JEE 2013 :: Advanced 2013 :: Mathematics 2013

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• Advanced 2013 - Mathematics 2013

For 3x3 matrices M and N  , which of the following statement(s)  is (are) not correct?

Answer:

Explanation:

(a)   $(N^{T}MN )^{T}=N^{T}M^{T}(N^{T})^{T}$

  =  $N^{T}M^{T}N $  , is symmetric  is M is symmetric and skew symmetric is M is skew-symmetric

(b)   $(MN-NM)^{T}=(MN)^{T}-(NM)^{T}$

  =NM-MN=-(MN-NM)

 $\therefore$  skew-symmetric , when M and N  are symmetric

(c)    $(MN)^{T}=N^{T}M^{T}=NM\neq MN$

 $\therefore$ not correct

 (d)   (adj MN)= (adj N)-(adj M)

 $\therefore$ not correct

Let  f(x)= x sin  $\pi$ x, x>0 . Then for all natural numbers n, f'(x) vanishes at 

Answer:

Explanation:

Concept involved

 This question is based on the concept of transcendal  equation and sholud be solved with the help of graph

f(x)= x sin  $\pi$ x, x>0

$f'(x)= sin\pi x+\pi xcos \pi x=0$

$\Rightarrow$   $ tan \pi x=-\pi x$

 let f(x) = tan $\pi$ x  and g(x)= - $\pi$ x

 which could be  plotted as,

 clearly , f(x) and g(x)  intersect when

 $\frac{1}{2}< x<1$

  or            $\frac{3}{2}< x<2$

 or     $\frac{5}{2}< x<3$

 $\therefore$  a unique point in   $(n+\frac{1}{2},n+1)$

 or  (n,n+1)

A line l passing through the origin is perpendicular to the lines

$l_{1}:(3+t)\hat{i}+(-1+2t)\hat{j}+(4+2t)\hat{k},-\infty<t<\infty$

$l_{2}:(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k},-\infty<t<\infty$

Then, the coordinate (s)  of the point(s)  on l₂ at a distance of   $\sqrt{17}$ from the point of intersection of l and l₁ is (are)

Answer:

Explanation:

Concept involved

Equation of straight line is.

  $l: \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$

 Since, l is perpendicular to l₁ and l₂

$\therefore$  its DR's are cross product of l₁  and l₂

Now, to find a point on l₂ whose  distance is given, assume a point and find its distance to obtain point

 Let    $l: \frac{x-0}{a}=\frac{y-0}{b}=\frac{z-0}{c}$

 which is perpendicular to 

$l_{1}:(3\hat{i}-\hat{j}+4\hat{k})+t(\hat{i}+2\hat{j}+2\hat{k})$

$l_{2}:(3\hat{i}+3\hat{j}+2\hat{k})+s(2\hat{i}+2\hat{j}+\hat{k})$

$\therefore$     DR's of l, is 

               $\begin{bmatrix}\hat{i} & \hat{j} &\hat{k}\\1 & 2& 2 \\2 &2&1 \end{bmatrix}$

$=-2\hat{i}+3\hat{j}-2 \hat{k}$

$\therefore$    $l:\frac{x}{-2}=\frac{y}{3}=\frac{z}{-2}=k_{1},k_{2}$

 Now, A(-2k₁,3k₁,-2k₁)

 and   B (-2k₂,3k₂,-2k₂)

 since , A lies on l₁

 $\therefore$   $(-2k_{1})\hat{i}+(3k_{1})\hat{j}-2k_{1})\hat{k}$

                    = $(3+t)\hat{i}+(-1+2t)\hat{j}+(4+2t)\hat{k}$

   $\Rightarrow$    $  3+t=-2k_{1},-1+2t=3k_{1}$

   4+2t=-2k₁

 $\therefore$   k₁=-1

$\Rightarrow  $    A(2,-3,2)

Let any point on l₂ (3+2s,3+2s,2+5)

 $\Rightarrow  $            $\sqrt{(2-3+2s)^{2}+(-3-3-2s)^{2}+(2+2+s)^{2}}=\sqrt{17}$

$\Rightarrow  $   $9s^{2}+28s+37=17$

$\Rightarrow  $   $9s^{2}+28s+20=0$

$\Rightarrow  $  $9s^{2}+18s+10s+20=0$

$\therefore$ (9s+10)(s+2) =0

$\Rightarrow  $     s=-2,-10/9

Hence (-1,-1,0) and   $(\frac{7}{9},\frac{7}{9},\frac{8}{9})$ are required points 

Let   $S_{n}=\sum_2^{4n}(-1)^{\frac{k(k+1)}{2}}k^{2}$.Then , S_n can take value (s)

Answer:

Explanation:

Concept involved

It is to convert into differences and using sum of n terms of AP

 i.e,   $S_{n}=\frac{n}{2}[2a+(n-1)d]$

    $S_{n}=\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^{2}$

 $=-(1)^{2}-2^{2}+3^{2}+4^{2}-5^{2}-6^{2}+7^{2}+8^{2}+....$

$=(3^{2}-1^{2})+(4^{2}-2^{2})+(7^{2}-5^{2})+(8^{2}-6^{2})+....$

= 2{(4+6+12+...)  +(6+14+22+...)}

    n terms                    n terms

$=2\left[\frac{n}{2}\left\{2\times 4+(n+1)8\right\}+\frac{n}{2}\left\{ 2\times 6+(n-1)8\right\}\right]$

 =2[n(4+4n-4)+n(6+4n-4)]

=2[4n²+4n²+2n] =4n(n+1)

Here, 1056=32 x33, 1088=32 x34, 1120=32 x35, 1332=36x37

 1056  and 1332 are possible  answers

A rectangular sheet of the fixed perimeter with sides having their lengths in the ratio 8:15 is converted into an open rectangular box by folding after removing squares of the equal area from all four corners.If the total area of removed squares is 100, the resulting box has maximum volume.The lengths of the sides of the rectangular sheet are 

Answer:

Explanation:

Concept involved

 The problem is based on the concept to maximize the volume of a cuboid

 i.e,  to form a function of volume , say f(x) find  f'(x)  and f"(x).

 Put f'(x) =0 and check f'(x)  to be +ve  or -ve for minimum and maximum, respectively

Here, l=15x-2a, b=8x-2a

 h=a

$\therefore$ Volume =(8x-2a)(15x-2a)a

 V=2a(4x-a)(15x-2a)....(i)

$\frac{dv}{da}=6a^{2}-46ax+60x^{2}$

$\frac{d^{2}v}{da^{2}}=12a-46x$

  Here,   $(\frac{dv}{da})=0$

$\Rightarrow $   $ 6x^{2}-23x+15=0$

 at   $a=5\Rightarrow x=3,\frac{5}{6}$

$\left(\frac{d^{2}v}{da^{2}}\right)=2(30-23x)$

at    x=3, $\left(\frac{d^{2}v}{da^{2}}\right)$ =2(30-69)<0

$\therefore$  Maximum when x=3,

 also at              $x=\frac{5}{6}\Rightarrow \left(\frac{d^{2}v}{da^{2}}\right)>0$

 $\therefore$  at   x=5/6 , volume is minimum

 Thus, sides are 8x=24  and 15 x=45

• Advanced 2013 - Physics 2013
• Advanced 2013 - Chemistry 2013
• Advanced 2013 - Mathematics 2013