1

### JEE Main 2018 (Online) 16th April Morning Slot

If an angle A of a $\Delta$ABC satiesfies 5 cosA + 3 = 0, then the roots of the quadratic equation, 9x2 + 27x + 20 = 0 are :
A
secA, cotA
B
sinA, secA
C
secA, tanA
D
tanA, cosA

## Explanation

Here, 9x2 + 27x + 20 = 0

$\therefore\,\,\,$ x = ${{ - b \pm \sqrt {{b^2} - 4ac} } \over {2a}}$

$\Rightarrow $$\,\,\, x = {{ - 27 \pm \sqrt {{{27}^2} - 4 \times 9 \times 20} } \over {2 \times 9}} \Rightarrow$$\,\,\,$ x = $-$ ${4 \over 3}$, $-$ ${5 \over 3}$

Given, cosA = $-$ ${3 \over 5}$

$\therefore\,\,\,$ sec A = ${1 \over {\cos A}}$ = $-$ ${5 \over 3}$

Here, A is an obtuse angle.

$\therefore\,\,\,$ tan A = $-$ $\sqrt {{{\sec }^2}A - 1} = - {4 \over 3}.$

Hence, roots of the equation are sec A and tan A.
2

### 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)
3

### 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
4

### 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