JEE Main 2020 (Online) 6th September Evening Slot | Functions Question 100 | Mathematics

JEE Main 2020 (Online) 6th September Evening Slot

MCQ (Single Correct Answer)

-1

For a suitably chosen real constant a, let a

function, $$f:R - \left\{ { - a} \right\} \to R$$ be defined by

$$f(x) = {{a - x} \over {a + x}}$$. Further suppose that for any real number $$x \ne - a$$ and $$f(x) \ne - a$$,

(fof)(x) = x. Then $$f\left( { - {1 \over 2}} \right)$$ is equal to :

$$ {1 \over 3}$$

–3

$$ - {1 \over 3}$$

JEE Main 2020 (Online) 6th September Morning Slot

MCQ (Single Correct Answer)

-1

If f(x + y) = f(x)f(y) and $$\sum\limits_{x = 1}^\infty {f\left( x \right)} = 2$$ , x, y $$ \in $$ N, where N is the set of all natural number, then the value of $${{f\left( 4 \right)} \over {f\left( 2 \right)}}$$ is :

$${2 \over 3}$$

$${1 \over 9}$$

$${1 \over 3}$$

$${4 \over 9}$$

JEE Main 2020 (Online) 5th September Morning Slot

MCQ (Single Correct Answer)

-1

A survey shows that 73% of the persons working in an office like coffee, whereas 65% like tea. If x denotes the percentage of them, who like both coffee and tea, then x cannot be :

JEE Main 2020 (Online) 5th September Morning Slot

MCQ (Single Correct Answer)

-1

If the minimum and the maximum values of the function $$f:\left[ {{\pi \over 4},{\pi \over 2}} \right] \to R$$, defined by
$$f\left( \theta \right) = \left| {\matrix{ { - {{\sin }^2}\theta } & { - 1 - {{\sin }^2}\theta } & 1 \cr { - {{\cos }^2}\theta } & { - 1 - {{\cos }^2}\theta } & 1 \cr {12} & {10} & { - 2} \cr } } \right|$$ are m and M respectively, then the ordered pair (m,M) is equal to :

$$\left( {0,2\sqrt 2 } \right)$$

(-4, 0)

(-4, 4)

(0, 4)

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