JEE 2017 :: Advanced 2017 :: Mathematics 2017

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

Let α and β be non zero real numbers such $2(\cos\beta-\cos\alpha)+\cos\alpha\cos\beta=1$ . Then which of the following is/are true?

Answer:

Explanation:

We have

$2(\cos\beta-\cos\alpha)+\cos\alpha\cos\beta=1$

 or    $4(\cos\beta-\cos\alpha)+2\cos\alpha\cos\beta=2$

$\Rightarrow$   $1-\cos\alpha+\cos\beta-\cos\alpha\cos\beta$

$=3+3\cos\alpha-3\cos\beta-3\cos\alpha\cos\beta$

$\Rightarrow$   $(1-\cos\alpha)(1+\cos\beta)$

                     =$3(1-\cos\alpha)(1+\cos\beta)$

$\Rightarrow$         $\frac{(1-\cos\alpha)}{(1+\cos\alpha)}=\frac{3(1-\cos\beta)}{(1+\cos\beta}$

$\Rightarrow$   $\tan^{2}\frac{\alpha}{2}=3\tan^{2}\frac{\beta}{2}$

$\therefore$     $\tan\frac{\alpha}{2}\pm \sqrt{3}\tan\frac{\beta}{2}=0$

If the line $x=\alpha$ divides the area of region R={(x,y) $\in$ R² : x³≤ y≤ x, o≤x≤1 } into two equal parts, then

Answer:

Explanation:

$\int_{0}^{1} (x-x^{3})dx=2\int_{0}^{\alpha} (x-x^{3})dx$

                    $\frac{1}{4}=2(\frac{\alpha^{2}}{2}-\frac{\alpha^{4}}{4})$

$2\alpha^{4}-4\alpha^{2}+1=0$

                       $\alpha^{2}=\frac{4-\sqrt{16-8}}{4}(\because\alpha\in(0,1))$

                  $\alpha^{2}=1-\frac{1}{\sqrt{2}}$

If f:R→ R  is differentiable function such that f'(x) >2f(x) for all x ε R, and f(0)=1 then

Answer:

Explanation:

f'(x) >2f(x)

$\Rightarrow$   $\frac{dy}{y}>2dx$

$\Rightarrow$   $\int_{1}^{f(x)} \frac{dy}{y}>2\int_{0}^{x} dx$

 $ln(f(x))>2x$

 $\therefore$  f(x) >e^2x

Also as f'(x)>2f(x)

 $\therefore$  f'(x)>2c^2x>0

The equation of the plane passing through the point(1,1,1) and perpendicular to the planes

2x+y-2z=5 and 3x-6y-2z=7

Answer:

Explanation:

Let the equation of plane be ax+by+cz=1, Then

 a+b+c=1

2a+b-2c=0

3a-6b-2c=0

$\Rightarrow$      a=7b

   $c=\frac{15b}{2}$

$b=\frac{2}{31}$ ,$a=\frac{14}{31}$, $c=\frac{15}{31}$

14x+2y+15z=31

Let O be the origin and let PQR be an arbitrary triangle. The point S is such that

OP.OQ+OR.OS=OR.OP+OQ.OS=OQ.OR+OP.OS

  Then the triangle PQR has S as its

Answer:

Explanation:

OP.OQ+OR.OS

=OR.OP+OQ.OS

$\Rightarrow$ OP(OQ-OR)+OS(OR-OQ)=0

$\Rightarrow$  (OP-OS)(OQ-OR)=0

$\Rightarrow$ SP.RQ=0

Similarly SR.PQ=0

and SQ.PR=0

 So   S is orthocentre

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