1
JEE Main 2019 (Online) 12th January Evening Slot
+4
-1
If the function f given by f(x) = x3 – 3(a – 2)x2 + 3ax + 7, for some a$$\in$$R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, $${{f\left( x \right) - 14} \over {{{\left( {x - 1} \right)}^2}}} = 0\left( {x \ne 1} \right)$$ is :
A
$$-$$ 7
B
5
C
7
D
6
2
JEE Main 2019 (Online) 11th January Evening Slot
+4
-1
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
3
JEE Main 2019 (Online) 11th January Morning Slot
+4
-1
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]
4
JEE Main 2019 (Online) 11th January Morning Slot
+4
-1
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
$${1 \over {12}}$$
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