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

• Mathematics (2018) - Solved questions (2018)

Let E₁ E₂ and F₁ F₂ be the chords of S passing through the point P₀ (1,1) and parallel to the X-axis and the Y-axis, respectively. Let G₁ G₂  be the chord of S passing through P₀ and having slope -1. Let the tangents to S at E₁ and E₂ meet at E_3, then tangents to S at F₁ and  F₂ meet at F_3, and the tangents to S at G₁ and G₂  meet at G3. Then, the points E₃, F_3, and G₃ lie on the curve

Answer:

Explanation:

Equation of tangent at E₁ $(-\sqrt{3},1)$ is 

                      $-\sqrt{3}x+y=4$  and at  E₂ $(-\sqrt{3},1)$  is

      $\sqrt{3}x+y=4$

Intersection point of tangent at E₁  and E₂  is (0,4)

            Coordinates of E₃ is (0,4)

Similarly, equation of tangent at F₁ (1,- $\sqrt{3}$)anf F₂ (1, $\sqrt{3}$) are x-$\sqrt{3}$y=4 and  x+$\sqrt{3}$y=4 respectively and intersection point is (4 ,0), i.e. , F₃ (4,0) and equation of tangent at G₁ (0,2) and G₂ (2,0) are 2y=4

and 2x=4, respectively and intersection point is (2,2) ie., G₃ (2,2).

      Point    E₃ (0,4) ,F₃ (4,0)   and  G₃ (2,2)  satisfies the x+y=4.

A farmer F₁  has a land in the shape of a triangle  with vertices  at P(0,0), Q(1,1), and R (2,0). From this land, a neighbouring farmer F₂  takes away the region  which lies between the sides PQ and a curve of the form y=xⁿ (n>1). If the area of the region taken away by the farmer F₂  is exactly 30% of the area Δ PQR , then the value of n is........

Answer:

Explanation:

We have

                    y= xⁿ  n>1

  $\because$  P(0,0), Q(1,1) and R(2,0) are vertices of  Δ PQR.

$\because$   Area of shaded region= 30% of area of   Δ PQR

   $\Rightarrow$    $\int_{0}^{1} (x-x^{n})dx=\frac{30}{100}\times\frac{1}{2}\times2\times1$

$\Rightarrow$        $[\frac{x^{2}}{2}-\frac{x^{n+1}}{n+1}]_0^1=\frac{3}{10}$

$\Rightarrow$       $[\frac{1^{}}{2}-\frac{1^{}}{n+1}]=\frac{3}{10}$

$\Rightarrow$     $\frac{1^{}}{n+1}=\frac{1}{2}-\frac{3}{10}=\frac{2}{10}=\frac{1}{5}$

$\Rightarrow$     n+1=5 $\Rightarrow$  n=4

Let a,b,c  be three non-zero  real  numbers such that the equation  $\sqrt{3} a \cos x+2b \sin x= c$ , $x\in[-\frac{\pi}{2},\frac{\pi}{2}]$, has two distinct real roots $\alpha$ and $\beta$  with $\alpha$ + $\beta$= $\frac{\pi}{3}$ , Then , the value of $\frac{b}{a}$ is.....

Answer:

Explanation:

We have $\alpha$ and $\beta$  are the roots of

            $\sqrt{3} a \cos x+2b \sin x= c$

$\therefore$    $\sqrt{3}a \cos \alpha +2b \sin \alpha=c$     .....(i)

 and       $\sqrt{3}a \cos \beta +2b \sin \beta=c$      .......(ii)

On substracting Eq. (ii) from Eq. (i) , we get

    $\sqrt{3}a ( \cos \alpha-\cos \beta)+2b(\sin\alpha-\sin \beta)=0$

$\Rightarrow$     $\sqrt{3}a (-2\sin(\frac{\alpha+\beta}{2}))\sin(\frac{\alpha-\beta}{2}) +2b (2\cos(\frac{\alpha+\beta}{2}))\sin(\frac{\alpha-\beta}{2})=0$

$\Rightarrow$   $\sqrt{3}a \sin(\frac{\alpha+\beta}{2})=2b\cos(\frac{\alpha+\beta}{2})$

 $\Rightarrow$  $\tan(\frac{\alpha+\beta}{2})=\frac{2b}{\sqrt{3}a}$

$\Rightarrow$  $\tan(\frac{\pi}{6})=\frac{2b}{\sqrt{3}a}$      [ $\therefore$      $\alpha$ + $\beta$= $\frac{\pi}{3}$, given ]

$\Rightarrow$   $\frac{1}{\sqrt{3}}=\frac{2b}{\sqrt{3}a}\Rightarrow \frac{b}{a}=\frac{1}{2}$

