JEE 2013 :: Advanced 2013 :: Mathematics 2013

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

Let f:[0,1]→ R (the set of all real numbers) be a function .Suppose  the function f is twice differentiable

 f(0)=f(1) =0 and satisfies

 $f '' (x)-2 f'(x)+ f(x) \geq e^{x}, x \in [0,1]$.

 If the function  e^-x f(x)  assumes its minimum in the interval [0,1] a  $x=\frac{1}{4}$, which of the following is true ?

Answer:

Explanation:

 Concept involved It is based on the concept of converting into total differential equation (i.e, completing the equation into differential ). So , as to check the function to be increasing or decreasing

 Here,  $\phi'(x)<0,x\in (0,\frac{1}{4})$

 and   $\phi'(x)>0,x\in (\frac{1}{4},1)$

  $\Rightarrow $    $e^{-x}f'(x) -e^{-x}f(x)<0, x\in (0,\frac{1}{4})$

$\Rightarrow$     $ f'(x) <f(x) , 0 <x<\frac{1}{4}$

Let f:[0,1]→ R (the set of all real numbers) be a function .Suppose  the function f is twice differentiable

 f(0)=f(1) =0 and satisfies

 $f '' (x)-2 f'(x)+ f(x) \geq e^{x}, x \in [0,1]$

 Which of the following is true for 0<x <1 

Answer:

Explanation:

 Concept involved It is based on the concept of converting into total differential equation (i.e, completing the equation into differential ). So , as to check the function to be increasing or decreasing

 Here   $f''(x) -2 f'(x) + f(x) \geq e^{x}$

$\Rightarrow $    $f''(x)e^{-x} - f'(x)e^{-x} - f'(x)e^{-x}+f(x)e^{-x} \geq0$

$\Rightarrow \frac{d}{dx}(f'(x)e^{-x})-\frac{d}{dx}(f(x)e^{-x})\geq1$

  $\Rightarrow \frac{d}{dx}(f'(x)e^{-x}-f(x)e^{-x})\geq1$

$\Rightarrow \frac{d^{2}}{dx^{2}}(e^{-x}f(x))\geq1$ for all x $\in$ [0,1]

 $\therefore$    $\phi(x) =e^{-x}f(x) $  is concave $n\phi$

 f(0)=f(1)=0

 $\Rightarrow$    $\phi(0)=0=\phi(1)$

 $\Rightarrow  $     $\phi(x)<0$

$\Rightarrow $       $e^{-x}f(x)<0$
$\therefore$     $ f(x) <0$

 The function f(x) = 2|x|+|x+2|-||x+2|-2|x||  has a local minimum or a local maximum at x is equal to 

Answer:

Explanation:

Concept involved

 We know, 

$|x|=\begin{cases}x & x \geq 0\\-x, & x < 0\end{cases}$
$\Rightarrow$      $ |x-a|=\begin{cases}x-a & x \geq0\\-(x-a) & x < a\end{cases}$

 and for non-differentiable continuous function the maximum or minimum Can be checked with graph as

Here  f(x) =2|x|+|x+2| -||x+2|-2|x||

$=\begin{cases}-2x-(x+2)+(x+2), & when x\leq-2\\-2x+x+2+3x+2, & when -2<x\leq-2/3\\ -4x&when- \frac{2}{3}<x\leq0\\ 4x,&when 0<x\leq2\\ 2x+4,&when x>2\end{cases}$

 $=\begin{cases}-2x-4, &  x\leq-2\\2x+4, &  -2<x\leq-2/3\\ -4x&- \frac{2}{3}<x\leq0\\ 4x,& 0<x\leq2\\ 2x+4,& x>2\end{cases}$

 Graph for y=f(x) is shown as

Let ω be a complex cube root of unity with ω ≠ 1 and P=[p_ij] be a n x n matrix with p_ij  =ω ^i+j. Then, P²≠ 0, when n is equal to 

Answer:

Explanation:

Here , P=[pij]_nxn with p_ij=w^i+j

 $\therefore$ when n=1

 $P=[p_{ij}]_{1\times1}=[w^{2}]\Rightarrow p^{2}=[w^{4}]\neq0$

 $\therefore$ when n=2

$P=[p_{ij]_{2\times2}}=\begin{bmatrix}p_{11} & p_{12} \\p_{21} & p_{22} \end{bmatrix}=\begin{bmatrix}w^{2} & w^{3} \\w^{3} & w^{4} \end{bmatrix}=\begin{bmatrix}w^{2} & 1 \\1 & w \end{bmatrix}$

$p^{2}=\begin{bmatrix}w^{2} & 1 \\1 & w \end{bmatrix}\begin{bmatrix}w^{2} & 1 \\1 & w \end{bmatrix}$

$p^{2}=\begin{bmatrix}w^{4}+1 & w^{2}+w \\w^{2}+w & 1+w^{2} \end{bmatrix}\neq0$

when n=3

$p=[p_{ij}]_{3\times3}=\begin{bmatrix}w^{2} & w^{3}&w^{4} \\w^{3} & w^{4}&w^{5}\\ w^{4}&w^{5}&w^{6} \end{bmatrix}$

    $=\begin{bmatrix}w^{2} & 1&w \\1& w&w^{2}\\ w&w^{2}&1 \end{bmatrix}$

 $p^{2}=\begin{bmatrix}w^{2} & 1&w \\1& w&w^{2}\\ w&w^{2}&1 \end{bmatrix}\begin{bmatrix}w^{2} & 1&w \\1& w&w^{2}\\ w&w^{2}&1 \end{bmatrix}$

$=\begin{bmatrix}0 & 0&0 \\0& 0&0\\ 0&0&0 \end{bmatrix}=0$

$\therefore$  $p^{2}$=0, when n  is mutilple of 3

$p^{2}\neq0$  , when n is not a multiple of 3

 $\Rightarrow$ n=57 is not possible

 $\therefore$   n=55,58,56 is possible

If   $3^{x}=4^{x-1}$    , then x is equal to

Answer:

Explanation:

$3^{x}=4^{x-1}$  , taking log3on both sides

$\Rightarrow$     $ x\log_3^3=(x-1)\log_3^4$

$\Rightarrow  x=2 \log_3^2.x-\log_3^4$

$\Rightarrow $     $x(1-2\log_3^2)=-2\log_3^2$

$\Rightarrow$    $ x=\frac{2\log_3^2}{2\log_3^2-1}$

$\Rightarrow x=\frac{1}{1-\frac{1}{2\log_3^2}}=\frac{1}{1-\frac{1}{\log_3^4}}=\frac{1}{1-\log_3^4}$

$\Rightarrow x=\frac{2}{2-\log_2^3}$

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