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JEE Advanced 2015Paper 1
Exam
JEE Advanced
Year
2015
Paper
Paper 1
Subject
Mathematics
Chapter
Properties of Triangles

JEE Advanced 2015 · Paper 1 · Mathematics · Properties of Triangles

Column I Column II (A) In a triangle ∆ XYZ let a, b and c be the...

Column I Column II
(A)\left(A\right)

In a triangle ∆XYZ∆ XYZ let a,bandca, b and c be the lengths of the angles X,YandZX, Y and Z, respectively. If 2(a2−b2)=c2and2\left({a}^{2} - {b}^{2}\right) = {c}^{2} and\lambda = \dfrac{\sin \left(X - Y\right)}{\sin Z}thenpossiblevaluesofthen possible values ofnforwhichfor which\cos \left(n\pi\lambda\right) = 0isare</p></td><td>(P)is are</p> </td> <td> \left(P\right)

11
\left(B\right)</td><td><p>Inatriangle</td> <td> <p> In a triangle\delta XYZ,letleta, b and cbethelengthsofthesidesoppositetotheanglesbe the lengths of the sides opposite to the anglesX, Y and Z,respectively.If, respectively. If1 + \cos 2x - 2\cos 2y= 2\sin x \sin y,thenpossiblevalue(s), then possible value(s) of \dfrac{a}{b}is(are)</p></td><td>(Q)is (are)</p> </td> <td> \left(Q\right) 22
\left(C\right)</td><td>In</td> <td> In{\mathbb{R}}^{2},letlet\sqrt{3}\overset{^}{i} + \overset{^}{j}, \overset{^}{i} + \sqrt{3}\overset{^}{j}andand\beta\overset{^}{i} + \left(1 - \beta\right)\overset{^}{j}bethepositionvectorsofbe the position vectors ofX, Y and Zwithrespecttotheoriginwith respect to the originO,respectively.Ifthedistanceof, respectively. If the distance ofZfromthebisectoroftheacuteangleoffrom the bisector of the acute angle of\overset{\to}{OX}withwith\overset{\to}{OY}isis\dfrac{3}{\sqrt{2}},thenpossiblevalue(s)of∣β∣, then possible value (s) of \left|\beta\right| is (are) \left(R\right)</td><td></td> <td>3</td></tr><tr><td>(D)</td> </tr> <tr> <td> \left(D\right)

Suppose that F(α)F ( \alpha ) denotes the area of the region bounded by x=0,x=2,y2=4xx = 0, x = 2, {y}^{2} = 4x and y=∣ax−1∣+∣\alphax−2∣+\alphax,y = \left|ax - 1\right| +\left|\alphax - 2\right| + \alphax, where α∈{0,1}.\alpha \in \left\{0, 1\right\} . Then the value (s)ofF(α)+832,of F\left(\alpha\right) + \dfrac{8}{3}\sqrt{2}, when α=0\alpha = 0 and α=1\alpha = 1, is (are)

\left(S\right)</td><td></td> <td>5</td></tr><tr><td></td><td></td><td>(T)</td> </tr> <tr> <td> </td> <td> </td> <td> \left(T\right) 66

Column I Column II (A) In a triangle ∆ XYZ let a, b and c be the lengths of the angles X, Y and Z , respectively. If 2(Column I Column II (A) In a triangle ∆ XYZ let a, b and c be the lengths of the angles X, Y and Z , respectively. If 2(Column I Column II (A) In a triangle ∆ XYZ let a, b and c be the lengths of the angles X, Y and Z , respectively. If 2(

Options

  1. Aa-p,r,s;b-p;c-p,q;d-s,t;
  2. Ba-q,r,s,t;b-q,r,s,t;c-r,s,t;d-q,r,s;
  3. Ca-q;b-r;c-r,s;d-r,s;
  4. Da-q,r,s,t;b-t;c-r,s,t;d-q,r,s,t;

Correct answer

A. a-p,r,s;b-p;c-p,q;d-s,t;

Step-by-step solution

\left(A\right) \lambda = \dfrac{\sin \left(X - Y\right)}{sin Z}<br><br >\dfrac{\sin X \cos Y - \cos X \sin Y}{\sin Z}<br><br >\dfrac{acosY - bcosX}{c}<br><br >\dfrac{a\left(\dfrac{{a}^{2} + {c}^{2} - {b}^{2}}{2ac}\right) - b\left(\dfrac{{b}^{2} + {c}^{2} - {a}^{2}}{2bc}\right)}{c} = \lambda<br><br >\dfrac{{a}^{2} - {b}^{2}}{{c}^{2}} = \lambda<br>As<br >As2\left({a}^{2} - {b}^{2}\right) = {c}^{2}<br><br >∴ \lambda = \dfrac{1}{2}<br><br >∴ \cos \left(\dfrac{n\pi}{2}\right) = 0<br><br >∴ n = 1, 3, 5<br>(B)1+cos⁡2X−2cos⁡2Y=2sinXsinY<br >\left(B\right) 1 + \cos 2X - 2\cos 2Y = 2sin Xsin Y
Solving
4sin⁡2Y−2sin⁡2X=2sin⁡Xsin⁡Y4\sin^{2}Y - 2\sin^{2}X = 2\sin X \sin Y
4b2−2a2=2ab4{b}^{2} - 2{a}^{2} = 2ab
4−2(ab)2=2ab4 - 2{\left(\dfrac{a}{b}\right)}^{2} = \dfrac{2a}{b}

Column I Column II (A) In a triangle ∆ XYZ let a, b and c be the lengths of the angles X, Y and Z , respectively. If 2(
ab=X\dfrac{a}{b} = X

X2+X−2=0{X}^{2} + X - 2 = 0

X=1X = 1

ab=1\dfrac{a}{b} = 1

x=−2x = - 2

ab=−2\dfrac{a}{b} = - 2 (rejected)

\left(C\right) OY ≡ y = \sqrt{3}x</p><p></p><p>OX ≡ y = \dfrac{1}{\sqrt{3}}x</p><p></p><p>∴EquationofbisectorofEquation of bisector ofOX, OY</p><p></p><p>y = x</p><p></p><p>∴ \left|\dfrac{x - y}{\sqrt{2}}\right| = \dfrac{3}{\sqrt{2}}</p><p></p><p>\dfrac{\beta - \left(1 - \beta\right)}{\sqrt{2}} = \pm \dfrac{3}{\sqrt{2}}</p><p></p><p>2\beta - 1 = \pm 3</p><p></p><p>\beta = 2</p><p></p><p>\beta = - 1</p><p>∣β∣=1,2</p><p>\left|\beta\right| = 1, 2

(D)( D )
Column I Column II (A) In a triangle ∆ XYZ let a, b and c be the lengths of the angles X, Y and Z , respectively. If 2(

∴∴ Area =6−∫022xdx= 6 - {∫}_{0}^{2}2\sqrt{x}dx

=6−823= 6 - \dfrac{8\sqrt{2}}{3}

∴∴ Required value =6−823+823=6= 6 - \dfrac{8\sqrt{2}}{3} + \dfrac{8\sqrt{2}}{3} = 6

Column I Column II (A) In a triangle ∆ XYZ let a, b and c be the lengths of the angles X, Y and Z , respectively. If 2(
Area= Shaded region −- area under parabola.

=5−823= 5 - \dfrac{8\sqrt{2}}{3}

∴∴ Required value =5= 5

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