  $\Rightarrow$     $\frac{b}{a}=0.5$

Let a and b be two unit vectors such that a.b=0. For some x,y $\in$ R, let c=xa+yb+ (a× b) . If  $\mid c\mid=2$ and the vector c is inclined at the same angle $\alpha$ to both a and b, then the value of $8\cos^{2}\alpha$  is

Answer:

Explanation:

We have

  $\overrightarrow{c}=x\overrightarrow{a}+y\overrightarrow{b}+\overrightarrow{a}\times\overrightarrow{b}$   and  $\overrightarrow{a}.\overrightarrow{b}=0$

    $\mid\overrightarrow{a}\mid=\mid\overrightarrow{b}\mid=1$ and $\mid\overrightarrow{c}\mid=2$

   Also , given $\overrightarrow{c}$ is inclined on $\overrightarrow{a}$ and $\overrightarrow{b}$ with same angle $\alpha$

$\therefore$     $\overrightarrow{a}$ . $\overrightarrow{c}$ = $x\mid\overrightarrow{a}\mid^{2}+y(\overleftarrow{a}.\overrightarrow{b})+\overrightarrow{a}.(\overrightarrow{a}\times\overrightarrow{b})$

  $\mid\overrightarrow{a}\mid^{}.\mid\overrightarrow{c}\mid^{}\cos\alpha=x+0+0$ 

                            x= $2\cos\alpha$

Similarly,

       $\mid\overrightarrow{b}\mid^{}.\mid\overrightarrow{c}\mid^{}\cos\alpha=0+y+0$

                     y = $2\cos\alpha$

 $\mid\overrightarrow{c}\mid^{2}=x^{2}+y^{2}+\mid\overrightarrow{a}\times\overrightarrow{b}\mid^{2}$

   $4= 8\cos^{2}\alpha+\mid a\mid^{2}\mid b\mid^{2}\sin^{2}90^{0}$

    $4= 8\cos ^{2}\alpha+1$

$\Rightarrow$ $8\cos^{2}\alpha=3$

For each positive integer n,  let $y_{n}=\frac{1}{n}((n+1)(n+2)...(n+n))^{\frac{1}{n}}. $. For x ε R, let [x] be the greatest integer less than or equal to x. If $ \lim_{n \rightarrow\infty} y_{n}=L$, then the value of [L] is ...........

Answer:

Explanation:

We have 

       $y_{n}=\frac{1}{n}((n+1)(n+2)...(n+n))^{\frac{1}{n}}$    and   

                                                                                                           $\lim_{n \rightarrow\infty}y_{n}=L$

   $\Rightarrow$    L=  $\lim_{n \rightarrow\infty}\frac{1}{n}[(n+1)(n+2)(n+3)....(n+n)]^{\frac{1}{n}}$

  $\Rightarrow$    L= $\lim_{n \rightarrow \infty}\frac{1}{n}[(1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})....(1+\frac{n}{n})]^{\frac{1}{n}}$

$\Rightarrow$    $\log_{}{L}=\lim_{n \rightarrow\infty}\frac{1}{n}[\log_{}{}(1+\frac{1}{n})+\log_{}{}(1+\frac{2}{n})+\log_{}{}(1+\frac{3}{n})....\log_{}{}(1+\frac{n}{n})]^{\frac{}{}}$

$\Rightarrow$     $\log_{}{L}= \lim_{n \rightarrow\infty}\frac{1}{n}\sum_r^n\log_{}{(1+\frac{r}{n})}$

  $\Rightarrow$  $\log_{}{L}= \int_{0}^{1}1 \times \log_{}{(1+x)dx} $

$\Rightarrow$   $\log_{}{L}= (x.\log_{}{(1+x)_0^1}-\int_{0}^{1}[\frac{\text{d}}{\text{d}x}(\log_{}{(1+x)\int_{}^{} dx]dx} $

                           [ by using integration by parts]

$\Rightarrow$   $\log_{}{L}= [x \log_{}{(1+x)_0^1}-\int_{0}^{1} \frac{x}{1+x}dx$

$\Rightarrow$     $\log_{}{L}= \log_{}{2}-\int_{0}^{1}(\frac{x+1}{x+1}-\frac{1}{x+1})dx $

 $\Rightarrow$   $\log_{}{L}= \log_{}{2}-[x]_0^1+[\log_{}{(x+1)]_0^1} $

$\Rightarrow$     $\log_{}{L}= \log_{}{2}-1+\log_{}{2}-0 $

$\Rightarrow$   $\log_{}{L}= \log_{}{4}-\log_{}{e}=\log_{}{\frac{4}{e}}$

$\Rightarrow$    $L=\frac{4}{e}$

$\Rightarrow$  [L]=[$\frac{4}{e}$]=1

• Mathematics (2018) - Solved questions (2018)