JEE 2014 :: Advanced 2014 :: Mathematics 2014

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Let a,r, s, t be non-zero real numbers. Let  P( at²,2at) , Q , R (ar², 2ar) and S( as², 2as) be distinct point on the parabola y²=4ax . Suppose  that PQ is the  focal  chord and  lines QR and PK are parallel , where K is the point (2a,0)

 If st=1 , then the tangent at P  and the normal at S to the parabola meet at a point  whose ordinate is 

Answer:

Explanation:

Plan  Equation  of tangent and normal at (at², 2at) are given by ty= x+at²  and y+t x= 2at+at³ , respectively

  Tangent at P :ty= x+at²

  or      $y= \frac{x}{t}+at$

 Normal at S:

  $y+ \frac{x}{t}=\frac{2a}{t}+\frac{a}{t^{3}}$

 Solving   $2y=at+\frac{2a}{t}+\frac{a}{t^{3}}$

 $\Rightarrow$     $ y=\frac{a(t^{2}+1)^{2}}{2t^{3}}$

Let a,r, s, t be non-zero real numbers. Let  P( at²,2at) , Q , R (ar², 2ar) and S( as², 2as) be distinct point on the parabola y²=4ax . Suppose  that PQ is the  focal  chord and  lines QR and PK are parallel, where K is the point (2a,0)

The value of r is 

Answer:

Explanation:

 Plan 

 (i)  P( at², 2at) is one end point of focal chord of parabola y²=4ax,

 then other end point is   $(\frac{a}{t^{2}},-\frac{2a}{t})$

 (ii)  Slope of line joining two points  (x₁,y₁) and (x₂,y₂) is given by  $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

If PQ is foacl  chord, then coordinates of Q will be $(\frac{a}{t^{2}},-\frac{2a}{t})$

 Now, slope of QR= slope of PK

  $\frac{2ar+2a/t}{ar^{2}-a/t^{2}}=\frac{2at}{at^{2}-2a}$

 $\Rightarrow$    $ \frac{r+1/t}{r^{2}-1/t^{2}}=\frac{t}{t^{2}-2}$

 $\Rightarrow$    $ \frac{1}{r-1/t}=\frac{t}{t^{2}-2}$

$\Rightarrow$     $ r-\frac{1}{t}=\frac{t^{2}-2}{t}=t-\frac{2}{t}$

  $\Rightarrow$     $ r=  t-\frac{1}{t}=\frac{t^{2}-1}{t}$

For  $x \epsilon  (0,\pi)$  the equation   sin x+ 2 sin 2x-sin 3x=3 has 

Answer:

Explanation:

 Plan for solving this type of questions , obtain the LHS and RHS in equation and examine , the two are equal  or not for a given interval

  Given, Trigonometrical Equation

         (sin x -sin 3x)+2 sin2x =3

  $\Rightarrow$     $ -2\cos 2x\sin x+4 \sin x \cos x=3$

$[\because $     $  \sin C-\sin D=2\cos\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right) $  and $\sin2\theta=2\sin\theta\cos\theta]$

$\Rightarrow $   $2\sin x(2\cos x-\cos 2x)=3$

$\Rightarrow $   $ 2\sin x(2\cos x-\cos^{2} x+1)=3$

$\Rightarrow $    $2\sin x\left(\frac{3}{2}-2\left(\cos x-\frac{1}{2}\right)^{2}\right)=3$

$\Rightarrow $    $3\sin x-3=4(\cos x-\frac{1}{2})^{2}\sin x$

 As  $x\epsilon (0,\pi)$

       $LHS \leq 0 $   and  $ RHS\geq0$

 for solution to exist LRS=RHS

 Now, LHS =0 $\Rightarrow $ 3 sin x-3=0

$\Rightarrow $    sin x=1

$\Rightarrow $    $x=\frac{\pi}{2}$

 For $x=\frac{\pi}{2}$

  RHS=   $4\left(\cos\frac{\pi}{2}-\frac{1}{2}\right)^{2}\sin\frac{\pi}{2}$

  = $4\left(\frac{1}{4}\right)(1)=1\neq0$

 $\therefore$  No solution  of the equation exists

 Coefficient of x¹¹  in the expansion of  (1+x²)⁴ (1+x³)⁷  (1+x⁴)¹²   is 

Answer:

Explanation:

 Plan   Coefficient of  x^r  in (1+x)ⁿ  is ⁿC_r

 In this type of questions, we find differemnt composition of terms where product  will give us x¹¹

   Coefficient  of x¹¹ in   $(1+x^{2})^{4}(1+x^{3})^{7}(1+x^{4})^{12}$

Now,consider the following cases for x¹¹ in  $(1+x^{2})^{4}(1+x^{3})^{7}(1+x^{4})^{12}$

 Coefficient of x⁰  x³   x⁸ : coefficient of x²  x⁹  x⁰

Coefficient of x⁴  x³   x⁴ : coefficient of x⁸  x³  x⁰

  = $^{4}C_{0}\times^{7}C_{1}\times^{12}C_{2}+^{4}C_{1}\times^{7}C_{3}\times^{12}C_{0}+$

     $^{4}C_{2}\times^{7}C_{1}\times^{12}C_{2}+^{4}C_{4}\times^{7}C_{1}\times^{12}C_{0}$

     = 462+140+504+7=1113

Let f:[0,2]→  R  be a function which is continuous on [0,2] and is differentiable on (0,2)  with f(0)=1.Let   $F(x)=\int_{0}^{x^{2}} f(\sqrt{t}) dt, for x\in [0,2]$ . If  F'(x)= f '(x) for all  x ε (0,2) , then F(2) equal to

Answer:

Explanation:

 Plan    Newton - Leibnitz.s formula

   $\frac{d}{dx}\left[\int_{\phi(X)}^{\psi (X)} f(t) dt \right]= f\left\{\psi(X)\right\}\left\{\frac{d}{dx}\psi(X)\right\}-f(\phi(X))\left\{\frac{d}{dx}\phi(X)\right\}$

 Given, $F(X)=\int_{0}^{x^{2}} f(\sqrt{t})dt$

 $\therefore$  F'(x)= 2x f(x)

 Also, F '(x)= f' (x)

$\Rightarrow$    $2x f(x)=f'(x)\Rightarrow \frac{f'(x)}{f(x)}=2x$

$\Rightarrow$  $\int_{}^{} \frac{f '(x)}{f(x)}dx=\int_{}^{} 2x dx$

$\Rightarrow$      ln f(x)= x²+ c

$\Rightarrow$     $f(x)= e^{x^{2}}+c$

$\Rightarrow$   $f(x)= K e^{x^{2}}$             $ [K=e^{c}]$

 Now, f(0)=1

                               1=K

 Hence     $f(x)=  e^{x^{2}}$

       $F(2)=\int_{0}^{4} e^{t} dt$

    $=[e^{t}]_{0}^{4}=e^{4}-1$

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