y mx 25m 25 4 circle. Then the perpendicular distance of tangent from the centre (0, 0) is the radius. Since tangent

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1 Mathematics. The sides AB, BC and CA of ABC have, 4 and 5 interior points respectively on them as shown in the figure. The number of triangles that can be formed using these interior points is () 80 () 85 () 0 (4) 05 C A B The number of ways of selecting points from among 4 5 ( ) points is C. But from among these, we have to discount collinear sets of points. Therefore the number of triangles is C C C C ! Hence, the correct option is (4). A tangent which is common to the circle x + y = 6 and the ellipse quadrant. If the tangent makes an intercept of length 4 value of k. () () () 4 (4) 5 x y is drawn in the first 5 4 k between the coordinate axes, then find the x y The equation of tangent to ellipse is y mx 5m 4, which is also the tangent to the 5 4 circle. Then the perpendicular distance of tangent from the centre (0, 0) is the radius. Since tangent y mx 5m 4 of ellipse is also the tangent to the circle x + y = 6, we have 5m 4 4 m m

2 Here, we ignore the positive sign in order to consider the tangent in the st quadrant. Therefore, the equation of common tangent is 7 y x 4 Therefore, this tangent meets the axes at 7 A( 7, 0) and B 0,4 Thus, the length of the intercepted portion of tangent between axes is 4 4 AB k Hence, the correct option is (). Let A, B, C and D be four sets, then ( A B) ( C D) is equal to () ( A B) ( C D) () ( A B) ( C D) () ( A C) ( B D) (4) ( A C) ( B D) Let A = {, }; B = {,, }; C = {4, 5, 6} and D = {6, 7}. Then ( A B) ( C D) {, } {6} {(,6),(,6)} A B {(,),(,),(,),(,),(,),(,)} C D {(4,6),(4,7),(5,6),(5,7),(6,6),(6,7)} ( A B) ( C D) {} Hence, option () is incorrect. ( A B) ( C D) {(,),(,),(,),(,),(,),(,),(4,6),(4,7), (5,6),(5,7),(6,6),(6,7)} Hence, option () is incorrect. ( A C) {(,4),(,5),(,6),(,4),(,5),(,6)} ( B D) {(,6),(,7),(,6),(,7),(,6),(,7)} Therefore, ( A C) ( B D) {(,6),(,6)} Hence, option () is correct. ( A C) ( B D) {(,4),(,5),(,6),(,7),(,4),(,5),(,6),(,7),(,6),(,7)} {(,6),(,6)} Hence, option (4) is incorrect. Hence, the correct option is () 4. The equation of the planes passing through the line of intersection of the planes x + y + z = and x + y z = 4 and parallel to x-axis is () y + 4z 7 = 0 () y 4z 7 = 0 () y z + 6 = 0 (4) y 4z + 7 = 0

3 Now required plane is x y z (x y z 4) 0 or x( ) y( ) z( ) (4 ) 0 () If this plane is parallel to x-axis then the product of sum of DRs of normal and x-axis must be 0. DRs of x-axis are (, 0, 0) we have, ( )() ( )(0) ( )(0) 0 or 0 From () x y z 4 0 or y z 0 or y z 6 0 Hence, the correct option is () 5. If h is the height and r is the radius of the base of a circular cylinder of greatest volume that can be inscribed in a given sphere, then h is equal to (as shown in Figure below) () r () r () r

4 (4) r Let R be the radius of the sphere. Then h r R Now V is the volume of the cylinder. So V r h Differentiating we get Now Also since h 0 dv dh h R R dv 0 h dh h d V h 0 dh So V is greatest, when h R/. In such a case Therefore, r R R So, h R r Hence, the correct option is (4) h 4 h 4 4 R R h 4R R r R R 4 4 () [From Eq. ()] 6. If a point P(, ) moves such that sum of its distance from the lines x y 0 and y x 5 is 0, then the locus of point P(, ) is () hyperbola () ellipse () parabola (4) square The given lines are perpendicular to each other because mm Let us consider that R( h, k) to be a moving point and sum of its distance from x-axis and y-axis be 0. We take x-axis and y-axis instead of given lines since the given lines are perpendicular to each other and the sum of distances = 0:

5 h k 0 which implies that the locus of point R is x y 0, which is a square as shown in the following figure: Hence, the correct option is (4) 7. If α, β and γ be the roots of the equation (x a)(x b)(x c) = d (where d 0), the roots of equation (x α)(x β)(x γ) + d = 0 are () a, b, d () b, c, d () a, b, c (4) a + d, b + d, c + d The roots of any function f(x) are the points where the function vanishes, that is, f(x) = 0 Since α, β and γ are the roots of equation ( x a)( x b)( x c) d, we can write ( x a)( x b)( x c) d ( x )( x )( x ) ( x )( x )( x ) d ( x a)( x b)( x c) which imply that the roots of the equation ( x )( x )( x ) d 0 are a, b, c. Hence, the correct option is () 8. If y ( x n ) n x ( n ) ; n W, then evaluate 9 () 4 () 40 () (4) 0 y dx.

