1

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

The value of k for which the function

$f\left( x \right) = \left\{ {\matrix{ {{{\left( {{4 \over 5}} \right)}^{{{\tan \,4x} \over {\tan \,5x}}}}\,\,,} & {0 < x < {\pi \over 2}} \cr {k + {2 \over 5}\,\,\,,} & {x = {\pi \over 2}} \cr } } \right.$

is continuous at x = ${\pi \over 2},$ is :
A
${{17} \over {20}}$
B
${{2} \over {5}}$
C
${{3} \over {5}}$
D
$-$ ${{2} \over {5}}$
2

### 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
3

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

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

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

$= \,\,\,\,120$
4

### 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

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