1

### 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
2

### 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
3

### JEE Main 2019 (Online) 11th January Morning Slot

If one real root of the quadratic equation 81x2 + kx + 256 = 0 is cube of the other root, then a value of k is
A
$-$ 81
B
$-$ 300
C
100
D
144

## Explanation

81x2 + kx + 256 = 0 ; x = $\alpha$, $\alpha$3

$\Rightarrow$  $\alpha$4 = ${{256} \over {81}}$ $\Rightarrow$   $\alpha$ = $\pm$ ${{4} \over {3}}$

Now   $-$ ${k \over {81}}$ = $\alpha$ + $\alpha$3 = $\pm$ ${{100} \over {27}}$

$\Rightarrow$  k = $\pm$300
4

### JEE Main 2019 (Online) 11th January Evening Slot

Let $\alpha$ and $\beta$ be the roots of the quadratic equation x2 sin $\theta$ – x(sin $\theta$ cos $\theta$ + 1) + cos $\theta$ = 0 (0 < $\theta$ < 45o), and $\alpha$ < $\beta$. Then $\sum\limits_{n = 0}^\infty {\left( {{\alpha ^n} + {{{{\left( { - 1} \right)}^n}} \over {{\beta ^n}}}} \right)}$ is equal to :
A
${1 \over {1 + \cos \theta }} + {1 \over {1 - \sin \theta }}$
B
${1 \over {1 - \cos \theta }} + {1 \over {1 + \sin \theta }}$
C
${1 \over {1 - \cos \theta }} - {1 \over {1 + \sin \theta }}$
D
${1 \over {1 + \cos \theta }} - {1 \over {1 - \sin \theta }}$

## Explanation

D = (1 + sin$\theta$ cos$\theta$)2 $-$ 4sin$\theta$cos$\theta$ = (1 $-$ sin$\theta$ cos$\theta$)2

$\Rightarrow$  roots are $\beta$ = cosec$\theta$ and $\alpha$ = cos$\theta$

$\sum\limits_{n = 0}^\infty {\left( {{\alpha ^n} + {{\left( { - {1 \over \beta }} \right)}^n}} \right)} = \sum\limits_{n = 0}^\infty {{{\left( {\cos \theta } \right)}^n}} + \sum\limits_{n = 0}^n {{{\left( { - \sin \theta } \right)}^n}}$

$= {1 \over {1 - \cos \theta }} + {1 \over {1 + \sin \theta }}$