1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Morning Slot

Let f : R $$ \to $$ R be defined by f(x) = $${x \over {1 + {x^2}}},x \in R$$.   Then the range of f is :
A
$$\left[ { - {1 \over 2},{1 \over 2}} \right]$$
B
$$R - \left[ { - {1 \over 2},{1 \over 2}} \right]$$
C
($$-$$ 1, 1) $$-$$ {0}
D
R $$-$$ [$$-$$1, 1]

Explanation

f(0) = 0 & f(x) is odd

Further, if x > 0 then

f(x) = $$f(x) = {1 \over {x + {1 \over x}}} \in \left( {0,{1 \over 2}} \right]$$

Hence,  $$f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$$
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Morning Slot

Let fk(x) = $${1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)$$ for k = 1, 2, 3, ... Then for all x $$ \in $$ R, the value of f4(x) $$-$$ f6(x) is equal to
A
$${1 \over 4}$$
B
$${5 \over {12}}$$
C
$${{ - 1} \over {12}}$$
D
$${1 \over {12}}$$

Explanation

f4(x) $$-$$ f6(x)

= $${1 \over 4}$$ (sin4 x + cos4 x) $$-$$ $${1 \over 6}$$ (sin6 x + cos6 x)

= $${1 \over 4}$$ (1$$-$$ $${1 \over 2}$$ sin2 2x) $$-$$ $${1 \over 6}$$ (1 $$-$$ $${3 \over 4}$$ sin2 2x) = $${1 \over 12}$$
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Evening Slot

Let a function f : (0, $$\infty $$) $$ \to $$ (0, $$\infty $$) be defined by f(x) = $$\left| {1 - {1 \over x}} \right|$$. Then f is :
A
not injective but it is surjective
B
neiter injective nor surjective
C
injective only
D
both injective as well as surjective

Explanation

$$f\left( x \right) = \left| {1 - {1 \over x}} \right| = {{\left| {x - 1} \right|} \over x} = \left\{ {\matrix{ {{{1 - x} \over x}} & {0 < x \le 1} \cr {{{x - 1} \over x}} & {x \ge 1} \cr } } \right.$$


$$ \Rightarrow $$  f(x) is not injective

but range of function is $$\left[ {0,\infty } \right)$$

Remarks :  If co-domain is $$\left[ {0,\infty } \right)$$, then f(x) will be surjective.
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 12th January Morning Slot

Let S = {1, 2, 3, … , 100}. The number of non-empty subsets A of S such that the product of elements in A is even is :
A
250 – 1
B
250 (250 $$-$$ 1)
C
2100 $$-$$ 1
D
250 + 1

Explanation

S = {1,2,3, . . . .100}

= Total non empty subsets-subsets with product of element is odd

= 2100 $$-$$ 1 $$-$$ 1[(250 $$-$$ 1)]

= 2100 $$-$$ 250

= 250 (250 $$-$$ 1)

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