Mathematics_Today_-_October_2016

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SECTION-I
Single Correct Answer Type


  1. A continuous function f : [0, 1] → R satisfies the


equation ∫f x dx() =+^1 ∫f ( ) ,x dx
3

22
0

1

0

1
then f^1
4


⎝⎜


⎠⎟=
(a) 1 (b) 1/2 (c) 1/4 (d) –1/4


  1. A continuous function f : [0, 1] → R satisfies the


condition ∫fx x fx dx()(− ()) =^1 ,
0 12

1
then f^1
4


⎝⎜


⎠⎟=
(a) 1/2 (b) 1/4 (c) 1/6 (d) 1/8


  1. Compute () 2332113 /
    0


1
∫ xxx dx−−+

(a) 0 (b) –1 (c) 1/8 (d) –1/8


  1. Consider a one-to-one differentiable solution y to


the equation dy
dx

dx
dy

2
2

2
+= 2 0 such that y(0) = 2 and
y(–1) = –1 then y(1) =
(a) –20 (b) 20 (c) –5 (d) 5


  1. Let f : [0, 1] → R be a continuous function with the
    property that x f(y) + y f(x) ≤ 1 for all x, y ∈ [0, 1]
    then


(a) ∫fxdx() =π
0 4

1
(b) ∫fxdx() ≤π
0 4

1

(c) ∫fxdx() ≥π 4
0

1
(d) ∫fxdx() ≥π 2
0

1


  1. For every θ ∈ (0, π], we have


(a)^1222
0

∫ +>+cos tdt θθsin


θ

(b)^1222
0

∫ +=+cos tdt θθsin


θ

(c)^1222
0

∫ +<+cos tdt θθsin


θ

(d)^1

1
2

222
0

∫ +=+cos tdt θθsin


θ


  1. y(x) satisfies the differential equation
    dy
    dx


=+yyyelog x. If y(0) = 1, then y(1) =
(a) ee (b) e–e (c) e1/e (d) e


  1. The function f dx
    x


()
cos

,(,)

/
λ
λ

λ

π
=

∫ ∈
1

01
0 2

2
is

(a) increasing (b) decreasing
(c) increasing on (0, 1/2) and decreasing on (1/2, 1)
(d) increasing on (1/2, 1) and decreasing on (0, 1/2)


  1. Let f satisfy the equation x = f(x)·ef(x) then
    fxdx


e
∫ () =
0
(a) 1 (b) e – 1 (c) e + 1 (d) 0


  1. Let a, b ∈ R, a < b and let f be a differentiable function


on the interval [a, b] then lim ( ) sin
n a

b
f x nxdx
→∞∫

=

(a) 0 (b) 1 (c) b (d) b – a


  1. The shortest possible length of an interval [a, b] for
    which^4
    1 +^2
    ∫ =
    x


dx
a

b
π is

(a) 2 (b) 22 (c) 22 2+ (d) 22 2−
SECTION-II
Comprehension Type
Paragraph for Question No. 12-14
Let θk(x) be 0 for x < k and 1 for x ≥ k. The dirac delta
function is defined to be δθkkx d
dx

()= ().x Suppose

 


 










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