JEE 2017 :: Advanced 2017 :: Mathematics 2017

• Advanced 2017 - Physics 2017
• Advanced 2017 - Chemistry 2017
• Advanced 2017 - Mathematics 2017

If 2x-y+1=0,  is a tangent to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{16}=1$  , then which of the following CANNOT be sides of a right angled triangle?

Answer:

Explanation:

Tangent=2x-y+1=0

Hyperbola= $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{16}=1$

It point = $(a\sec\theta,4\tan\theta)$

$tangent=\frac{x\sec\theta}{a}-\frac{y\tan\theta}{4}=1$

  On comparing we get,

$\sec\theta=-2a$

$\tan\theta=-4$

$\Rightarrow$    $4a^{2}-16=1$

        $a=\frac{\sqrt{17}}{2}$

  Subsuitute the value of a in option (a).(b), (c)and (d)

Let a,b,x and y be real numbers such that a-b=1 and y≠ o. If the complex number z=x+iy satisfies  $Im(\frac{az+b}{z+1})=y$ , then which of the following is(are) possible value(s) of x?

Answer:

Explanation:

$\frac{az+b}{z+1}=\frac{ax+b+aiy}{(x+1)+iy}$

$= \frac{(ax+b+aiy)((x+1)-iy)}{(x+1)^{2}+y^{2}}$

$\because$   $Im(\frac{az+b)}{z+1})=\frac{-(ax+b)y+ay(x+1)}{(x+1)^{2}+y^{2}}$

$\Rightarrow$     $\frac{(a-b)y}{(x+1)^{2}+y^{2}}=y$

$\because$    a-b=1

$\because$   $(x+1)^{2}+y^{2}=1$

$\because$    $x=-1\pm \sqrt{1-y^{2}}$

Let f: R→ (0,1) be a continuous function. Then, which of the following function (s) has (have) the value zero at some point in the interval (0,1)?

Answer:

Explanation:

(a)   $\because e^{x}\epsilon(1,e)in(0,1)$ and

   $\int_{0}^{x} f(t)\sin t dt\in (0,1)$  in (0,1)

$\therefore e^{x}-\int_{0}^{x} f(t) \sin t dt$ cannot be zero

so, option (a) is incorrect

(b)   $f(x)+\int_{0}^{\frac{\pi}{2}}f(t) \sin t dt $ always postive

 option b is incorrect

(c)   Let   $h(x)=x-\int_{0}^{\frac{\pi}{2}-x} f(t) \cos t dt$

$h(0)=-\int_{0}^{\frac{\pi}{2}} f(t) \cos t dt<0$

$h(1)=1-\int_{0}^{\frac{\pi}{2}-1} f(t) \cos t dt>0$

 Option C is correct

(d) Let g(x)= x⁹-f(x)

  g(0)=-f(0)<0, g(1)=1-f(1)>0

 Option d is correct

Let X and Y the two events such  $P(X)=\frac{1}{3}$ , $P(X\mid Y)= \frac{1}{2}$, $P(Y\mid X)= \frac{2}{5}$    Then

Answer:

Explanation:

$P(X)=\frac{1}{3}$

$P(\frac{X}{Y})=\frac{P(X\cap Y)}{P(Y)}=\frac{1}{2}$

$P(\frac{Y}{X})=\frac{P(X\cap Y)}{P(X)}=\frac{2}{5}$

$P(X\cap Y)=\frac{2}{15}$

$P(Y)=\frac{4}{15}$

  $P(\frac{X'}{Y})=\frac{P(Y)-P(X\cap Y)}{P(Y)}$

$=\frac{\frac{4}{15}-\frac{2}{15}}{\frac{4}{15}}=\frac{1}{2}$

$P(X\cup Y)=\frac{1}{3}+\frac{4}{15}-\frac{2}{15}$

   = $\frac{7}{15}=\frac{7}{15}$

Let O be the origin and OX, OY, OZ be three unit vectors in the directions of the sides QR,RP,PQ respectively of triangle PQR.

I f the triangle PQR varies, then the minimum value of $\cos(P+Q)+\cos(Q+R)+\cos(R+P)$ is

Answer:

Explanation:

$\cos(P+Q)+\cos(Q+R)+\cos(R+P)=-(cosR+\cos P+\cos Q)$

Max of $\cos P+\cos Q +\cos R=\frac{3}{2}$

Min of  $\cos(P+Q)+\cos(Q+R)+\cos(R+P) is =-\frac{3}{2}$

• Advanced 2017 - Physics 2017
• Advanced 2017 - Chemistry 2017
• Advanced 2017 - Mathematics 2017