1

### JEE Main 2019 (Online) 9th January Morning Slot

Let f : R $\to$ R be a function defined as
$f(x) = \left\{ {\matrix{ 5 & ; & {x \le 1} \cr {a + bx} & ; & {1 < x < 3} \cr {b + 5x} & ; & {3 \le x < 5} \cr {30} & ; & {x \ge 5} \cr } } \right.$

Then, f is
A
continuous if a = 0 and b = 5
B
continuous if a = –5 and b = 10
C
continuous if a = 5 and b = 5
D
not continuous for any values of a and b

## Explanation

Checking

if f(x) is continuous at x = 1 :

f(1$-$) = 5

f(1) = 5

f(1+) = a + b

if f(x) is continuous at x = 1,

then

f(1$-$) = f(1) = f(1+)

$\Rightarrow$  5 = 5 = a + b

$\therefore$  a + b = 5 . . . . . . . . (1)

checking if f(x) is continuous at x = 3 :

f(3$-$) = a + 3b

f(3) = b + 15

f(3+) = b + 15

if   f(x) = is continuous at x = 3

then,

f(3$-$) = f(3) = f(3+)

$\Rightarrow$  a + 3b = b + 15 = b + 15

$\Rightarrow$  a + 2b = 15 . . . . . (2)

checking if f(x) is continuous at x = 5 :

f(5$-$) = b + 25

f(5) = 30

f(5+) = 30

if f(x) is continuous at x = 5 then,

f(5$-$) = f(5) = f(5+)

$\Rightarrow$   b + 25 = 30 = 30

$\Rightarrow$   b = 5

By putting this value in equation (2), we get,

a + 2(5) = 15

$\Rightarrow$  a = 5

when a = 5 and b = 5 then equation (1)

a + b = 5 does not satisfy.

$\therefore$  f is not continuous for any value of a and b.
2

### JEE Main 2019 (Online) 9th January Morning Slot

$\mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$
A
exists and equals ${1 \over {2\sqrt 2 }}$
B
exists and equals ${1 \over {4\sqrt 2 }}$
C
exists and equals ${1 \over {2\sqrt 2 (1 + \sqrt {2)} }}$
D
does not exists

## Explanation

$\mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$

If you put y = 0 at ${{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$ it is in ${0 \over 0}$ form. So we can use L' Hospital's Rule.

= $\mathop {\lim }\limits_{y \to 0} {{{1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times \left( {{1 \over {2\sqrt {1 + {y^4}} }}} \right) \times 4{y^3}} \over {4{y^3}}}$

= $\mathop {\lim }\limits_{y \to 0} {1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times {1 \over {2\sqrt {1 + {y^4}} }}$

= ${1 \over {4\sqrt 2 }}$
3

### JEE Main 2019 (Online) 9th January Evening Slot

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

Then $\mathop {\lim }\limits_{x \to {0^ - }} \,\,{{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}$ is equal to :
A
$-$ sin 1
B
1
C
sin 1
D
0

## Explanation

$\mathop {\lim }\limits_{x \to {0^ - }} {{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}$

$= \mathop {\lim }\limits_{h \to 0} {{\left( {0 - h} \right)\left( {\left[ {0 - h} \right] + \left| {0 - h} \right|} \right)\sin \left[ {0 - h} \right]} \over {\left| {0 - h} \right|}}$

$= \mathop {\lim }\limits_{h \to 0} {{ - h\left( { - 1 + h} \right)\sin \left( { - 1} \right)} \over h}$

$= \mathop {\lim }\limits_{h \to 0} \left( { - 1 + h} \right)\sin \left( 1 \right) = - \sin 1$
4

### JEE Main 2019 (Online) 9th January Evening Slot

Let f be a differentiable function from

R to R such that $\left| {f\left( x \right) - f\left( y \right)} \right| \le 2{\left| {x - y} \right|^{{3 \over 2}}},$

for all  $x,y \in$ R.

If   $f\left( 0 \right) = 1$

then   $\int\limits_0^1 {{f^2}} \left( x \right)dx$  is equal to :
A
1
B
2
C
${1 \over 2}$
D
0

## Explanation

$\left| {f(x) - f(y)} \right| \le 2{\left[ {x - y} \right]^{3/2}}$

$\left| {{{f(x) - f(y)} \over {x - y}}} \right| \le 2{\left| {x - y} \right|^{1/2}}$

$\mathop {\lim }\limits_{y \to x} \left| {{{f(x) - f(y)} \over {x - y}}} \right| \le \mathop {\lim }\limits_{y \to x} 2{\left| {x - y} \right|^{1/2}}$

$\Rightarrow \left| {f'\left( x \right)} \right| \le 0$  $\Rightarrow f'\left( x \right) = 0$

$\Rightarrow f\left( x \right) =$ constant

as  $f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = 1$

$\int\limits_0^1 {{f^2}} \left( x \right)dx = 1$