If are complex numbers such that are vertices of a triangle ABC,
and then
If then the triangle is

The general procedure for solving inequation containing modulus functions is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach.
For example:
i) The inequation has no solution.
ii) Solving the inequaiton is equivalent to solving the equations
iii) Solving the inequation is equivalent to solving the equations
iv) Solving the inequation is equivalent to solving the equations
v) The inequation is true for all domain of
vi) The inequation is true for all domain of
Set of all real values of x for which is defined, is

Ten different letters of English alphabet are given. Words with five letters are formed from these given letters. Then, the number of words which have at least one letter repeated is

If two real valued functions f(x) & g(x) are given, for fog(x), g(x) is substituted in place of x in f(x) taking care of the domain of fog(x) which is subset of Domain of g(x), only those values of x such that g(x) lies in the Domain of f. Now, If then g of (x) is

where x, y, t R If t (5, k], then greatest value of k for which x is one-one function of t is

STATEMENT-I : If [4x] = [x], then the largest set of values of x is. and
STATEMENT-II : If [nx] = [x], then the largest set of values of x is

has

ClCH₂COOH is heated with fuming HNO₃ in the presence of AgNO₃ in carius tube After filtration and washing a white precipitate is obtained The precipitate is of

Sodium extract gives blood red colour when treated with FeCl₃ Formation of blood red colour confirms the presence of

STATEMENT-I : The graph of consists of only two line segments. Where [.] represent greatest integer function and
STATEMENT-II : The graph of is

Statement 1: Let be a continuous and onto function, then it cannot be one-one. And
Statement 2: If for a continuous and differentiable function f, then f(x) is one-one

Number of solutions of the equation |3x 2| = 3[x] + 2{x}, where [.] and {.} denote the integral and fractional parts of x respectively, is

A mixture of benzene and toluene can be separated by

The solution set of the inequality log_1/5 (2x + 5) + log₅ (16 x 2) 1 is

Let f be an injective map with domain and range such that exactly one of the following statements is correct and the remaining are false The value ofis

For such questions, we should know concept of injective function.

Hyper conjugation effect is not possible in which of the following species