1

### JEE Main 2017 (Online) 9th April Morning Slot

The sum of all the real values of x satisfying the equation
2(x$-$1)(x2 + 5x $-$ 50) = 1 is :
A
16
B
14
C
$-$4
D
$-$ 5

## Explanation

We know, 2x = 1 only when x = 0.

Similarly, 2(x$-$1)(x2 + 5x $-$ 50) = 1 when

(x$-$1)(x2 + 5x $-$ 50) = 0

$\Rightarrow$ (x - 1)(x + 10)(x - 5) = 0

$\therefore$ x = 1, -10, 5

Sum of real values of x = 1 + (-10) + 5 = -4
2

### JEE Main 2018 (Offline)

Let S = { $x$ $\in$ R : $x$ $\ge$ 0 and

$2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$}. Then S
A
contains exactly four elements
B
is an empty set
C
contains exactly one element
D
contains exactly two elements

## Explanation

Given,

$2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$

Case 1 :

When $\sqrt x - 3 \ge 0,$ then equation becomes

$2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$

$\Rightarrow \,\,\,\,2\sqrt x - 6 + x - 6\sqrt x + 6 = 0$

$\Rightarrow \,\,\,\,x\, - 4\sqrt x = 0$

$\Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 4} \right) = 0$

$\therefore\,\,\,$ $\sqrt x = 0,4$

but as $\sqrt x - 3 \ge 0$ or $\sqrt x \ge 3$ then $\sqrt x \ne 0$

$\therefore\,\,\,$ $\sqrt x = 4$ value is possible.

Case 2 :

When $\sqrt x - 3 < 0$ or $\sqrt x < 3.$ There equation becomes

$- 2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$

$\Rightarrow \,\,\,\,\, - 2\sqrt x + 6 + x - 6\sqrt x + 6 = 0$

$\Rightarrow \,\,\,\,x - 8\sqrt x + 12 = 0$

$\Rightarrow \,\,\,\,x - 6\sqrt x - 2\sqrt x + 12 = 0$

$\Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 6} \right) - 2\left( {\sqrt x - 6} \right) = 0$

$\Rightarrow \,\,\,\,\left( {\sqrt x - 2} \right)\left( {\sqrt x - 6} \right) = 0$

$\therefore\,\,\,$ $\sqrt x = 2,6$

as $\sqrt x < 3$ so $\sqrt x \ne 6$

$\therefore\,\,\,$ $\sqrt x = 2$ is possible.

So, total possible value of $\sqrt x = 2,4$

or for x possible values are 4, 16.

$\therefore\,\,\,$ Set S contains exactly two elements 4 and 16.
3

### JEE Main 2018 (Online) 15th April Morning Slot

If $\lambda$ $\in$ R is such that the sum of the cubes of the roots of the equation,
x2 + (2 $-$ $\lambda$) x + (10 $-$ $\lambda$) = 0 is minimum, then the magnitude of the difference of the roots of this equation is :
A
$4\sqrt 2$
B
$2\sqrt 5$
C
$2\sqrt 7$
D
20

## Explanation

Let $\alpha$, $\beta$ are the roots of the equation,

$\therefore$ $\alpha$ + $\beta$ = $\lambda$ $-$ 2 and $\alpha $$\beta = 10 - \lambda {\alpha ^3} + {\beta ^3} = (\alpha + \beta )3 - 3\alpha$$\beta$ ($\alpha$ + $\beta$)

= ($\lambda$ $-$ 2)3 $-$ 3(10 $-$ $\lambda$)($\lambda$ $-$ 2)

= $\lambda ^3$ $-$ 3$\lambda ^2$ $-$ 24$\lambda$ + 52

Let $f(\lambda$) = $\lambda ^3$ $-$ 3$\lambda ^2$ $-$ 24$\lambda$ + 52

$\therefore$ ${{df(\lambda )} \over {d\lambda }}$ = 3$\lambda ^2$ $-$ 6$\lambda$ $-$ 24

$\therefore$ at maximum of minimum ${{df(\lambda )} \over {d\lambda }}$ = 0

$\therefore$ $\lambda ^2$ $-$ 2$\lambda$ $-$ 8 = 0

$\Rightarrow$ ($\lambda$ + 2) ($\lambda$ $-$ 4) = 0

$\Rightarrow$ $\lambda$ = $-$2, 4

${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$ = 2$\lambda$ $-$ 2

When $\lambda$ = $-$2

${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$ = $-$ 6 < 0

$\therefore$ at $\lambda$ = $-$2, f($\lambda$) has maximum value.

When $\lambda$ = 4

${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$ = 6 > 0

$\therefore$ at $\lambda$ = 4, f($\lambda$) has minimum value.

$\therefore$ When $\lambda$ = 4 equation is,

x2 $-$ 2x + 6 = 0

$\therefore$ ($\alpha$ $-$ $\beta$)2 = ($\alpha$ + $\beta$)2 $-$ 4$\alpha \beta$

$\Rightarrow$ x2 $-$ 4 $\times$ 6

= $-$ 20

$\Rightarrow$ ($\alpha$ $-$ $\beta$) = $2\sqrt 5 i$

$\Rightarrow$ $\left| {\alpha - \beta } \right|$ = $2\sqrt 5$ (ans)
4

### JEE Main 2018 (Online) 15th April Morning Slot

If tanA and tanB are the roots of the quadratic equation, 3x2 $-$ 10x $-$ 25 = 0, then the value of 3 sin2(A + B) $-$ 10 sin(A + B).cos(A + B) $-$ 25 cos2(A + B) is :
A
$-$ 10
B
10
C
$-$ 25
D
25

## Explanation

As tan A and tan B are the roots of 3x2 $-$ 10x $-$ 25 = 0,

So, tan(A + B) = ${{\tan A + \tan B} \over {1 - \tan A\tan B}}$

= ${{{{10} \over 3}} \over {1 + {{25} \over 3}}}$ = ${{10/3} \over {28/3}}$ = ${5 \over {14}}$

Now, cos2 (A + B) = $-$ 1 + 2 cos2 (A + B)

= ${{1 - {{\tan }^2}(A + B)} \over {1 + {{\tan }^2}(A + B)}}\,$ $\Rightarrow$ cos2(A + B) = ${{196} \over {221}}$

$\therefore\,\,\,$ 3sin2(A + B) $-$ 10sin(A + B)cos(A + B) $-$ 25 cos2(A + B)

= cos2(A + B) [ 3tan2(A + B) $-$ 10tan(A + B) $-$ 25]

= ${{75 - 700 - 4900} \over {196}} \times {{196} \over {221}}$

= $-$ ${{5525} \over {196}}$ $\times$ ${{196} \over {221}}$ = $-$ 25