1

### JEE Main 2017 (Online) 9th April Morning Slot

Let f be a polynomial function such that

f (3x) = f ' (x) . f '' (x), for all x $\in$ R. Then :
A
f (2) + f ' (2) = 28
B
f '' (2) $-$ f ' (2) = 0
C
f '' (2) $-$ f (2) = 4
D
f (2) $-$ f ' (2) + f '' (2) = 10
2

### JEE Main 2018 (Offline)

For each t $\in R$, let [t] be the greatest integer less than or equal to t.

Then $\mathop {\lim }\limits_{x \to {0^ + }} x\left( {\left[ {{1 \over x}} \right] + \left[ {{2 \over x}} \right] + ..... + \left[ {{{15} \over x}} \right]} \right)$
A
does not exist in R
B
is equal to 0
C
is equal to 15
D
is equal to 120

## Explanation

Given,

$\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left( {\left[ {{1 \over x}} \right] + \left[ {{2 \over x}} \right]} \right. +$ $\left. {\,.\,.\,.\,.\,.\, + \left[ {{{15} \over x}} \right]} \right)$

as we know that

${1 \over x} = \left[ {{1 \over x}} \right] + \left\{ {{1 \over x}} \right\}$

$\Rightarrow \,\,\,\,\left[ {{1 \over x}} \right] = {1 \over x} - \left\{ {{1 \over x}} \right\}$

$= \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\,x\left[ {{1 \over x} - \left\{ {{1 \over x}} \right\} + {2 \over 2} - \left\{ {{2 \over x}} \right\} + ........{{15} \over x} - \left\{ {{{15} \over x}} \right\}} \right]$

$= \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\left[ {x.{1 \over x} + x.{2 \over x} + .....x.{{15} \over x}} \right]$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\left[ {x.\left\{ {{1 \over 2}} \right\} + .. + x.\left\{ {{{15} \over x}} \right\}} \right]$

We know $\left\{ {{1 \over x}} \right\}$ is fractional part of ${1 \over x}.$

So, the range of $\,\left\{ {{1 \over x}} \right\}$ is $0 \le \left\{ {{1 \over x}} \right\} < 1$

So, $\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left\{ {{1 \over x}} \right\} = 0.$ (finite no) $=0$

Similarly $\mathop {\lim }\limits_{x \to {0^ + }} x.\left\{ {{2 \over x}} \right\} = 0$

$= \,\,\,\,\left( {1 + 2 + ... + 15} \right) - \left( {0 + 0...} \right)$

$= \,\,\,\,{{15 \times 16} \over 2}$

$= \,\,\,\,120$
3

### JEE Main 2018 (Online) 15th April Morning Slot

If $f\left( x \right) = \left| {\matrix{ {\cos x} & x & 1 \cr {2\sin x} & {{x^2}} & {2x} \cr {\tan x} & x & 1 \cr } } \right|,$ then $\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}$
A
does not exist.
B
exists and is equal to 2.
C
existsand is equal to 0.
D
exists and is equal to $-$ 2.

## Explanation

Given,

$f\left( x \right) = \left| {\matrix{ {\cos x} & x & 1 \cr {2\sin x} & {{x^2}} & {2x} \cr {\tan x} & x & 1 \cr } } \right|$

= cosx(x2 - 2x2) - x(2 sinx - 2x tanx) + (2x sinx - x2 tanx)
= x2 (tanx - cosx)
$\therefore$ ${f^{'}}(x)$ = 2x (tanx - cosx) + x2(sec2x + sinx)

$\therefore$ $\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}$

= $\mathop {\lim }\limits_{x \to o} {{2x(\tan x - \cos x) + {x^2}({{\sec }^2}x + \sin x)} \over x}$

= $\mathop {\lim }\limits_{x \to o} \,\,2(\tan x - \cos x) + x({\sec ^2}x + \sin x)$

= 2 (0-1) + 0

= -2
4

### JEE Main 2018 (Online) 15th April Morning Slot

Let S = {($\lambda$, $\mu$) $\in$ R $\times$ R : f(t) = (|$\lambda$| e|t| $-$ $\mu$). sin (2|t|), t $\in$ R, is a differentiable function}. Then S is a subset of :
A
R $\times$ [0, $\infty$)
B
[0, $\infty$) $\times$ R
C
R $\times$ ($-$ $\infty$, 0)
D
($-$ $\infty$, 0) $\times$ R

## Explanation

S = {($\lambda$, $\mu$) $\in$ R $\times$ R : f(t) = $\left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)$ sin $\left( {2\left| t \right|} \right),$ t $\in$ R

f(t) = $\left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)\sin \left( {2\left| t \right|} \right)$

= $\left\{ {\matrix{ {\left( {\left| \lambda \right|{e^t} - \mu } \right)\sin 2t,} & {t > 0} \cr {\left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( { - \sin 2t} \right),} & {t < 0} \cr } } \right.$

f'(t) = $\left\{ {\matrix{ {\left( {\left| \lambda \right|{e^t}} \right)\sin 2t + \left( {\left| \lambda \right|{e^t} - \mu } \right)\left( {2\cos 2t} \right),\,\,t > 0} \cr {\left| \lambda \right|{e^{ - t}}\sin 2t + \left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( {-2\cos 2t} \right),t < 0} \cr } } \right.$

As, f(t) is differentiable

$\therefore$  LHD = RHD at t = 0

$\Rightarrow$   $\left| \lambda \right|$ . sin2(0) + $\left( {\left| \lambda \right|{e^0} - \mu } \right)$2cos(0)

= $\left| \lambda \right|{e^{ - 0}}\,$ . sin 2(0) $-$ 2 cos (0) $\left( {\left| \lambda \right|{e^{ - 0}} - \mu } \right)$

$\Rightarrow$$\,\,\,$ 0 + $\left( {\left| \lambda \right| - \mu } \right)$2 = 0 $-$ 2$\left( {\left| \lambda \right| - \mu } \right)$

$\Rightarrow$  4$\left( {\left| \lambda \right| - \mu } \right)$ = 0

$\Rightarrow$  $\left| \mu \right|$ = $\mu$

So, S $\equiv$ ($\lambda$, $\mu$) = {$\lambda$ $\in$ R & $\mu$ $\in$ [0, $\infty$)}

Therefore set S is subset of R $\times$ [0, $\infty$)