JEE 2017 :: Main 2017 :: Mathematics 2017

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For a positive integer n, if the quadratic equation, $x(x+1)+(x+1)(x+2)+.....+(x+\overline{n-1}) (x+n)=10n$  has two consecutive integral solutions, then n is equal to

Answer:

Explanation:

Given quadratic equations is $x(x+1)+(x+1)(x+2)+.....+(x+\overline{n-1}) (x+n)=10n$

       $\Rightarrow$    $( x^{2}+x^{2}+....+x^{2})+[1+3+5+...+(2n-1)]x $

                                                       $   +[1.2+2.3+...+(n-1)n]=10n$

$\Rightarrow$   $nx^{2}+n^{2}x+\frac{n(n^{2}-1)}{3}-10n=0$

$\Rightarrow$  $x^{2}+nx+\frac{n^{2}-1}{3}-10=0$

$\Rightarrow$    3x²+3nx+n²-31=0

   Let  $\alpha$ and β be the roots

   Since,  $\alpha$ and β are consecutive

$\therefore$        | $\alpha$-β|=1

$\Rightarrow$   ( $\alpha$ -β)²=1

Again   $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4\alpha\beta$

$\Rightarrow$  $1=\left(\frac{-3n}{3}\right)^{2}-4\left(\frac{n^{2}-31}{3}\right)$

$\Rightarrow$   $1= n^{2}-\frac{4}{3}(n^{2}-31)$

$\Rightarrow$    3= $3n^{2}-4n^{2}+124$

$\Rightarrow$   n²=121

 $\Rightarrow$   n=± 11

$\therefore$    n=11              [ n>0]

The radius of a circle having minimum area, which touches the curve y=4-x² and the line y=|x| , is

Answer:

Explanation:

Let the radius circle with least area be r

                  Then, then coordinates of centre= (0,4,-r)

Since, circle touches the line y=x in first quadrant

   $\therefore$    

                        $\mid\frac{0-(4-r)}{\sqrt{2}}\mid=r$

$\Rightarrow$    $r-4=\pm r\sqrt{2}$

$\Rightarrow$         $r=\frac{4}{\sqrt{2}+1}$   or   $\frac{4}{1-\sqrt{2}}$

But                    $r\neq\frac{4}{1-\sqrt{2}}$        $[\because\frac{4}{1-\sqrt{2}}<0]$

  $\therefore$      $r=\frac{4}{\sqrt{2}+1}=4(\sqrt{2}-1)$

The value of $(^{21}C_{1}-^{10}C_{1})+(^{21}C_{2}-^{10}C_{2})+(^{21}C_{3}-^{10}C_{3})+.....+(^{21}C_{10}-^{10}C_{10})$ is

Answer:

Explanation:

$(^{21}C_{1}-^{10}C_{1})+(^{21}C_{2}-^{10}C_{2})+(^{21}C_{3}-^{10}C_{3})+.....+(^{21}C_{10}-^{10}C_{10})$

    =   $(^{21}C_{1}+^{21}C_{2}+....+^{21}C_{10})-(^{10}C_{1}+^{10}C_{2}+.......+^{10}C_{10})$

= $\frac{1}{2}(^{21}C_{1}+^{21}C_{2}+.....+^{21}C_{20})-(2^{10}-1)$

= $\frac{1}{2}(^{21}C_{1}+^{21}C_{2}+.....+^{21}C_{21}-1)-(2^{10}-1)$

=  $\frac{1}{2}(2^{21}-2)=2^{20}-1-2^{10}+1-(2^{10}-1)$

   = $2^{20}-2^{10}$

The normal to the curve y(x-2) (x-3)=x+6  at the point, where the curve intersects the Y-axis passes through the point

Answer:

Explanation:

Given curve is

    y(x-2)(x-3)=x+6   .........(i)

put x=0 in Eq.(i) , we get

$y(-2)(-3)=6\Rightarrow y=1$

   So, point of intersection is (0,1)

  Now,   $y= \frac{x+6}{(x-2)(x-3)}$

$\Rightarrow$ $\frac{\text{d}y}{\text{d}x}$=  $\frac{1(x-2)(x-3)-(x+6)(x-3+x-2)}{(x-2)^{2}(x-3)^{2}}$

$=(\frac{\text{d}y}{\text{d}x})_{(0,1)}=\frac{6+30}{4\times9}=\frac{36}{36}=1$

  $\therefore$     Equation of normal at (0,1) is given by

                             $y-1   =\frac{-1}{1}(x-0)$

  $\Rightarrow$     x+y-1=0

which passes through the point $(\frac{1}{2},\frac{1}{2})$

$\lim_{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2x)^{3}}$ equals

Answer:

Explanation:

$\lim_{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2x)^{3}}$

      =  $\lim_{x \rightarrow \frac{\pi}{2}}\frac{1}{8} \frac{\cos x(1-\sin x)}{\sin x(\frac{\pi}{2}-x)^{3}}$

  = $\lim_{h \rightarrow 0}\frac{1}{8} \frac{\cos (\frac{\pi}{2}-h)(1-\sin (\frac{\pi}{2}-h))}{\sin(\frac{\pi}{2}-h) (\frac{\pi}{2}-\frac{\pi}{2}+h)^{3}}$

    =  $\frac{1}{8}\lim_{h \rightarrow 0}\frac{\sin h(1-\cosh)}{\cosh.h^{3}}$

 = $\frac{1}{8}\lim_{h \rightarrow 0}\frac{\sin h(2\sin^{2}\frac{h}{2})}{\cosh.h^{3}}$

  = $\frac{1}{4}\lim_{h \rightarrow 0}\frac{\sin h.\sin^{2}(\frac{h}{2})}{\cosh.h^{3}}$

=    $\frac{1}{4}\lim_{h \rightarrow 0}(\frac{\sinh}{h})(\frac{\sin\frac{h}{2}}{\frac{h}{2}})^{2}.\frac{1}{\cosh}.\frac{1}{4}$

=   $\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}$

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