Recently Active Calculus Questions

let g(x) be a continous function for all x ,and f(x) = f(a) +(x-a)g(x) for all x ε R , then a. f(x) is necessarily differentiable at x =a b. f(x) is not necessarily differentiable at x=a c. f(x) is not necessarily continous ...

f(x)= Sin-1(Cos-1[x]) ...

f(x)= (2x4 - 14x2 - 8x +49)/(x4-7x2-4x+23) ...

can someone plz tell me how to do these types: 1. is the function f:R→R 3x(x-3) invertible?? if not change its domain so that it becomes so. 2.let g(x)=x2-4x-5 a) g is one-one on R b)g is one-one on (-∞,2] c)g is not one- ...

The eqn-- ex-ax-b=0 has A. 1 real root if a<0 B. 1 real root if b>0 & a<=0 C. 2 real roots if a>0 & a.ln a>=(a-b) D. No real root if a ln a<a-b Multiple Ans correct ...

let Pn = aPn-1 -1 for all n= 2,3,4,.. and let P1= ax-1 where a ε R+ then evaluate the limit lim ( x → 0 ) Pn/x ...

if y=sin-1(cos x) , then dy/dx=? a)sin2x b) sinx c)-1 d)none ...

find the fifth derivative of y= 1/(1+x2) please give a good method ...not the tedious one !! ...

∫(x3m+x2m+xm)(2x2m+3xm+6)1/mdx note: do without using integration by parts ...

hi! i am new to this topic , so pls help ∫(x2tan-1x3)dx/(1 + x6) note: do without using integration by parts ...

The funtion f(x) = [x]2 - [x2] (where [] is GIF) is discontinuous at a.All integers b.All integers exvept 0 c.All integers except 1 d.All integers except 0 and 1 ...

solve question of bitsat i m not able to provide option *Image* ...

Lim ( n → infinity) { (1/2) tan x/2  + (1/22) tan x/22 + (1/23) tan x/23 .............+(1/2n)tan x/2n) } ...

THE VALUE OF LIM ( N→ INFINTIY ) (N ! / NN)(2N4+1)/(5N5+1) IS A. 1 B. 0 C. e-2/5 d. e2/5 ...

let f(x) and g(x) are two differential functions and the limit of the function lim ( n → infinity ) { (x2nf(x) + x 100g(x))/(x2n +1)} exists at x= 1 , then the equation f(x)=g(x) has a. at least one real root in (0,2) b. no ...

limn→∞(nαsin2n!/n+1), where 0<α<1 ...

f(x) = lim (n→ infinity) n( x1/n -1) , x>0 then f(xy) is a. f(x)f(y) b. f(x)+f(y) c. y f(x) + x f(y) d. N.O.T ...

What is the value of limn→∞Πn=0n(1+1/22n)?? ...

limx→∞( (x+a)(x+b)(x+c) - x) = ?? ...

Let f:(4,6)--> (6,infinity) be a funtion defined by f(x) = x +[x/2], where [] is GIF.Then f-1(x) = a. x-[x/2] b.-x-2 c.x-2 d. NOT How to find the inverse of such funtions involving GIF, fractional part etc.. And soln and e ...

If funtion :R-->R be such that f(x)= x - [x], where [] is GIF..Then f-1(x)= a. 1/x-[x] b.[x]-x c.not defined d.NOT ...

The value of 2*a and 2*b such that (where a and b satisfies) Ltx-->0 x(1+acosx)-b sinx =1 ------------------------- x3 a. 5,3 b. 5,-3 c.-5,-3 d.NOT ...

The value of limx-->0(cosx + a sinbx)1/x Ans was e^ab The method was using the expansion...but pls explain how the higher powers gets cancelled out..(because denominator of cos and sin goes as 2! , 3! respectively) ...

The value of Ltn->infinity(1/1.3 +1/3.5+....n terms) a. 1/4 b.1/2 c. 1 d. NOT ...

∫(tan x)1/3 dx ...

If\ f(x)=\sqrt{1-sin(2x)},f'(x) equals\\\ \\ a)-(cos x + sin x) for\ x\in(0,\pi/4)\\b)(cos x + sin x) for\ x\in(\pi/4,\pi/2)\\c)-(cos x + sin x) for\ x\in(\pi/4,\pi/2)\\d)(cos x + sin x) for\ x\in(0,\pi/4)\\ ...

f(x) is a discontinous function only at x=0 such that ( f(x) )2=1 for all values of x ε R , then the numbers of such functions is : A. 3 B. 2 C. 6 D. 4 Actually i am getting it as 4 but the answer given is 6 .... now is the ...

\int \frac{cos dx}{sinx+cosx} ...

lim x→infinity ( (x+a) (x+b) (x+c) -x) = a. abc b.(a+b+c)/3 c.abc d. abc ...

pls plot this graph-- x + Sin x ...