1

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

If both the roots of the quadratic equation x2 $-$ mx + 4 = 0 are real and distinct and they lie in the interval [1, 5], then m lies in the interval :
A
($-$5, $-$4)
B
(4, 5)
C
(5, 6)
D
(3, 4)

## Explanation

x2 $-$mx + 4 = 0 Case-I :

D > 0

m2 $-$ 16 > 0

$\Rightarrow$  m $\in$ ($-$ $\infty$, $-$ 4) $\cup$ (4, $\infty$)

Case-II :

$\Rightarrow \,\,1 < {{ - b} \over {2a}} < 5$

$\Rightarrow \,\,1 < {m \over 2} < 5 \Rightarrow \,m \in \left( {2,10} \right)$

Case-III :

f(1) > 0   and f(5) > 0

1 $-$ m + 4 > 0   and 25 $-$ 5m + 4 > 0

m < 5   and m < ${{29} \over 5}$

Case-IV :

Let one root is x = 1

1 $-$ m + 4 = 0

m = 5

Now equation x2 $-$ 5x + 4 = 0

(x $-$ 1) (x $-$ 4) = 0

x = 1 i.e. m = 5 is also included

hence m $\in$ (4, 5]

So given option is (4, 5)
2

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

The number of all possible positive integral values of $\alpha$  for which the roots of the quadratic equation, 6x2 $-$ 11x + $\alpha$ = 0 are rational numbers is :
A
3
B
2
C
4
D
5

## Explanation

For rational D must be perfect square

D = 121 $-$ 24$\alpha$

for 121 $-$ 24$\alpha$ to be perfect square a must be 3, 4, 5

So, ans $\alpha$ = 3
3

### JEE Main 2019 (Online) 10th January Morning Slot

Consider the quadratic equation (c – 5)x2 – 2cx + (c – 4) = 0, c $\ne$ 5. Let S be the set of all integral values of c for which one root of the equation lies in the interval (0, 2) and its other root lies in the interval (2, 3). Then the number of elements in S is -
A
12
B
18
C
10
D
11

## Explanation Let f(x) = (c $-$ 5)x2 $-$ 2cx + c $-$ 4

$\therefore$  f(0)f(2) < 0      . . . . .(1)

& f(2)f(3) < 0      . . . . .(2)

from (1) and (2)

(c $-$ 4)(c $-$ 24) < 0

& (c $-$ 24)(4c $-$ 49) < 0

$\Rightarrow$  ${{49} \over 4}$ < c < 24

$\therefore$  s = {113, 14, 15, . . . . . 23}

Number of elements in set S = 11
4

### JEE Main 2019 (Online) 10th January Evening Slot

The value of $\lambda$ such that sum of the squares of the roots of the quadratic equation, x2 + (3 – $\lambda$)x + 2 = $\lambda$ has the least value is -
A
1
B
2
C
${{15} \over 8}$
D
${4 \over 9}$

## Explanation

$\alpha$ + $\beta$ = $\lambda$ $-$ 3

$\alpha $$\beta = 2 - \lambda \alpha 2 + \beta 2 = (\alpha + \beta )2 - 2\alpha$$\beta$ = ($\lambda$ $-$ 3)2 $-$ 2$\left( {2 - \lambda } \right)$

= $\lambda$2 + 9 $-$ 6$\lambda$ $-$ 4 + 2$\lambda$

= $\lambda$2 $-$ 4$\lambda$ + 5

= ($\lambda$ $-$ 2)2 + 1

$\therefore$  $\lambda$ = 2