JEE Main 2019 (Online) 9th January Evening Slot | Functions Question 137 | Mathematics

JEE Main 2019 (Online) 9th January Evening Slot

MCQ (Single Correct Answer)

-1

Let A = {x $$ \in $$ R : x is not a positive integer}.

Define a function $$f$$ : A $$ \to $$ R as $$f(x)$$ = $${{2x} \over {x - 1}}$$,

then $$f$$ is :

not injective

neither injective nor surjective

surjective but not injective

injective but not surjective

JEE Main 2019 (Online) 9th January Morning Slot

MCQ (Single Correct Answer)

-1

For $$x \in R - \left\{ {0,1} \right\}$$, Let f₁(x) = $$1\over x$$, f₂ (x) = 1 – x

and f₃ (x) = $$1 \over {1 - x}$$ be three given

functions. If a function, J(x) satisfies

(f₂ o J o f₁) (x) = f₃ (x) then J(x) is equal to :

f₁ (x)

$$1 \over x$$ f₃ (x)

f₂ (x)

f₃ (x)

JEE Main 2018 (Online) 16th April Morning Slot

MCQ (Single Correct Answer)

-1

Let N denote the set of all natural numbers. Define two binary relations on N as _R = {(x, y) $$ \in $$ N $$ \times $$ N : 2x + y = 10} and R₂ = {(x, y) $$ \in $$ N $$ \times $$ N : x + 2y = 10}. Then :

Range of R₁ is {2, 4, 8).

Range of R₂ is {1, 2, 3, 4}.

Both R₁ and R₂ are symmetric relations.

Both R₁ and R₂ are transitive relations.

JEE Main 2018 (Online) 15th April Evening Slot

MCQ (Single Correct Answer)

-1

Let f : A $$ \to $$ B be a function defined as f(x) = $${{x - 1} \over {x - 2}},$$ Where A = R $$-$$ {2} and B = R $$-$$ {1}. Then f is :

invertible and $${f^{ - 1}}(y) = $$ $${{3y - 1} \over {y - 1}}$$

invertible and $${f^{ - 1}}\left( y \right) = {{2y - 1} \over {y - 1}}$$

invertible and $${f^{ - 1}}\left( y \right) = {{2y + 1} \over {y - 1}}$$

not invertible