6 y x 0 x, when n = 0 ( x ) x 4, when n = ( x 4 ) 4 x 6, when n = Hence, the correct option is (). 0 y dx y dx y dx y dx x dx 0 0 x x 5 5 [ ] y dx x dx 9. Let A be a non-singular square matrix. If B is a square matrix such that ( A B) is equal to () A B () A B () O (4) I B A BA, then the matrix We have AB AA BA BA AB BA O Now ( A B) A AB BA B A O B A B Hence, the correct option is () 0. The number of common tangent to the circles x + y = 4 and x + y 6x 8y 4 = 0 is () 0 () () (4) 4

7 Equation of circle is Equation of circle is Now, Here, x y 4 c (0,0); r x y 6x 8y 4 0 c (,4); r 7 cc 4 5 cc r r Therefore, the circles touch internally, which implies that the number of common tangent is. Hence, the correct option is (). Find the value of the series 4 4 () log 5 log () log () log log (4) log This problem is based on the concept of expansion of the logarithmic series. 4 x x x log( x) x 4 Substituting x on both sides, we have log 4 4 log 4 4 By using the results of exponential and logarithmic series, some complex series can be easily solved. Using the logarithmic series expansion, we calculated the value of the given series as log log log Hence, the correct option is (). The term independent of x in the expansion of x x x x x x 0

8 is () 0 () 90 () 0 (4) 00 Put x y and x z so that and The given expansion is Therefore T is independent of x implies r Hence, 0 5r 0 or r 4 Therefore independent term value is Hence, the correct option is () x y x x x x x z 0 y x z x 0 0 (0 r)/ r r/ r Tr C x ( ) x 0 r r C ( ) 0. There are three boxes B, B and B. B contains white and black balls. B contains white, black ball and B contain white and black balls. If a ball is taken from each box then among the balls drawn, the probability that there are two black and white ball is () 7 () 60 () 7 60 (4) 5 Let E be the event that the draw contains black and one white ball. Let w i be the extent of drawing a white ball from the B i and b j be the event of drawing a black ball from the box, ). Therefore, E ( b b w ) ( b w b ) ( w b b ). Each of the events in the union are mutually exclusive and independent. B j (i =,,, and j =,

9 P( E) P( b b w ) P( b w b ) P( w b b ) P( b ) P( b ) P( w ) P( b ) P( w ) P( b ) P( w ) P( b ) P( b ) Hence, the correct option is (4) 4. In ABC, if BC = a, CA = b and AB = c and the internal bisector of the angle A meets the side BC in L, then AL is equal to a bc () bc ( b c) abc () bc ( b c) ab () a bc ( b c) bc (4) a bc ( b c) It is known that BL :LC = c : b. Take A as origin, let AB and AC so that c, b and the angle between and is A. BL :LC = c : b the position vector b c AL of L is given by AL. b c

10 Hence, the correct option is () 5. ~(X Y) is equivalent to () X ~Y () ~ X Y () ~X Y (4) X Y We have ~ (X Y) Applying De Morgan s Law, AL [ b c bc( )] ( b c) ( b c) [ b c b c b c cos A] ( b c a ) b c b c ( b c) bc b c bc b c a ( b c) bc a bc b c a bc bc [( ) ]. ( b c) ( b c) (X Y) = ~X Y ~ (X Y) = ~ (~X Y) Hence, the correct option is () X ~Y 6. The mean of n items is x. If each of the n th item is increased by n, then the new mean is () () ( n ) x ( n ) x () x n (4) n( n ) x Let x, x... x n be n items

11 Now x becomes x + x becomes x + and so on. So new mean (... ) n x x x x n x ( x x x... xn ) (... n ) n ( x x... xn ) (... n) n n n x n ( n ) x Hence, the correct option is () 7. If A is a square matrix such that () I A () ( I A) () I A (4) ( I A) A O, then I A A (I is the corresponding unit matrix) is We have Therefore Hence, the correct option is (4) ( I A)( I A A ) I A I I A A ( I A) 8. Let f ( ) sin sin sin (sin sin 4 ) R. Then () f ( ) 0 R () f ( ) 0 R () f ( ) 0 0 (4) f ( ) 0 R f ( ) sin sin sin (sin sin 4 ) sin sin sin (sin cos ) (sin )(sin )(sin )(sin ) Hence, the correct option is () [(sin )(sin )] 0 R

12 9. Let f ( x) x[ x] where [ x ] denotes the integral part of x. If a is not an integer, then f ( a h) f ( a) lim h0 h () a () [a] () [a] (4) does not exist We have Hence, the correct option is () f ( a h) f ( a) ( a h)[ a h] a[ a] lim lim h h ( a h)[ a] a[ a] lim h0 h ( a is not an integer, [a + h] = [a] for small values of h) lim[ a] [ a] h0 h0 h0 0. Parabola y x c touches the circle x y 5 ( c 5). The area (in sq. units) bounded by the parabola and y-axis is 0 0 () () () 0 0 (4) (0) 0 4 For the points of intersection, we have x ( x c) 5 0 For curves to touch each other, discriminant D = 0: b 4ac 0 4( c 5) 0 4c 0 0 c 4 The required area is 0 0 / 0 [ x (0/ 4) ] ydx x dx 4 / / sq. units Hence, the correct option is ()

