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Practice Online Test Series for JEE Main 2018 & NEET 2018
Let A = [-1, 1] = B then which of the following function from A to B is objective function
If(x) is a polynomial in `x (> 0)` satisfying the equation `f(x)+f(1/x)=f(x).f(1/x)`and `f(2)=-7`,then `f(3) =`
Let `f(x) = (2x(sin x+ tan x))/(2[(x+21pi)/pi]-41), x != npi`, then f is (where [.] represent greater integer function)
The fundamental period of the function `f(x) = 4cos^4((x-pi)/(4pi^2))-2cos((x-pi)/(2pi^2))` is equal to
The domain of `f(x)= log_2log_3log_4 x` is
The domain of `f(x)=log_10 (x-5)/(x^2-10x+24)-root3 (x+5)` is
Domain of `sqrt(x(1-e^x)(x+2)(x-3)^2)`
The range of `f(x)= log_e (3x^2-4x+5)` is
The range of `f(x)=[sin x+[cos x+[tan x+[sec x]]]], x in (0, pi/4)`, where [.] denotes the greatest integer function `<=x,` is
The functions `f:R -> R` is defined by
`f(x)=|(x-1)(x-2)|` is
Total no of ordered pairs (x, y) satisfying `x(sin^2 x+1/x^2)=2sinx.sin^2y`, where `x in (-pi,o)uu(0,pi)` and `y in [0,2pi] ` is / are
Number of solution of `e^(cot^2theta) + sin^2theta-2cos^2 2theta-4sintheta+4=0` in `[0, 10pi]` is.
If `sin2A + sin2B +sin2C+2cos (A-B)+2cos(A+B)=4`, then `triangle` ABC is
Total no. of triplets `(alpha, beta, gamma)` satisfying `cos(alpha-beta)+ cos(beta-gamma)+cos(gamma-alpha)=3` and `cos(alpha+beta)=cos(beta+gamma)= cos(gamma+alpha)=1/sqrtpi, (0<=alpha, beta, gamma <=2pi)`
If `theta_1, theta_2, theta_3, theta_4, theta_5 in (0, pi/2)`, satisfying `tan theta_1 tan theta_2 tan theta_3 tan theta_4 tan theta_5=1` then maximum value of `cos theta_1 cos theta_2 cos theta_3 cos theta_4 cos theta_5` is
A triangle has circum radius R and sides a, b, c with `4R (b^2+c^2)=a(b+c)^2`, then triangle is
`(int_0^(pi/2)x/sinx dx)/(int_0^1(tan^(-1)x^50)/x dx) =`
If `f(x)` is a periodic function with fundamental period 2 satisfies `f(1+x)+f(1-x)=0; AA|x|<=1 int_(-2n)^(2n)[f(x)]dx` equals (Here [x] is greatest integer of x)
If `f(x)=sin x-x`, then `int_(-2pi)^(2pi)|f^(-1)(x)|dx` equals
`int((cos x+xsin x))/(x(cos x-x)) dx` equals
