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1.

If `A(5,2),B(10,12)` and `P(x,y)` is such that `(AP)/(PB) = (3)/(2)`, then then internal bisector of `/_APB` always passes throughA. `(20,32)`B. `(8,8)`C. `(8,-8)`D. `(-8,-8)`

Answer» Correct Answer - B
Let internal bisector of `/_APB` meets the side AB at C. We know that `(AC)/(BC) =(AP)/(PB)`
`:. (AC)/(BC) =(3)/(2)`
`:.` coordinates of C are `((3(10)+2(5))/(3+2),(3(12)+2(2))/(3+2))` or (8,8) Hence the internal bisector of `/_APB` always passes through point (8,8)
Discussion
2.

Let `A(0,beta), B(-2,0)` and `C(1,1)` be the vertices of a triangle. Then All the values of `beta` for which angle A of triangle ABC is largest lie in the intervalA. `(-2,1)`B. `(-2,(2)/(3))uu((2)/(3),1)`C. `(-2,(2)/(3))uu((2)/(3),sqrt(6))`D. none of these

Answer» Correct Answer - C
Whenever A is obtuse is the largest angle also. But there many be more values of `beta` for which A is largest but is not obtuse. Angle A will be largest if `BC gt AC, BC gt AB, A,B,C` are not collinear
`rArr 10 gt 1 +(beta -1)^(2),10 gt 4 +beta^(2), beta ne (2)/(3)`
`rArr beta^(2)-2beta -8 lt 0, beta^(2) lt 6`
`rArr-2 lt beta lt 4, -sqrt(6) lt beta sqrt(6)`
`beta in (-2,(2)/(3))uu((2)/(3),sqrt(6))`
Discussion
3.

`A(x_1,y_1), B(x_2,y_2),C(x_3,y_3)` are three vertices of a triangle ABC, `lx+my+n=0` is an equation of line L. If L intersects the sides BC,CA and AB of a triangle ABC at P,Q,R respectively, then `(BP)/(PC)xx(CQ)/(QA)xx(AR)/(RB)` is equal toA. `-1`B. `-(1)/(2)`C. `(1)/(2)`D. 1

Answer» Correct Answer - A
Let `BP: PC = alpha: 1`
`:. P ((alpha x_(3)+x_(2))/(alpha+1),(alphay_(3)+y_(2))/(alpha+1))`
P lies on `lx +my +n = 0`
`:. L ((alpha x_(3)+x_(2))/(alpha+1)) +m ((alphay_(3)+y_(2))/(alpha+1)) +n = 0`
`:. alpha =(-(lx_(2)+my_(2)+n))/((lx_(3)+my_(3)+n))`
Similarly, if `CQ: QA = beta:1`
`:. beta = (-(lx_(3)+my_(3)+n))/((lx_(1)+my_(1)+n))`
and if `AR: RB = gamma:1`
`:. gamma =(-(lx_(1)+my_(1)+n))/((lx_(2)+my_(2)+n))`
`:. (BP)/(PC).(CQ)/(QA).(AR)/(RB) =-1`
Discussion
4.

Number of values of `alpha` such that the points `(alpha,6),(-5,0)` and (5,0) form an isosceles triangle isA. 4B. 5C. 6D. 7

Answer» Correct Answer - B
We have `A(alpha,6), B(-5,0)` and `C(5,0)`
If `AB =AC`, we get `alpha = 0`. Hence, we get one value of c If `AC = BC`, we get `alpha = - 3,13`
`alpha` can take 5 units: `0,3,-3,13,-13`
Discussion
5.

The maximum value of `y = sqrt((x-3)^(2)+(x^(2)-2)^(2))-sqrt(x^(2)-(x^(2)-1)^(2))` isA. 3B. `sqrt(10)`C. `2sqrt(5)`D. none of these

Answer» Correct Answer - B
`y = f(x) = sqrt((x^(2)-2)^(2)+(x-3)^(2)) -sqrt(x^(2)+(x^(2)-1)^(2))`
Note that the first radical sign describes the distance between `P(x,x^(2))` and `A(3,2)` whereas the second radical sign describes the distance between `P(x,x^(2))` and `B(0,1)`. Now `PA -PB le AB` for possible positions of P. Hence `f(x)]_(max) =` distance between `AB = sqrt(9+1) = sqrt(10)`
Discussion
6.

If `m_(1),m_(2)` be the roots of the equation `x^(2)+(sqrt(3)+2)x+sqrt(3)-1 =0`, then the area of the triangle formed by the lines `y = m_(1)x,y = m_(2)x` and `y = 2` isA. `sqrt(33)-sqrt(11)` sq. unitsB. `sqrt(11) +sqrt(33)` sq. unitsC. `2sqrt(33)` sq. unitsD. 121 sq. units

Answer» Correct Answer - B
Sides are along lines `y = m_(1)x,y = m_(2)x` and `y = 2`
`:.` Vartices of the triangle are `(0,0), ((2)/(m_(1)),2),((2)/(m_(2)),2)`
Area `=(1)/(2) |quad{:(0,0),((2)/(m_(1)),2),((2)/(m_(2)),2),(0,0):}|`
` =2|(m_(2)-m_(1))/(m_(1)m_(2))|`
`:. |m_(1)-m_(2)| =sqrt((m_(1)+m_(2))^(2)-4m_(1)m_(2))`
`= sqrt((sqrt(3)+2)^(2)-4(sqrt(3)-1))`
`= sqrt(11)`
`:.` Area `= 2|(sqrt(11))/(sqrt(3)-1)| =sqrt(33)+sqrt(11)`
Discussion