JEE 2014 :: Advanced 2014 :: Mathematics 2014

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If a,b,c be positive integers such that b/a is an integer. If a,b,c are  in geometric  progression and the arithmetic mean of a,b,c is b+2, then the value of   $\frac{a^{b}+a-14}{a+1}$  is 

Answer:

Explanation:

Plan 

(i)  a,b,c are in G.P ,

 then they can be taken as a, ar, ar² where r,(r ≠ 0)  is the common ratio

 (ii) Arithmetic mean of x₁,x₂,............x_n

= $\frac{x_{1}+x_{2}+....+x_{n}}{n}$

  Let a,b,c are a,ar, ar²  , where rε N

Also,    $\frac{a+b+c}{3}=b+2$

$\Rightarrow$              a+ar+ar²= 3(ar)+6

$\Rightarrow$           ar²-2ar+a =6

$\Rightarrow$                    (r-1)² =  $\frac{6}{a}$

 $\because$             6/a must be perfect square   a ε N

  $\therefore$   a can be 6 only.

 $\Rightarrow$          $ r-1=\pm1\Rightarrow r=2$

 and            $\frac{a^{b}+a-14}{a+1}=\frac{36+6-14}{7}=4$

Let a,b and c be three  non-coplanar unit vectors such that  the  angle betwwen every  pair of them is   $\pi$ /3, if a x b +b x c  =pa+qb+rc, where p,q, r are scalars , then  the value of   $\frac{p^{2}+2q^{2}+r^{2}}{q^{2}}$ is

Answer:

Explanation:

Plan

(i)a.b= |a|.|b| cosθ

(ii)   [a b c]²=  $\begin{bmatrix}a.a & a.b &a.c\\b.a & b.b &b.c \\ c.a &c.b&c.c\end{bmatrix}$

a.a=|a|² =1,similarly b.b=c.c=1

a.b=|a||b|cos $\left(\frac{\pi}{3}\right)=\left(\frac{1}{2}\right)$similarly

b.c=c.a=$\frac{1}{2}$

  [a b c]²= $\begin{bmatrix}a.a & a.b &a.c\\b.a & b.b &b.c \\ c.a &c.b&c.c\end{bmatrix}$

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

         $\therefore$             $[ a.b.c]=\frac{1}{\sqrt{2}}$         .........(i)

As , given a x b+b x c=pa+qb+rc

  Take dot  product  with a

a.(axb)+a.(bxc)  = p a²+q b²+r c.a

 $\Rightarrow$                                  $  0+\frac{1}{\sqrt{2}}=p+\frac{q}{2}+\frac{r}{2}$

$\therefore$                   [a a b]=0

Now, take dot product  with b and c

           $0=\frac{p}{2}+q+\frac{r}{2}$         ......(iii)

 and            $\frac{1}{\sqrt{2}}=\frac{p}{2}+\frac{q}{2}+r$        .......(iv)

 On subtracting Eq(ii) from Eq. (iv), we get

  $\frac{p}{2}-\frac{r}{2}=0\Rightarrow p=r$

$\Rightarrow$ p+r=0   [By Eq.(iii)]

$\therefore$   $\frac{p^{2}+2q^{2}+r^{2}}{q^{2}}$

                                                   $=\frac{p^{2}+2p^{2}+p^{2}}{p^{2}}=4$

The value of   $\int_{0}^{1}4x^{3} \left\{\frac{d^{2}}{dx^{2}}(1-x^{2})^{5}\right\} dx $  is

Answer:

Explanation:

Plan Intergration by parts

$\int_{}^{} f(x)g(x)dx=f(x)\int_{}^{} g(x) dx-\int_{}^{}\left(\frac{d}{dx} [f(x)]\int_{}^{} g(x) dx\right) dx$

 Given   $I= \int_{0}^{1}  4x^{3}\frac{d^{2}}{dx^{2}}(1-x^{2})^{5}dx$

   $=\left[4x^{3}\frac{d}{dx}(1-x^{2})^{5}\right]^{1}_{0}-\int_{0}^{1}12x^{2} \frac{d}{dx}(1-x^{2})^{5}dx$

    $=\left[4x^{3}\times 5(1-x^{2})^{4}(-2x)\right]^{1}_{0}$   

                                      $-12 \left[(x^{2}(1-x^{2})^{5}\right]_0^1-\int_{0}^{1}2x(1-x^{2})^{5} dx]$

                                    =0-0-12(0-0)

                                          +   $12\int_{0}^{1} 2x(1-x^{2})^{5}dx$

                            = $12\times\left[-\frac{(1-x^{2})^{6}}{6}\right]_0^1$

                      =   $12\left[0+\frac{1}{6}\right]=2$

Let   $n_{1}$<$n_{2}$ <$n_{3}$<$n_{4}$  < $n_{5}$  be positive integers such that $n_{1}$+$n_{2}$+$n_{3}$+$n_{4}$+$n_{5}$=20. The numbers of such  distinct arrangements ( $n_{1}$,$n_{2}$,$n_{3}$,$n_{4}$,$n_{5}$)    is 

Answer:

Explanation:

Plan, Reducing the equation to a newer question, where sum of variables is less. Thus , find ing the number of arrangements  becomes easier

 As, $n_{1}\geq 1$ , $n_{2}\geq 2$, $n_{3}\geq 3$, $n_{4} \geq 4$,$n_{5}\geq 5$ 

 Let  n₁ -1= x₁ $\geq$ 0,  n₂-2 =x₂  $\geq$ 0, ....... n₅-5 =x₅  $\geq$ 0

New equation will be

x₁+1+x₂+2+.........x₅+5=20

    x₁+x₂+x₃+x₄+x₅

                    =20-15=5

  Now,    x₁  $\leq$  x₂ $\leq$ x₃   $\leq$ x₄   $\leq$ x₅

 So, 7 possible  cases will be there

For a point P in the plane, let d₁(P)   and d₂(P)  be the distances of the point P from the lines x-y=0 and x+y=0, respectively. The area of the region R consisting of all points P lying  in the first quadrant  of the plane and satisfying   $2\leq d_{1}(P)+d_{2}(P)\leq 4.$   is   

Answer:

Explanation:

Plan Distance of a point (x₁,y₁) from  ax+bx+c=0 is given by

 $|\frac{ax_{1}+by_{1}+c}{\sqrt{a^{2}+b^{2}}}|$

Let P(x,y)  is the point in first quadrant.

 Now,   $2\leq |\frac{x-y}{\sqrt{2}}|+|\frac{x+y}{\sqrt{2}}|\leq 4$

 $2\sqrt{2}\leq |x-y|+|x+y|\leq4\sqrt{2}$

 Case I    $x \geq y$

    $2\sqrt{2}\leq (x-y)+(x+y)\leq4\sqrt{2}$

    $\Rightarrow$     $ x \in (\sqrt{2},2\sqrt{2})$

Case II     x<y

$2\sqrt{2}\leq y-x+(x+y)\leq4\sqrt{2}$

  $y \in (\sqrt{2},2\sqrt{2})$

 $\Rightarrow$       $A=(2\sqrt{2})^{2}-(\sqrt{2})^{2}=6$

• Advanced 2014 - Physics 2014
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• Advanced 2014 - Mathematics 2014