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

• Mathematics (2018) - Solved questions (2018)

A bag contains 4 red and 6 black balls, a ball is drawn at random from the bag. Its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is

Answer:

Explanation:

Key Idea Use the theorem of total probability

Let E₁ = Event that first ball drawn is red

E₂ = Event that first ball drawn is black

  A= Event that second ball drawn is red

 P( E₁)= $\frac{4}{10}$, $P[\frac{A}{E_{1}}]=\frac{6}{12}$

$\Rightarrow$    $P(E_{2})=\frac{6}{10}, P(\frac{A}{E_{2}})=\frac{4}{12}$

By law of total probability

  $P(A_{})=P( E_{1})\times P(\frac{A}{E_{1}})+P(E_{2})\times P(\frac{A}{E_{2}})$

$=\frac{4}{10}\times \frac{6}{12}+\frac{6}{10}\times\frac{4}{12}$

$=\frac{24+24}{120}=\frac{48}{120}=\frac{2}{5}$

The value  of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^{2}x}{1+2^{x}}$ is

Answer:

Explanation:

Key Idea 

 Use property = $\int_{a}^{b} f( x) dx=\int_{a}^{b} f( a+b-x)dx$

   Let $I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^{2}x}{1+2^{x}} dx$

$\Rightarrow$   I $= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin^{2}(\frac{-\pi}{2}+\frac{\pi}{2}-x)}{1+2^{\frac{-\pi}{2}+\frac{\pi}{2}-x}} dx  $

                                               [ $\because $  $\int_{a}^{b} f( x) dx=\int_{a}^{b} f( a+b-x)dx$ ]

$\Rightarrow$      $I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^{2}x}{1+2^{-x}}dx$

$\Rightarrow$   $I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2^{x}\sin^{2}x}{2^{x}+1^{}}dx$

$\Rightarrow$    $2I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{2}x(\frac{2^{x}+1}{2^{x}+1})dx$

$\Rightarrow$     $2I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{2}x dx$

$\Rightarrow$   $2I=2\int_{0}^{\frac{\pi}{2}}\sin^{2}x dx $       [ $\because\sin^{2}x$ is an even function ]

$\Rightarrow$  $I= \int_{0}^{\frac{\pi}{2}} \sin^{2}x dx$

$\Rightarrow$  $I= \int_{0}^{\frac{\pi}{2}} \cos^{2}x dx$

                                              [ $\because \int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx  $]

$\Rightarrow$   2I=$\int_{0}^{\frac{\pi}{2}} dx$

$\Rightarrow$  $2I= [ x] _{0}^{\frac{\pi}{2}}=I=\frac{\pi}{4}$

Two sets A and B are as  under A= { (a,b)} $\in$ RxR: $\mid a-5\mid<1$ and $\mid b-5\mid<1$;  B={(a,b) $\in$ RxR:

 $4(a-6)^{2}+9(b-5)^{2}\leq 36$ } , Then

Answer:

Explanation:

We have

$\mid a-5\mid<1$ and $\mid b-5\mid<1$

$\therefore$    -1 < a - 5< 1 and -1 < b-5 < 1   

$\Rightarrow$      4<a<6 and 4<b<6

  Now ,  $4(a-6)^{2}+9(b-5)^{2}\leq 36$

$\Rightarrow$   $\frac{(a-6)^{2}}{9}+\frac{(b-5)^{2}}{4}\leq1$

   Taking axes as a-axis and b-axis

The set A represents square PQRS inside set B representing ellipse and hence $A\subset B$

If the system of linear equation x+ky+3z=0;3x+ky-2z=0; 2x+4y-3z=0  has a non-zero solution ( x,y,z), then $\frac{xz}{y^{2}}$ is equal to

Answer:

Explanation:

We have, x+ky+3z=0; 3x+ky-2z=0;2x+4y-3z=0

System of the equation has non-zero solution, If

 $\begin{bmatrix}1 & k & 3 \\3 & k & -2 \\2 & 4 & -3 \end{bmatrix}=0$

$\Rightarrow$   (-3k+8)-k(-9+4)+3(12-2k)=0

$\Rightarrow$   -3k+8+9k-4k+36-6k=0

$\Rightarrow$    -4k+44=0 $\Rightarrow$ k=11

Let z=λ , then we get

        x+11y+3λ =0      ...... (i)

       3x+11y-2λ=0         .......(ii)

and   2x+4y-3λ =0      ........(iii)

Solving Eqs. (i) and (ii) , we get

   $x=\frac{5\lambda}{2},y=\frac{-\lambda}{2}, z=\lambda$

$\Rightarrow$   $\frac{xz}{y^{2}}=\frac{5\lambda^{2}}{2\times (-\frac{\lambda}{2})^{2}}=10$

Let $g(x)=\cos x ^{2}, f(x)=\sqrt{x}$ and $\alpha ,\beta (\alpha <\beta)$ be the roots of the quadratic equation  

$18x^{2}-9\pi x +\pi^{2}=0$ .Then , the area (in sq units) bounded by the curve  $y= (gof) (x)$ and the lines x=α, x=β

 and y=0, is 

Answer:

Explanation:

We have 

$\Rightarrow$    $18x^{2}-9\pi x +\pi^{2}=0$

$\Rightarrow$   $18x^{2}-6\pi x-3\pi x +\pi^{2}=0$

 $(6x-\pi) (3x-\pi)=0$

  $\Rightarrow$     $x=\frac{\pi}{6},\frac{\pi}{3}$

 Now ,  $\alpha<\beta =\frac{\pi}{6},\beta=\frac{\pi}{3}$

  Given  $g(x)=\cos x^{2} $ and  $f(x)=\sqrt{x} $

       $y=g Of(x) $

       $\therefore$    $y=g (f(x) )=\cos x$

    Area of region bounded by x=α,

  x=β , y=0 and curve  $y=g (f(x) )$ is

    $A=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos x dx$

    $A=[\sin x]_\frac{\pi}{6}^\frac{\pi}{3}$

    $A=\sin \frac{\pi}{3}-\sin \frac{\pi}{6}=\frac{\sqrt{3}}{2}-\frac{1}{2}$

      $A=(\frac{\sqrt{3}-1}{2})$

• Mathematics (2018) - Solved questions (2018)