13 . The DRs of the line perpendicular to the plane determined by the lines x y z and x y z are () (5, 4, ) () (5, 4, ) () ( 5, 4, ) (4) (5, 4, ) We have n (,, ) and n (,,) as the DRs of the two given lines, respectively. Hence, n n is the normal to the plane determined by the given lines is i j k = 5i 4 j + k Therefore, the DRs of the normal are ( 5, 4, ) or (5, 4, ) Hence, the correct option is (). If the roots of the equation 0x Kx 54x 7 0 are in HP, then K is equal to () () 6 () 9 (4) Let,, be the roots of the given equation. Then K, , 7 Now,, are in HP, and hence /, /, / are in AP. This gives 54 7 Since is a root of the given equation, substituting the value of in it we get 0 K K K K 9 Hence, the correct option is (). The function

14 cos 4x if x0 x f ( x) A if x 0 x if x 0 6 x 4 is continuous at x = 0 for () A = 6 () A = 8 () A = 4 (4) no value of A f is continuous at x = 0 if Now Therefore lim f ( x) lim f (0 h) x00 h0 h0 lim h0 h0 lim f ( x) lim f ( x) f (0) A. x00 x00 6 h 4 h ( 6 h 4) lim h0 6 h 6 h lim (( 6 h 4)) (4 4) 6 So f ( x ) is not continuous for any value of A. Hence, the correct option is (4) lim f ( x) lim f ( x) x00 x00 4. A ray of light coming from the point (, 5) is reflected at a point P on the line mirror at x-axis and then passes through the point (7, 8). Then the coordinates of the point P is given by () () 59 () 6 5 (4) 59

15 Let the given points be A(, 5) and B(7, 8) and suppose the coordinates of point P is x, 0. So we will have, 5 0 Slope of AP x 8 0 Slope of BP 7 x From these two slopes, we have x 7 x 5 5x 4 8x x x Hence, the correct option is (4) B 5. Let A and B be two events such that P (A) = 0., P(B) = 0.6 and P 0.5. Then A A P B equals () 4 () 5 8 () 9 40 (4) 4 B P( A B) P( A) P (0.)(0.5) 0.5 A Now P( A B) P( A) P( B) P( A B) Also P A P( A B) P( A B) P( A B) B P ( B ) P ( B ) P ( B ) (0.6) Hence, the correct option is () 6. Find the equation of parabola if the coordinates of the focus are (, ) and the coordinates of the vertex are (, ).

16 () x 6x 6y 0 () x 6x 6y 0 () y 6y 6x 0 (4) None of these From the following figure, the equation of the axis is x =. The intersection of axis and the directrix is P, whose coordinates are (, 6). d We have the eccentricity e, where d is the distance of the parabola from the focus and d is the d distance of the directrix from variable point. The eccentricity of the parabola is, which implies that d d. Therefore, k 6 ( h ) ( k ) h 6h 9 k d d 4k 4 k 6 k h 6h 6k 0 () Since the eccentricity of the parabola is, we obtain the equation of the parabola by considering d d and changing h and k as the coordinates of x and y in Eq. (), we get the equation of the parabola as follows: x 6x 6y 0 Hence, the correct option is () x x x 7. If y x x x, then d y at x = is dx () 5 () 5

17 () 5 (4) 5 Therefore, Thus, Hence, the correct option is () ( x )( x x ) y ( x )( x x ) x y xr x x x x, ( 0 for any real ) dy ( x ) () ( x )() dx ( x ) ( x ) d y ( ) dx ( x ) d y 8. If z x iy is such that z 4 z, then () x 0, y 0 () x 0, y 0 () x, y (4) x and y is any real number. We have Hence, the correct option is (4) dx x 5 z 4 z z 4 z ( x 4) y ( x ) y 4x x 9. A principal solution of the equation tan x cot x cosec x is () 7 8 () 4 () 5

18 (4) tan x cot x cosec x sin x cos x cos x sin x sin x sin x cos x cos x cos x cos x n, n Hence /, ( / ) 5 / are the principal solutions. Hence, the correct option is () 0. () () () (4) dx x x 6x 4 is equal to x 6x 4 c 5 x x 6x 4 c 5 x x 6x 5 c 5 x x 6x 5 c 5 x Let Put x / t so that dx (/ t ) dt I I. Therefore dx ( x ) x 6x 4 t dt t 6 4 t t t dt ( t) 6 t( t) 4t t 5t dt

19 where t x Hence, the correct option is () 5 5 t c

QUESTION BANK ON. CONIC SECTION (Parabola, Ellipse & Hyperbola)

QUESTION BANK ON. CONIC SECTION (Parabola, Ellipse & Hyperbola) QUESTION BANK ON CONIC SECTION (Parabola, Ellipse & Hyperbola) Question bank on Parabola, Ellipse & Hyperbola Select the correct alternative : (Only one is correct) Q. Two mutually perpendicular tangents