1

### 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
2

### 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
3

### 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$\theta$

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

$= {1 \over {1 - \cos \theta }} + {1 \over {1 + \sin \theta }}$
4

### 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$

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

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

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