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1

### AIEEE 2004

If $$x = {e^{y + {e^y} + {e^{y + .....\infty }}}}$$ , $$x > 0,$$ then $${{{dy} \over {dx}}}$$ is
A
$${{1 + x} \over x}$$
B
$${1 \over x}$$
C
$${{1 - x} \over x}$$
D
$${x \over {1 + x}}$$

## Explanation

$$x = {e^{y + {e^{y + .....\infty }}}}\,\, \Rightarrow x = {e^{y + x}}.$$

Taking log.

$$\log \,\,x = y + x$$

$$\Rightarrow {1 \over x} = {{dy} \over {dx}} + 1$$

$$\Rightarrow {{dy} \over {dx}} = {1 \over x} - 1 = {{1 - x} \over x}$$
2

### AIEEE 2003

Let $$f\left( x \right)$$ be a polynomial function of second degree. If $$f\left( 1 \right) = f\left( { - 1} \right)$$ and $$a,b,c$$ are in $$A.P,$$ then $$f'\left( a \right),f'\left( b \right),f'\left( c \right)$$ are in
A
Arithmetic -Geometric Progression
B
$$A.P$$
C
$$G.P$$
D
$$H.P$$

## Explanation

$$f\left( x \right) = a{x^2} + bx + c$$

$$f\left( 1 \right) = f\left( { - 1} \right)$$

$$\Rightarrow a + b + c = a - b + c$$

or $$b = 0$$

$$\therefore$$ $$f\left( x \right) = a{x^2} + c$$

or $$f'\left( x \right) = 2ax$$

Now $$f'\left( a \right);f'\left( b \right);$$

and $$f'\left( c \right)$$ are $$2a\left( a \right);2a\left( b \right);2a\left( c \right)$$

i.e.$$\,2{a^2},\,2ab,\,2ac.$$

$$\Rightarrow$$ If $$a,b,c$$ are in $$A.P.$$ then

$$f'\left( a \right);f'\left( b \right)$$ and

$$f'\left( c \right)$$ are also in $$A.P.$$
3

### AIEEE 2003

If $$f\left( x \right) = {x^n},$$ then the value of

$$f\left( 1 \right) - {{f'\left( 1 \right)} \over {1!}} + {{f''\left( 1 \right)} \over {2!}} - {{f'''\left( 1 \right)} \over {3!}} + ..........{{{{\left( { - 1} \right)}^n}{f^n}\left( 1 \right)} \over {n!}}$$ is

A
$$1$$
B
$${{2^n}}$$
C
$${{2^n} - 1}$$
D
$$0$$

## Explanation

$$f\left( x \right) = {x^n} \Rightarrow f\left( 1 \right) = 1$$

$$f'\left( x \right) = n{x^{n - 1}} \Rightarrow f'\left( 1 \right) = n$$

$$f''\left( x \right) = n\left( {n - 1} \right){x^{n - 2}}$$

$$\Rightarrow f''\left( 1 \right) = n\left( {n - 1} \right)$$

$$\therefore$$ $${f^n}\left( x \right) = n!$$

$$\Rightarrow {f^n}\left( 1 \right) = n!$$

$$= 1 - {n \over {1!}} + {{n\left( {n - 1} \right)} \over {2!}}{{n\left( {n - 1} \right)\left( {n - 2} \right)} \over {3!}}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + .... + {\left( { - 1} \right)^n}{{n!} \over {n!}}$$

$$= {}^n\,{C_0} - {}^n\,{C_1} + {}^n\,{C_2} - {}^n\,{C_3}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ...... + {\left( { - 1} \right)^n}\,{}^n{C_n} = 0$$
4

### AIEEE 2003

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

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

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

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

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

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

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

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

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