JEE Advanced Mix Test 33

Statement-1 : If a 4^th degree polynomial function has four distinct real roots, then its graph have exactly two inflection points.

Statement-2 : If the equation f"(x) = 0 has distinct real roots, then the equation f(x) = 0 has atleast four real roots.

Statement-1 : Let 0 < a < b. Then Rolle's theorem is applicable to the function f(x) =  in [a,b] and Rolle's constant is the GM of a and b.

Statement-2 : All the three conditions of Rolle's theorem are satisfied by the above function f(x).

Statement-1 All points of intersection of y = f (x) & y = f ^{– 1} (x) lies on y = x only.

Statement-2 If point P (α, β) lies on y = f (x) then Q (β, α) lies on y = f ^{– 1} (x).

The general procedure for solving equation containing modulus function 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, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.

Since LHS > ⇒ f(x) – g(x) > 0. Now squaring both sides, we get f² + g² + 2|f.g| = f² + g² – 2fg ⇒ |fg| = – fg. The equation can hold if f.g < 0 and f > g. This can be simplified to f > 0, g < 0.

Answer the following questions on the basis of this method

The complete solution of the equation |x³ – x| + |2 – x| = x³ – 2 is

The general procedure for solving equation containing modulus function 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, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.

Since LHS > ⇒ f(x) – g(x) > 0. Now squaring both sides, we get f² + g² + 2|f.g| = f² + g² – 2fg ⇒ |fg| = – fg. The equation can hold if f.g < 0 and f > g. This can be simplified to f > 0, g < 0.

Answer the following questions on the basis of this method

The complete solution set of the equation |x² – x| + |x + 3| = |x² – 2x – 3| is

The general procedure for solving equation containing modulus function 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, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation.

Since LHS > ⇒ f(x) – g(x) > 0. Now squaring both sides, we get f² + g² + 2|f.g| = f² + g² – 2fg ⇒ |fg| = – fg. The equation can hold if f.g < 0 and f > g. This can be simplified to f > 0, g < 0. Answer the following questions on the basis of this method

The solution set belonging to (0, 2π) of the equation |sin x – cos x| = |sin x| + |cosx| is

Let f be a function defined so that every element of the codomain has at most two pre-images and there is at least one element in the co-domain which has exactly two pre-images we shall call this function as “two-one” function. A two-one function is definitely a many one function but vice-versa is not true. For example, y = |e^x – 1| is a “two-one” function. y = x³ – x is a many one function but not a “two-one” function. In the light of above definition answer the following questions:

In the following functions which one is a “two-one” function:-

Let f be a function defined so that every element of the codomain has at most two pre-images and there is at least one element in the co-domain which has exactly two pre-images we shall call this function as “two-one” function. A two-one function is definitely a many one function but vice-versa is not true. For example, y = |e^x – 1| is a “two-one” function. y = x³ – x is a many one function but not a “two-one” function. In the light of above definition answer the following questions:

Let f(x) = {x} be the fractional part function. For what domain is the function “two-one”?

Let f be a function defined so that every element of the codomain has at most two pre-images and there is at least one element in the co-domain which has exactly two pre-images we shall call this function as “two-one” function. A two-one function is definitely a many one function but vice-versa is not true. For example, y = |e^x – 1| is a “two-one” function. y = x³ – x is a many one function but not a “two-one” function. In the light of above definition answer the following questions:

The values of ‘a’ for which the function f(x) = x⁴ – ax² is a “two-one” function are