JEE 2012 :: General Questions :: Mathematics

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The value of the integral 

$\int_{-\pi /2}^{\pi /2} \left(x^{2}+\log\frac{\pi-4}{\pi+4}\right)\cos x dx$ is 

Answer:

Explanation:

$\int_{-\pi /2}^{\pi /2} \left(x^{2}+\log\frac{\pi-x}{\pi +x}\right)\cos x dx$

 As   $\int_{-a}^{a} f(x) dx=0$

 when    f(-x) =-f(x)

$\therefore$   $I=\int_{-\pi/2}^{\pi/2} x^{2} \cos x dx+0$

 = $=2\int_{0}^{\pi/2}( x^{2} \cos x) dx$

=$2\left\{\left(x^{2} \sin x\right)_{0}^{\pi/2}-\int_{0}^{\pi/2} 2x.\sin x dx\right\}$

=$2\left[\frac{\pi^{2}}{4}-2\left\{- x \cos x)_{0}^{\pi/2}-\int_{0}^{\pi/2} 1.(-\cos x)dx\right\}\right]$

 =$2\left[\frac{\pi^{2}}{4}-2(\sin x)_{0}^{\pi/2}\right]$

=$2\left[\frac{\pi^{2}}{4}-2\right]=\left(\frac{\pi^{2}}{2}-4\right)$

Four fair dice $D_{1},D_{2},D_{3}$ and $D_{4}$ each having  six faces numbered 1, 2,3, 4, 5 and 6 are rolled simultaneously. The  probability that $D_{4}$  shows a number

appearing on one of  $D_{1},D_{2}$ and $,D_{3}$ is 

Answer:

Explanation:

 Concept  involved  As one of the dices shows a number appearing on one of $P_{1},P_{2} $ and $P_{3}$ 

Thus, three cases arise

 (i) All shows the same number

 (ii)  Number appearing on $D_{4}$ appears on any one of $D_{1},D_{2}, D_{3}$

 (iii) Number appearing on $D_{4}$  appears on any two  of $D_{1},D_{2},D_{3}$

 sol .Sample  space= $ 6 \times 6 \times 6 \times 6$

  =$6^{4}$  favourable events.

 = case I or case II  or case III

 Case I.  First, we should select one number for $D_{4}$ which appears on all i.e.,

 $^{6} C_{1} \times 1.$

Case II. for $D_{4}$ there are $^{6}C_{1}$ ways .Now, it appears on any one of $D_{1},D_{2},D_{3}$ 

i.e, $^{3}C_{1} \times 1$ for other two there are 5 x 5 ways

 $\Rightarrow$ $^{6}C_{1} \times ^{3}C_{1} \times 1 \times 5 \times 5$

 Case III

 For $D_{4}$ there are $^{6}C_{1}$  ways  now it appears on any two of $D_{1},D_{2},D_{3} \Rightarrow  ^{3}C_{2} \times 1^{2}$

 for other one there are 5 ways

$\Rightarrow$  $^{6}C_{1} \times ^{3}C_{2} \times 1^{2} \times 5 $

 Thus probability

 $=\frac{^{6}C_{1}+^{6}C_{1} \times^{3}C_{1}\times 5^{2}+^{6}C_{1}\times ^{3}C_{2}\times 5}{6^{4}}$

 $=\frac{6(1+75+15)}{6^{4}}=\frac{91}{216}$

Let PQR  be a triangle of area $\triangle$ with a=2,b=$\frac{7}{2}$ and c=$\frac{5}{2}$ where a,b and c  the length of the sides of the triangle opposite to the angle s  at P, Q and R respectively.Then $\frac{2 \sin P-\sin 2P}{2 \sin P+\sin 2P}$ equal to 

Answer:

Explanation:

 Concept involved if $\triangle ABC$ has sides a, b, c then, 

 $\tan (A/2)=\sqrt{\frac{(s-b)(s-c)}{a(s-a)}}$

 where, $s=\frac{a+b+c}{2}$

 Sol ,$s= \frac{2+\frac{7}{2}+\frac{5}{2}}{2}=4$

  $\therefore$      $\frac{2 \sin P-\sin 2P}{2 \sin P+\sin 2p}=\frac{2 \sin P(1-\cos P)}{2 \sin P(1+\cos P)}$

