JEE 2017 :: Advanced 2017 :: Mathematics 2017

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

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.

(ox × oy)= 

Answer:

Explanation:

$\sin R=\sin(P+Q)$

Let p,q be integers and let α ,β be the roots of the equation $x^{2}-x-1=0$  where α ≠β, For n=0,1,2...... Let  $a_{n}=p\alpha^{n}+q\beta^{n}$  ( If a and b are rational numbers and  $a+b\sqrt{5}=0$, then a=0=b)

If a₂₄=28   , then p+2q=

Answer:

Explanation:

$\alpha =\frac{1+\sqrt{5}}{2}$

$\beta =\frac{1-\sqrt{5}}{2}$

$a_{4}=a_{3}+a_{2}$

 $=2a_{2}+a_{1}$

 $=3a_{1}+2a_{0}$

$28=p(3\alpha+2)+q(3\beta+2)$

$28=p+q(\frac{3}{2}+2)+p-q(\frac{3\sqrt{5}}{2})$

$\therefore$ p-q=0

and      $(p+q)\times\frac{7}{2}=28$

$\Rightarrow$   p+q=8

$\Rightarrow$   p=q=4

    p+2q=12

Let p,q be integers and let α ,β be the roots of the equation $x^{2}-x-1=0$  where α ≠β, For n=0,1,2...... Let  $a_{n}=p\alpha^{n}+q\beta^{n}$  ( If a and b are rational numbers and  $a+b\sqrt{5}=0$, then a=0=b)

a₁₂=

Answer:

Explanation:

$\alpha^{2}=\alpha+1$

$\beta^{2}=\beta+1$

$a_{n}=p\alpha^{n}+q\beta^{n}$

$=p(\alpha^{n-1}+\alpha^{n-2})+q(\beta^{n-1}+\beta^{n-2})$

$=a_{n-1}+a_{n-2}$

$\therefore$   $a_{12}=a_{11}+a_{10}$

If $g(x)=\int_{\sin x}^{\sin(2x)} \sin^{-1}(t)dt$ , then

Answer:

Explanation:

$g(x)=\int_{\sin x}^{\sin(2x)} \sin^{-1}(t)dt$ 

$g'(x)=2\cos 2x\sin^{-1}(\sin2x)-\cos x\sin^{-1}(\sin x)$

$g'(\frac{\pi}{2})=-2\sin^{-1}(0)=0$

$g'(-\frac{\pi}{2})=-2\sin^{-1}(0)=0$ 

no option is matching

Let 

   $f(x)= \frac{1-x(1+\mid1-x\mid)}{\mid1-x\mid}\cos(\frac{1}{1-x})$   for x≠ 1, then

Answer:

Explanation:

$f(x)= \frac{1-x(1+\mid1-x\mid)}{\mid1-x\mid}\cos(\frac{1}{1-x})$

Now, 

     $\lim_{x \rightarrow 1^{-}}f(x)$

$=  \lim_{x \rightarrow 1^{-}}\frac{1-x(1+1-x)}{\mid1-x\mid}\cos(\frac{1}{1-x})$

$=  \lim_{x \rightarrow 1^{-}}(1-x)\cos(\frac{1}{1-x})=0 and \lim_{x \rightarrow 1^{+}}f(x)=\lim_{x \rightarrow 1^{+}}$

$\frac{1-x(1-1+x)}{x-1}\cos(\frac{1}{1-x})$

$=\lim_{x \rightarrow 1^{+}}-(x+1) cos (\frac{1}{x+1}) $ where x≠ 1 doesnot exist

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