JEE 2018 :: Mathematics (2018) :: Solved questions (2018)

• Mathematics (2018) - Solved questions (2018)

From 6 different novels  and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf,

so that the dictionary is always in the middle. The number of such arrangement is  

Answer:

Explanation:

Given 6 different novels and 3 different dictionaries.

Number of ways of selecting 4 novels from 6 novel is $^{6}C_{4}=\frac{6!}{2!4!}=15$

  Number of ways of selecting 1 dictionary is from 3 dictionaries is $^{3}C_{1}=\frac{3!}{1!2!}=3$

$\therefore$     Total number of arrangement of 4 novels and 1 dictionary where dictionary is always in the middle is

                      $15\times 3\times 4!=45\times 24=1080$

If α,β $\in$ C are the distinct roots of the equation $x^{2}-x+1=0, $ , then $\alpha^{101}+\beta^{107}$  is equal to 

Answer:

Explanation:

We have,

α,β are the roots of the x²-x+1=0,

$\therefore$        Roots of  $x^{2}-x+1=0$ and $-\omega,-\omega^{2}$

$\therefore$        Let $\alpha =-\omega $ and  $\beta =-\omega^{2} $

$\Rightarrow$      $\alpha^{101}+\beta^{107}=(-\omega)^{101}+(-\omega^{2})^{107}$

                                                                       = $-(\omega^{101}+\omega^{214})$

                                                                       =  $-(\omega^{2}+\omega^{}) $     ( $\because$  $\omega^{3}=1$)

                                                                      = - ( -1)       [ $\because$  $1+\omega+\omega^{2}=0$ ]

                                                                      = 1

Let S= {$x\in R:x\geq0$ and $2\mid \sqrt{x}-3\mid +\sqrt{x}(\sqrt{x}-6)+6=0$. Then , S

Answer:

Explanation:

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

Let    $\sqrt{x}-3=y$

$\Rightarrow$        $\sqrt{x}=y+3$

                        $2\mid y\mid+(y+3)(y-3)+6=0$

$\Rightarrow$       $2\mid y\mid+y^{2}-3=0$

$\Rightarrow$      $\mid y \mid^{2} +2\mid y\mid-3=0$

$\Rightarrow$     $( \mid y\mid +3)( \mid y\mid-1)=0$

$\Rightarrow$     $ \mid y\mid \neq-3 \Rightarrow \mid y\mid=1$

$\Rightarrow$     $y=\pm 1\Rightarrow\sqrt{x}-3=\pm 1$

$\Rightarrow$     $\sqrt{x}=4,2   \Rightarrow x=16,4$

• Mathematics (2018) - Solved questions (2018)