1
MCQ (Single Correct Answer)

### 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
2
MCQ (Single Correct Answer)

### 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$\theta$

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

$= {1 \over {1 - \cos \theta }} + {1 \over {1 + \sin \theta }}$
3
MCQ (Single Correct Answer)

### JEE Main 2019 (Online) 12th January Morning Slot

If $\lambda$ be the ratio of the roots of the quadratic equation in x, 3m2x2 + m(m – 4)x + 2 = 0, then the least value of m for which $\lambda + {1 \over \lambda } = 1,$ is
A
$- 2 + \sqrt 2$
B
4$-$3$\sqrt 2$
C
2 $-$ $\sqrt 3$
D
4 $-$ 2$\sqrt 3$

## Explanation

3m2x2 + m(m $-$ 4) x + 2 = 0

$\lambda + {1 \over \lambda } = 1,{\alpha \over \beta } + {\beta \over \alpha } = 1,{\alpha ^2} + {\beta ^2} = \alpha \beta$

($\alpha$ + $\beta$)2 = 3$\alpha$$\beta$

${\left( { - {{m\left( {m - 4} \right)} \over {3{m^2}}}} \right)^2} = {{3\left( 2 \right)} \over {3{m^2}}},{{{{\left( {m - 4} \right)}^2}} \over {9{m^2}}} = {6 \over {3m}}$

${\left( {m - 4} \right)^2} = 18,m = 4 \pm \sqrt {18,} \,\,4 \pm 3\sqrt 2$
4
MCQ (Single Correct Answer)

### JEE Main 2019 (Online) 12th January Evening Slot

The number of integral values of m for which the quadratic expression, (1 + 2m)x2 – 2(1 + 3m)x + 4(1 + m), x $\in$ R, is always positive, is :
A
7
B
8
C
3
D
6

## Explanation

Expression is always positive it

2m + 1 > 0 $\Rightarrow$ m > $-$ ${1 \over 2}$ &

D < 0 $\Rightarrow$ m2 $-$ 6m $-$ 3 < 0

3 $-$ $\sqrt {12}$ < m < 3 + $\sqrt {12}$ . . . . (iii)

$\therefore$  Common interval is

3 $-$ $\sqrt {12}$ < m < 3 + $\sqrt {12}$

$\therefore$  Intgral value of m {0, 1, 2, 3, 4, 5, 6}

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