 $= \frac{2 \sin^{2}(P/2)}{2 \cos^{2}(P/2)}=\tan^{2}(P/2)$

 $\Rightarrow$      $\frac{(s-b)(s-c)}{s(s-a)} \times \frac{(s-b)(s-c)}{(s-b)(s-c)}$

 $\Rightarrow$     $\frac{((s-b)(s-c)^{2})}{\triangle^{2}}= \frac{\left(4+\frac{7}{2}\right)^{2}\left(4-\frac{5}{2}\right)^{2}}{\triangle^{2}}$

 =$(\frac{3}{4 \triangle})^{2}$

lf a and b are vectors such that |a+b|=$\sqrt{29}$ and $a \times (2 i+3j+4k)=(2i+3j+4k) \times b$ the  a possible value of $(a+b).(-7 i+2j+3k)$ is 

Answer:

Explanation:

 Concept involved  if a x b= a x c

 $\Rightarrow$  a x b-a  x c=0

$\Rightarrow$ a x (b-c)=0 ie, a || (b-c) 

 or b-c= $\lambda $ a

 Sol. Here, $a \times (2i+3j+4k)=(2i+3j+4k) \times b$ 

$\Rightarrow$   $a \times (2i+3j+4k)-(2i+3j+4k) \times b$

$\Rightarrow$   (a +b) x (2i+3j+4k)=0

$\Rightarrow$   $a+b=\lambda (2i+3j+4k)$.....(i)

 since,  $|a+b|= \sqrt{29}$

 $\Rightarrow$ $\pm \lambda \sqrt{4+9+16}=\sqrt{29}$

 $\Rightarrow$    $\lambda =\pm 1$

 $\therefore$      a+b=$\pm (2i+3j+4k)$

 Now, (a+b)(-7i+2j+3k)

 =$ \pm (-14+6+-12)=\pm 4$

The equation of a plane passing through the line of intersection of the plane x+2y+3z=2 and x-y+z=3 and at a distance $2/\sqrt{3}$ from the point (3,1m-1) is 

Answer:

Explanation:

 Concept involved (i) equation of the plane through the intersection of two planes.

 i.e,   $(a_{1}x+b_{1}y+c_{1}z+d_{1})+\lambda (a_{2}x+b_{2}y+c_{2}z+d_{2})=0$

 (ii)  Distance of a point $(x_{1},y_{1},z_{1})$ from ax+by+cz+d=0

 $\Rightarrow$    $\frac{|ax_{1}+by_{1}+cz_{1}+d|}{\sqrt{a^{2}+b^{2}+c^{2}}}$

 Sol. Equation of plane passing through intersection of two planes

 $x+2y+3z=2$ and x-y+z=3 is 

$\Rightarrow$    (x+2y+3z-2)+$\lambda (x-y+z-3)=0$

$\Rightarrow$    $(1+\lambda)x+(2- \lambda)y+(3+\lambda)z-(2+3 \lambda)=0$

 whose distance from (3,1,-1) is    $\frac{2}{\sqrt{3}}$

   $\Rightarrow$    $\frac{ |3(1+\lambda )+1.(2-\lambda)-1(3+\lambda)-(2+3 \lambda)|}{\sqrt{(1+\lambda)^{2}+(2-\lambda)^{2}+(3+\lambda})^{2}}=\frac{2}{\sqrt{3}}$

  $\Rightarrow$  $\frac{|-2 \lambda|}{\sqrt{3 \lambda^{2}+4 \lambda+14}}=\frac{2}{\sqrt{3}}$

 $\Rightarrow$  $3 \lambda^{2}=3 \lambda^{2}+4 \lambda+14 \Rightarrow \lambda =-\frac{7}{2}$

$\therefore$   $\left(1-\frac{7}{2}\right)x+\left(2+\frac{7}{2}\right)y+\left(3-\frac{7}{2}\right)z-\left(2-\frac{21}{2}\right)=0$

 $\Rightarrow$    $-\frac{5x}{2}+\frac{11}{2}y-\frac{1}{2} z+\frac{17}{2}=0$

 or    5x+11y+z-17=0

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