JEE 2014 :: Main 2014 :: Mathematics 2014

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Three positive numbers from an increasing GP. If the middle term on this GP is doubled, then new numbers are in AP. Then, the common ratio of the GP is 

Answer:

Explanation:

Let a,ar, ar² are in GP (r>1)

 On multiplying middle team by 2, a, 2ar, ar² are in AP.

  $\Rightarrow$   $4ar=a+ar^{2}$

$\Rightarrow$     $r^{2}-4r+1=0$

$\Rightarrow$        $r=\frac{4\pm\sqrt{16-4}}{2}= 2\pm\sqrt{3}$

$\Rightarrow$   $r=2+\sqrt{3}$   [ $\because$  AP is increasing ]

If   $(10)^{9}+2(11)^{1}(10)^{8}+3(11)^{2}(10)^{7}+....+10(11)^{9}=k(10)^{9}$    then k is equal to

Answer:

Explanation:

$k.10^{9}=10^{9}+2(11)^{1}(10)^{8}+3((11)^{2}(10)^{7}+....+10(11)^{9}$

   $k=1+2\left(\frac{11}{10}\right)+3\left(\frac{11}{10}\right)^{2}+...+10\left(\frac{11}{10}\right)^{9}$    .......(i)

$\left(\frac{11}{10}\right)k=1\left(\frac{11}{10}\right)+2\left(\frac{11}{10}\right)^{2}+...+9\left(\frac{11}{10}\right)^{9}+10\left(\frac{11}{10}\right)^{10}$        ..........(ii)

  On subtracting Eq,(ii) from Eq.(i) , we get

$k\left(1-\frac{11}{10}\right)=1+\left(\frac{11}{10}\right)+\left(\frac{11}{10}\right)^{2}+...+\left(\frac{11}{10}\right)^{9}-10\left(\frac{11}{10}\right)^{10}$

$\Rightarrow$     $k\left(\frac{10-11}{10}\right)=\frac{1\left[\left(\frac{11}{10}\right)^{10}-1\right]}{\left(\frac{11}{10}-1\right)}-10\left(\frac{11}{10}\right)^{10}$

[  $\because$    In GP sum of n terms =  $\frac{a(r^{n}-1)}{r-1}$  when r>1]

  $\Rightarrow$     -k

.$=10\left[10\left(\frac{11}{10}\right)^{10}-10-10 \left(\frac{11}{10}\right)^{10}\right]$

                 k=100

 The angle between the lines whose direction cosines satisfy  the equations l+m+m=0  and l² =m²+n²  is

Answer:

Explanation:

We konow  that angle berween two lines is 

$\cos \theta=\frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$

              l+m+n=0

$\Rightarrow$         $l=-(m+n)$

$\Rightarrow$       $(m+n)^{2}=l^{2}$

   $\Rightarrow$      $m^{2}+n^{2}+2mn=m^{2}+n^{2}$

                                                          [ $\because l^{2}=m^{2}+n^{2},given$]

   $\Rightarrow$               2mn=0

  when m=0 $\Rightarrow$ l=-n

 Hence, (l,m,n) is (1,0,-1)

 when n=0, then l=-m

 Hence , (l,m,n) is (1,0,-1)

 $\therefore$        $\cos\theta=\frac{1+0+0}{\sqrt{2}\times\sqrt{2}}$

  =   $\frac{1}{2}$

   $\Rightarrow$    θ=$\frac{\pi}{3}$

If the  coefficients of x³ and x⁴ in the expansion of   $(1+ax+bx^{2})$  (1-2x)¹⁸ in powers of x are both zero, then (a,b) is equal to 

Answer:

Explanation:

 To find the coefficient of x³ and x⁴, use the formula of coefficient of x^f  in (1-x)ⁿ is (-1)^r  nC_r   and then simplify.

 In expansion of    $(1+ax+bx^{2})$ (1-2x)¹⁸

 Coefficient of x³  = Coefficient of x³  in (1-2x)¹⁸+ coefficient of x² ina  (1-2x)¹⁸ + Coefficient of x in b(1-2x)¹⁸.

=  $-^{18}C_{3}.2^{3}+a^{18}C_{2}.2^{2}-b^{18}C_{1}.2$

Given, coefficient of x³ =0

=$^{18}C_{3}.2^{3}+a^{18}C_{2}.2^{2}-b^{18}C_{1}.2$  =0

$\Rightarrow$    $-\frac{18\times 17 \times 16}{3\times 2}.8+a.\frac{18\times17}{2}.2^{2}-b.18.2=0$

    $\Rightarrow$    $17a-b=\frac{34\times16}{3}$     .......(i)

Simolarly coefficient of x⁴

    $^{18}C_{4}.2^{4}-a^{18}C_{3}.2^{3}-b^{18}C_{2}.2^{2}=0$

$\therefore$      $32a-3b=240$     ..........(ii)

On solving      Eqs (i) and (ii), we get,

   a=16,  $b=\frac{272}{3}$

 If A is a 3 x3   non-singular matrix such that AA^T =A^TA  and B=A^-1A^T. then BB^T is equal to

Answer:

Explanation:

 Use the following properties of transpose  (AB)^T=B^TA^T.(A^T)^T =A   and A^-1A= I and simplify  

If A is a non-singular  matrix, then |A|≠ 0

                     AA^T=A^TA   and B=A^-1A^T

       BB^T=(A^-1A^T)(A^-1A^T)^T=  A^-1 A^T A(A^-1)^T                                                (  $\therefore$   (AB)^T=B^TA^T)

                =   A^-1AA^T(A^-1)^T                                            ($\therefore$  AA^T=A^T.A)

                ⁼ IA^T (A^-1)^T                                                     ($\therefore$    A^-1A=1)

               =A^T(A^-1)^T= (A^-1A)^T                                         ($\therefore$  (AB)^T=B^TA^T)

                       = I^T=1

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• Main 2014 - Mathematics 2014