JEE Mains Previous Years Questions with Solutions Android App

Download our App

JEE Mains Previous Years Questions with Solutions

4.5 
Star 1 Star 2 Star 3 Star 4
Star 5
  (100k+ )
1

AIEEE 2003

MCQ (Single Correct Answer)
If $$f\left( y \right) = {e^y},$$ $$g\left( y \right) = y;y > 0$$ and
$$F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)dy,} $$ then
A
$$F\left( t \right) = t{e^{ - t}}$$
B
$$F\left( t \right) = 1t - t{e^{ - 1}}\left( {1 + t} \right)$$
C
$$F\left( t \right) = {e^t} - \left( {1 + t} \right)$$
D
$$F\left( t \right) = t{e^t}$$.

Explanation

$$F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)} dy$$

$$ = \int\limits_0^t {{e^{t - y}}} ydy = {e^t}\int\limits_0^t {{e^{ - y}}} \,ydy$$

$$ = {e^t}\left[ { - y{e^{ - y}} - {e^{ - y}}} \right]_0^t$$

$$ = - {e^t}\left[ {y{e^{ - y}} + {e^{ - y}}} \right]_0^t$$

$$ = - {e^t}\left[ {t\,{e^{ - t}} + {e^{ - t}} - 0 - 1} \right]$$

$$ = - {e^t}\left[ {{{t + 1 - {e^t}} \over {{e^t}}}} \right]$$

$$ = {e^t} - \left( {1 + t} \right)$$
2

AIEEE 2002

MCQ (Single Correct Answer)
If $$y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n},$$ then $$\left( {1 + {x^2}} \right){{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}}$$ is
A
$${n^2}y$$
B
$$-{n^2}y$$
C
$$-y$$
D
$$2{x^2}y$$

Explanation

$$y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n}$$

$${{dy} \over {dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 + {1 \over 2}{{\left( {1 + {x^2}} \right)}^{ - 1/2}}.2x} \right);$$

$${{dy} \over {dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\left( {\sqrt {1 + {x^2}} + x} \right)} \over {\sqrt {1 + {x^2}} }}$$

$$ = {{n{{\left( {\sqrt {1 + {x^2}} + x} \right)}^n}} \over {\sqrt {1 + {x^2}} }}$$

or $$\sqrt {1 + {x^2}} {{dy} \over {dx}} = ny$$

or $$\sqrt {1 + {x^2}} {y_1} = ny$$

$$\left( {{y_1} = {{dy} \over {dx}}} \right)$$

Squaring, $$\left( {1 + {x^2}} \right){y_1}^2 = {n^2}{y^2}$$

Differentiating, $$\left( {1 + {x^2}} \right)2{y_1}{y_2} + {y_1}^2.2x$$

$$ = {n^2}.2y{y_1}$$

or $$\left( {1 + {x^2}} \right){y_2} + x{y_1} = {n^2}y$$

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12