Question.1
For any t ∈ R and f be a continuous function,
let I1 =
and I2 =
. Then
is
(A) 0
(B) 1
(C) 2
(D) 3
Solution
![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image004.gif)
I1 =![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image005.gif)
= 2![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image006.gif)
2I1 = 2![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image007.gif)
&⇒
2I1 = 2I2
&⇒
= 1
Question.2
If
= 4 and
= 7, then the value of
is
(A) 2
(B) –3
(C) –5
(D) none of these
Solution
Given ![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image012.gif)
![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image013.gif)
&⇒
,
![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image015.gif)
Now
=
–1 –4 = –5
Question.3
Let f (x) be a
function satisfying f¢ (x) = f (x) with f
(0) = 1 and g be the function satisfying f (x) + g (x) = x2. The
value of the integral
is
(A)
(e –7)
(B)
(e –2)
(C)
(e –3)
(D) none of these
Solution
f¢ (x) = f (x)
= 1, on integrating
= ![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image022.gif)
ln |f (x)| + c = x,
given f (0) = 1
&⇒ c = 0
\ f (x) = ex, g (x) = x2
–ex
Now I =
= ![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image024.gif)
= ![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image025.gif)
Question.4
The value of ![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image026.gif)
is
(A) ln 2
(B) 1 + ln 2
(C) 1 – ln 2
(D) 0
Solution
![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image028.gif)
=
![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image028.gif)
![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image030.gif)
=
=
–
= 1 – ln 2.
Question.5
The value of the integral
is
(A) 1
(B)
(C)
(D) none of these
Solution
Using the
property
f (x) d x =
f (a + b – x) dx, the given
integral
I =
=
=
![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image040.gif)
Here 2 I = ![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image041.gif)
&⇒ I = ![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image042.gif)
=
.
Question.6
If f is an
odd function, then I =
dx
(A) can’t be
evaluated
(B) I = 0
(C) I =
(D) none of these
Solution
Since the
denominator is even and the numerator f(sin (– x)) = f (– sin x ) = – f (sin
x) is odd, (as f is odd function), the integral is equal to zero.
Question.7
If f (a + b –x) = f (x) then
f (x) dx is equal to
(A) ![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image048.gif)
d
x
(B) ![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image050.gif)
d
x
(C) 0
(D) none of these
Solution
I =
f
(x) dx =
(a + b – x) f (a + b – x) dx
= (a + b)
(a + b – x)dx –
x f(a + b – x) dx
= (a + b)
f (x) dx –
x f (x) d x
Hence I = ![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image054.gif)
f(x)
dx
Question.8
If f (x) =
dt (x > 0) then
is
(A)
(B) ![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image058.gif)
(C)
(D) none of these.
Solution
= cos ![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image061.gif)
=
=
![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image063.gif)
Question.9
The value of
is
(A) p/4
(B) 0
(C) p/2
(D) none of these
Solution
![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image065.gif)
Let f(x) = ![](http://www.quizsolver.com/radix/dth/notif/DEFI_OBJ_SOL_1_files/image065.gif)
Here f ¢(x) = 0
&⇒
f(x) = c (a constant)
But f(p/4) =
=
=
p/4.
Question.10
For an
integer n, the integral
cos3
(2n +1) x dx has the value
(A) p
(B) 1
(C) 0
(D) none of these
Solution
I =
cos3 (2n +1) x dx =
cos3 (2 n + 1) (p
– x) dx
=
cos3 (2n p + p –
(2 n + 1) x ) dx
= –
cos3 (2n + 1) x dx =
– I
Hence 2I = 0
&⇒ I
= 0