JEE 2012 :: General Questions :: Mathematics

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let $f(x)=(1-x)^{2}\sin^{2} x+x^{2}$ for all x $ \epsilon R $ and 

let $g(x)=\int_{1}^{x} \left(\frac{2(t-1)}{t+1}-ln t\right) f(t) dt,$ for all x $ \epsilon (1,\infty)$ 

consider the statements

P:  There exists some $x \epsilon R$ such that 

  $f(x)+2x=2(1+x^{2})$

Q:There exists some  $x \epsilon R$ such that 

   2f(x)+1=2x(1+x)

 Then 

Answer:

Explanation:

 Concept involved use of Newton leibnits formula

 i.e, $\frac{d}{dx}\left\{\int_{g(x)}^{f(x)} \phi(t) dt\right\}=\phi (f(x)).f'(x)-\phi (g(x0).g'(x)$

  Sol. Here , $f(x)(1-x)^{2} \sin^{2}x+x^{2},$ for all x

     $g(x)=\int_{1}^{x} \left(\frac{2(t-1)}{t+1}-\log(t)\right)f(t)dt$

$\forall x \epsilon (1, \infty)$

Here, $f(x)+2x=(1-x)^{2}.\sin^{2} x+x^{2}+2x$.............(i)

 where $P:f(x)+2x=2(1+x)^{2}$ ...............(ii)

 $\therefore$  $2(1+x^{2})=(1-x)^{2}\sin^{2}x+x^{2}+2x$

$\Rightarrow$  $(1-x)^{2} \sin^{2}x=x^{2}-2x+2$

$\Rightarrow$ $(1-x)^{2}\sin^{2}x=(1-x)^{2}+1$

$\Rightarrow$   $(1-x)^{2} \cos^{2}x=-1$

 which is never possible

$\therefore$  p is false

 Again h(x)=2f(x)+1-2x (1+x)

            h(1)=2f(1)+1-4=-3

 as  h(0) h(1)<0

 $\Rightarrow$  h(x) must have a solution

$\therefore$ Q is true

let $f(x)=(1-x)^{2}\sin^{2} x+x^{2}$ for all x $ \epsilon R $ and 

let $g(x)=\int_{1}^{x} \left(\frac{2(t-1)}{t+1}-ln t\right) f(t) dt,$ for all x $ \epsilon (1,\infty)$ 

which of the following is true?

Answer:

Explanation:

 Concept involved use of Newton leibnits formula

 i.e, $\frac{d}{dx}\left\{\int_{g(x)}^{f(x)} \phi(t) dt\right\}=\phi (f(x)).f'(x)-\phi (g(x)).g'(x)$

  Sol. Here , $f(x)(1-x)^{2} \sin^{2}x+x^{2},$ for all x

     $g(x)=\int_{1}^{x} \left(\frac{2(t-1)}{t+1}-\log(t)\right)f(t)dt$

$\forall x \epsilon (1, \infty)$

Here, 

$f(x)=(1-x)^{2}\sin^{2} x+x^{2}$ for all x 

 and  $g(x)=\int_{1}^{x} \left(\frac{2(t-1)}{t+1}-\log t \right) f(t) dt,$ 

 $\Rightarrow$   $g'(x)=\left\{\frac{2(x-1)}{x+1}-\log x\right\}f(x) (+ve)....(i)$

 for   g'(x) to be increasing or decreasing 

Let $\phi(x)=\frac{2(x-1)}{(x+1)}-\log x$

 $ \phi'(x)= \frac{4}{(x+1)^{2}}-\frac{1}{x}=\frac{-(x-1)^{2}}{x(x+1)^{2}}$

 $\phi '(x) <0,$ for x >1

 $\Rightarrow$  $\phi (x) < \phi(1) \Rightarrow \phi (x) <0$......(ii)

 For  Eqs.(i) and (ii) , we get

 $g'(x) <0$ for $ x \epsilon (1 ,\infty)$

 $\therefore$ g(x) is decreasing for $ x \epsilon (1 ,\infty)$

Let $\alpha(a) $ and $\beta(a)$ be the roots of the equation

$(\sqrt[3]{1+a}-1)x^{2}-(\sqrt{1+a}-1)x$

+$(\sqrt[6]{1+a}-1)=0$ where a>-1,

 Then , $\lim_{a \rightarrow {0^{+}}}\alpha(a)$ and $\lim_{a \rightarrow {0^{+}}}\beta(a)$ are 

Answer:

Explanation:

Concept Involved To make the quadratic into the simple form we should eliminate radical sign 

Description of Situation As forgiven equation when  $a\rightarrow 0$  equation reduces to identity in x.

i.e, $ax^{2}+bx+c=0$ for all $x \epsilon  R $ or a=b=c $\rightarrow 0$

Thus, first we should make above equation indepenrlent from coefficients as 0.

 Sol. Let $(a+1=t^{6}$ Thus, when $a \rightarrow 0,t \rightarrow 1$

 $\therefore$  $(t^{2}-1)x^{2}+(t^{3}-1)x+(t-1)=0$

$\Rightarrow$  $(t-1){(t+1)x^{2}+(t^{2}+t+1)x+1}=0$

 As $t \rightarrow 1$ 

                     $2x^{2}+3x+1=0$

 $2x^{2}+2x+x+1=0$

$\Rightarrow$   $(2x+1) (x+1)=0$

 Thus, x=-1, -1/2

 or  $\lim_{a \rightarrow {0^{+}}}\alpha(a)=-\frac{1}{2}$

 and   $\lim_{a \rightarrow {0^{+}}}\beta(a)=-1$

Let $a_{1},a_{2},a_{3}.......$ be in a harmonic progression with  $a_{1}=5$ and $a_{20}=25$. The least  positive integer n for which $a_{n} <0$ is 

Answer:

Explanation:

 Concept involved $n^{th}$ term of HP 

  $t_{n}=\frac{1}{a+(n-1)d}$

 Sol. Here , $a_{1}=5, a_{20}=25$ for HP

 $\therefore$   $\frac{1}{a}=5$ and $\frac{1}{a+19d}=25$ 

$\Rightarrow$ $\frac{1}{5}+19d=\frac{1}{25}$

$\Rightarrow$   $19d=\frac{1}{25}-\frac{1}{5}=-\frac{4}{25}$

 $\therefore$   $d= \frac{-4}{19 \times 25}$

 Since, $a_{n}$ <0

 $\therefore$  $\frac{1}{5}+(n-1)d <0$

$\Rightarrow$  $\frac{1}{5}-\frac{4}{19 \times 25}(n-1) <0$

 $\Rightarrow$  $(n-1) > \frac{95}{4}$

 $\Rightarrow$  $n > 1+\frac{95}{4}$ or $n > 24.75$

 least positive  value of n=25

If P is a 3x3 matrix such that $P^{T} =2P+I$, where $p^{T}$  is the transpose of P and I is the 3 x 3 identity matrix, then there exists a column matrix

$X=\begin{bmatrix}x \\y \\z \end{bmatrix}\neq\begin{bmatrix}0 \\0 \\0 \end{bmatrix}$ such that

Answer:

Explanation:

 Given, $P^{T}=2P+I$ ...........(i)

  $\therefore$  $(P^{T})^{T}=(2P+I)^{T}=2P^{T}+I$

 $\Rightarrow$    $P=2P^{T}+I$

$\Rightarrow$   $P=2(2P+I)+I$

$\Rightarrow$   $P=4P+3I$ or $3P=-3I$

$\Rightarrow$  PX=-IX

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