/2s qdotu n i t s (d) none of these

If the extremities of the base of an isosceles triangle are the points (2a ,0) and (0, a), and the equation of one of the side is x=2a , then the area of the triangle is 5a^2s qdotu n i t s (b) (5a^2)/2s qdotu n i t s (25 a^2)/2s qdotu n i t s (d) none of these

If sintheta=(a^2-b^2)/(a^2+b^2), find the value of t a ntheta,\ s e ctheta\ a n d\ cos e ctheta