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Numerical Valued Answer Type Jai Baba Ki3
(
bxc if x ≤ 2
19. Let f : R → R be a function defined by f(x) = , where bxc is the greatest integer
0 if x > 2
2
Z 2 x f(x )
less than or equal to x. If I = dx, then the value of (4I − 1) is [JEE 2015]
2 + f(x + 1)
−1
20. A cylindrical container is to be made from certain solid material with the following constraints: It has
3
a fixed inner volume of V mm , has a 2 mm thick solid wall and is open at the top. The bottom of
the container is a solid circular disc of thickness 2 mm and is of radius equal to the outer radius of the
container. If the volume of the material used to make the container is minimum when the inner radius
V
of the container is 10 mm, then the value of is [JEE 2015]
250π
x +π/6 1
Z 2
2
21. Let F(x) = 2 cos t dt for all x ∈ R and f : 0, → [0, ∞) be a continuous function. For
x 2
1
0
a ∈ 0, , if F (a) + 2 is the area of the region bounded by x = 0, y = 0, y = f(x) and x = a, then
2
f(0) is [JEE 2015]
Z 1 12 + 9x 2
−1
22. If α = e 9x+3 tan −1 x , where tan x takes only principal values, then the value of
1 + x 2
0
3π
log |1 + α| − is [JEE 2015]
e
4
1
23. Let f : R → R be a continuous odd function, which vanishes exactly at one point and f(l) = .
2
x x
Z Z
Suppose that F(x) = f(t) dt for all x ∈ [−1, 2] and G(x) = t|f(f(t))| dt for all x ∈ [−1, 2].
−1 −1
F(x) 1 1
If lim = , then the value of f is [JEE 2015]
x→1 G(x) 14 2
3
24. Suppose that ~p, ~q and ~r are three non-coplanar vectors in R . Let the components of a vector ~s along
~ p, ~q and ~r be 4, 3 and 5, respectively. If the components of this vector ~s along (−~p+~q +~r), (~p−~q +~r)
and (−~p − ~q + ~r) are x, y and z, respectively, then the value of 2x + y + z is [JEE 2015]
2
25. Let f : R → R and g : R → R be respectively given by f(x) = |x| + 1 and g(x) = x + 1. Define
h : R → R by
(
max{f(x), g(x)} if x ≤ 0
h(x) = .
min{f(x), g(x)} if x > 0
The number of points at which h(x) is not differentiable is [JEE 2014]
26. The value of
1 d 2
Z
2 5
4x 3 (1 − x ) dx
0 dx 2
is [JEE 2014]
5 2
2 2
27. The slope of the tangent to the curve (y − x ) = x(1 + x ) at the point (1, 3) is [JEE 2014]