JEE 2014 :: Advanced 2014 :: Mathematics 2014

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

Let x,y and z be three vectors each of magnitude  $\sqrt{2}$ and the angle between each pair of them is   $\frac{\pi}{3}$. If a is a   non-zero  vector perpendicular to x and y $\times$ z  and b is a non-zero vector perpendicular to y and z $\times$ x, then

Answer:

Explanation:

Plan  

(i) Dot product let a and be be two non-zero vectors inclined at an angle θ . Then a.b=|a| |b| cos θ

(ii) The direction of a x b  is perpendicular to the plane of a and b

 (iii) Vector triple product

   a x (b x c)= (a.c) b-  (a.b) c

 Given  that  |x|=|y|=|z|=   $\sqrt{2}$

 and angle between each pair  is   $\frac{\pi}{3}$

   $\therefore$       $  x.y=|x||y|\cos\frac{\pi}{3}$

 $=(\sqrt{2})(\sqrt{2})(\frac{1}{2})=1$

Similarly, y.z=z.x=1

also, x . x=|x|² =2

similarly,  y.y=z.z= 2

now, $a \bot $ x  and y $\times$ x 

Let,                    a= $\lambda $ [x $\times$ (y $\times$ z)]

                                   a= $\lambda $ [(x.z)y-(x.y)z]

     a=$\lambda $ (y-z)

                              (as x.z =x.y=1)

       a.y=  $\lambda $ (y-z).y

  a.y= $\lambda $(y.y-y.z)=$\lambda $ (2-1)

                              ( as y.y=2=y.z=1)

     a.y= $\lambda $

    $\Rightarrow$                           a= (a.y) (y-z)         .......(i)

 Also , $b \bot $ y and z $\times$ x

                        b= $\mu$  { y x (z x x)]

                              = $\mu$ [(y .x)z-(y.z)x]

                                      = $\mu$   (z-x)

                                                             [as    y.x=y.z=1)

          b.z= $\mu$ (z-x).z

               = $\mu$  (z.z -x.z]

                 = $\mu$ (2-1)=  $\mu$

 $\Rightarrow$               b= (b.z) (z-x)           ...........(ii)

  Now using Eqs.( i) and (ii), we get

               a.b= (a.y)(y-z).(b.z)(z-x)

                     =  (a.y)(b.z)(y.z-y.x. -z.z+z.x)

                     =(a,y)(b.z)(1-1-2+1)

                        a.b= -(a.y)(b.z)

For every pair of continuous function f,g:[0,1] → R such that max { f(x):xε [0,1]}=max {g(x): x ε [0,1]}.

The correct statement(s) is (are)

Answer:

Explanation:

 Plan if a continuous  function has values of opposite sign inside an interval, then it has a root in that  interval

 f,g:[0,1] → R

 We take two cases.

 Let f and g attain their common maximum value at p

   $\Rightarrow$ f(p)=g(p)

 where $p\in [0,1]$

 Let  f and g attain their common maximum value at different points

$\Rightarrow$            f(a)=M   and g(b)= M

$\Rightarrow$      f(a)-g(a)  >0

 and f(b)-g(b) <0

$\Rightarrow$   f(c)-g(c)=0 for some   $c \in [0,1]$ as f and g are continuous functions.

$\Rightarrow$   f(c)-g(c)=0 for some    $c \in [0,1]$ for all cases.              ........(i)

 Option

 (a)  $\Rightarrow$   f²(c)-g²(c)+3[f(c)-g(c)]=0

  which is true from Eq.(i).

 Option (d)  $\Rightarrow$   f²(c)-g² c)=0 which is true from Eq.(i)

 Now, if we take 

f(x)=1 and g(x)=1,  $\forall x \in [0,1]$

 Option (b) and (c) does not hold.

 Hence, option (a) and (d)  are correct

Let M be a 2x2 symmetric matrix with integer entries. Then, M is invertible, if

Answer:

Explanation:

Plan  A square matrix M  is invertible if det(M)   or |M|  $\neq$ 0,

  Let   $M=\begin{bmatrix}a & b \\b & c \end{bmatrix}$

 (a) Given that    $\begin{bmatrix}a  \\b  \end{bmatrix}=\begin{bmatrix}b  \\c  \end{bmatrix}$

 $\Rightarrow$         a=b=c= $\alpha$              (let)

$\Rightarrow$             M= $\begin{bmatrix}\alpha & \alpha \\\alpha & \alpha \end{bmatrix}$

$\Rightarrow$                       |M|=0

$\Rightarrow$                   Mis non-invertible

(b) Given that [b c]=[a b]

$\Rightarrow$                     a=b=c =  $\alpha$              (let)

                   Again  |M|=0

$\Rightarrow$       M is non-invertible

 (c)    As given   $M=\begin{bmatrix}a & 0 \\0 & c \end{bmatrix}$

  $\Rightarrow$                $|M|=ac\neq0$

                                      (   $\because$   a and c are non-zero)

$\Rightarrow$                 Mis invertiable

(d)        $M=\begin{bmatrix}a & b \\b & c \end{bmatrix}\Rightarrow|M|=ac-b^{2}\neq 0$

  $\because$ ac is not equal to square of an integer.

 $\because$   M is invertiable

Let   $f:[a,b]\rightarrow[1,\infty]$   be a continuous function and g:R → R be defined as $\begin{cases}0 & if & x <a &\\\int_{a}^{x}f(t)dt,  &if& a\leq x\leq b & \\\int_{a}^{b}f(t)dt  &if & x>b\end{cases}$Then,

Answer:

Explanation:

Plan  A function f(x) is continuous at x=a, if 

$\lim_{x \rightarrow a^{-}}$ f(x)= $\lim_{x \rightarrow a^{+}}$ f(x=f(a))

  $\lim_{x \rightarrow a^{-}}\frac{f(x)-f(a)}{x-a}=\lim_{x \rightarrow a^{+}}\frac{f(x)-f(a)}{x-a}$

 i.e, f '(a^-)=f '(a⁺)

 Given that   $f:[a,b]\rightarrow[1,\infty]$

   and     $\begin{cases}0 & if & x <a &\\\int_{a}^{x}f(t)dt,  &if& a\leq x\leq b & \\\int_{a}^{b}f(t)dt  &if & x>b\end{cases}$

 Now, g(a^-)=0=g (a⁺)=g(a)

[as     $g(a^{+})\lim_{x \rightarrow a^{+}}\int_{a}^{x} f(I) dt=0$

  and  $g(a)=\int_{a}^{a} f(t) dt=0$ ]

$g(b^{-})=g(b^{+})=g(b)=\int_{a}^{b}  f(t) dt\Rightarrow $

   is continuous   , $\forall x\epsilon R$

 Now,     $g'(x)=\begin{cases}0 &  & x <a &\\f(x),  && a< x< b & \\0 & & x>b\end{cases}$

 g'(a^-)=0 but g'(a⁺)= $f(a)\geq1$ 

 [  $\because $  Range of f(x)  is   $[1,\infty), \forall x\in[a,b]]$

$\Rightarrow$  g is non-differentiable at x=a

 and g'(b⁺ )=0

 but   $g'(b^{-})=f(b)\geq 1$

$\Rightarrow$  g is not differentiable at x=b

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