1
JEE Main 2019 (Online) 8th April Evening Slot
+4
-1
Let $$f(x) = \int\limits_0^x {g(t)dt}$$ where g is a non-zero even function. If ƒ(x + 5) = g(x), then $$\int\limits_0^x {f(t)dt}$$ equals-
A
5$$\int\limits_{x + 5}^5 {g(t)dt}$$
B
$$\int\limits_{x + 5}^5 {g(t)dt}$$
C
$$\int\limits_{5}^{x+5} {g(t)dt}$$
D
2$$\int\limits_{5}^{x+5} {g(t)dt}$$
2
JEE Main 2019 (Online) 8th April Morning Slot
+4
-1
If $$f(x) = {{2 - x\cos x} \over {2 + x\cos x}}$$ and g(x) = logex, (x > 0) then the value of integral

$$\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {g\left( {f\left( x \right)} \right)dx{\rm{ }}}$$ is
A
loge3
B
loge2
C
loge1
D
logee
3
JEE Main 2019 (Online) 12th January Evening Slot
+4
-1
The integral $$\int\limits_1^e {\left\{ {{{\left( {{x \over e}} \right)}^{2x}} - {{\left( {{e \over x}} \right)}^x}} \right\}} \,$$ loge x dx is equal to :
A
$$- {1 \over 2} + {1 \over e} - {1 \over {2{e^2}}}$$
B
$${3 \over 2} - e - {1 \over {2{e^2}}}$$
C
$${1 \over 2} - e - {1 \over {{e^2}}}$$
D
$${3 \over 2} - {1 \over e} - {1 \over {2{x^2}}}$$
4
JEE Main 2019 (Online) 12th January Evening Slot
+4
-1
Out of Syllabus
$$\mathop {\lim }\limits_{x \to \infty } \left( {{n \over {{n^2} + {1^2}}} + {n \over {{n^2} + {2^2}}} + {n \over {{n^2} + {3^2}}} + ..... + {1 \over {5n}}} \right)$$ is equal to :
A
tan–1 (2)
B
tan–1 (3)
C
$${\pi \over 4}$$
D
$${\pi \over 2}$$
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