+ {JEE Advace 03} Sept 0 Name: Batch (Day) Phoe No. IT IS NOT ENOUGH TO HAVE A GOOD MIND, THE MAIN THING IS TO USE IT WELL Marks: 00. If A (α, β) = (a) A( α, β) = A( α, β) (c) Adj (A ( α, β)) = Sol : We have A (α, β) = = A ( α, β) [Each right aswer carries 4 marks ad wrog ] cosα siα 0 siα cosα 0, the β 0 0 e β e A(- α, - β) cosα siα 0 siα cosα 0 β 0 0 e Also, A( α, β) A ( α, β) cosα siα 0 cosα siα 0 siα cosα 0 siα cosα 0 β β 0 0 e 0 0 e A (α, β) - = A ( α, β) Net, Adj A ( α, β) = A ( α, β) A( α, β) - e -β A( α, β). = (b) A( α, β) = A ( α, β) (d) A ( α, β) = A( α, β) cos( α) si( α) 0 si( α) cosα 0 β 0 0 e Time: hr. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH =

cos α siα. Let α = π/5 ad A = siα cosα, the B = A+ A + A 3 + A 4 is (a) sigular (b) o-sigular (c) skew-symmetric (d) B = Sol : We have cosα siα 3 cos3α si3α 4 cos 4α si 4α A, A ad A siα cosα si3α cos3α si 4α cos 4α we have cosα cosα cos3α cos 4α = cosα cosα cos(πα) cos(π α) [ 5α π] cosα cosα cosα cosα 0 ad siα siα si3α si 4α = siα siα si(πα) si(π α) = [siα siα] = = 3α α 3π π si cos 4si cos 0 0 π π π 4si cos 0 5 0 = 4 cos Thus, B = π π cos 5 0 = a (say) 0 a a 0 B is skew-symmetric. Also, B = a = 6 cos π 5 cos π 0 > 0 B is o-sigular. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH

3. The system of equatios: has y z a y z b y z c (a) o solutio if a + b + c 0 (b) uique solutio if a + b + c = 0 (c) ifiite umber of solutios if a + b + c = 0 Sol : Whe we add three equatios, we get 0 = a + b + c. Thus, the system does ot have a solutio if a + b + c 0. (d) oe of these If a + b + c = 0, the system has ifiite umber of solutios give by = 3 (3k a b) y= (3k a b) 3 z = k where k C. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 3

C 4. Let Ck = k 0 Ck for 0 k ad Ak = for k, ad 0 Ck A + A +. +A = k 0, the 0 k (a) k = k (b) k + k = C + (c) k = C (d) k = C + Sol : (a, c) Use k 0 c 5. Let A k C 0 siα siα siβ siα 0 cosα siβ, the siα siβ cosα cosβ 0 (a) A is idepedet of α ad β (c) A - depeds oly o β Sol : (a) Use A is a skew symmetric matri of odd order. 6. If a + b + c = -, ad A = (b) A - depeds oly o α (d) oe of these a ( b ) ( c ) (a ) b ( c ), the A is o-sigular if ( a ) ( b ) c (a) = 0 (b) (c) (d) Sol : (a, b, d) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 4

7. Suppose A ad B are two osigular matrices such that AB = BA ad B 5 = I, the (a) A 3 =I (b) A 3 = I (c) A 30 = I (d) A 50 = I Sol : (b) 8. Let A = (a) f( λ ) = a c b d ad f ( λ )= det (A - λ I), the λ - (a + d) λ + ad bc (b) f (a) = 0 (c) A = 0 implies A r = 0 r Sol : (a, b, c) (d) oe of these 9. If α, β, γ are the roots of 3 + a + b = 0. The the determiat α β γ β γ α equals γ α β (a) a 3 (b) a 3-3b (c) a 3b (d) a 3 Sol : (d) We have α + β + γ = a ad β γ + γ α + α β = 0. Usig C C + C3, we get α β γ β γ β γ β γ α β γ γ α a γ α a 0 γ β α γ α β γ α β α β 0 α β β γ a[ (γ β) (α γ)(α β) a[α β γ (βγ αγ αβ)] = a [( α + β + γ ] -3 (β γ + γ α + α β)] = a [(-a) 3 (0)] = a 3. usig C C C ad C C C 3 3 0. The umber of positive itegral solutios of the equatio =, where y z z y z z y y is PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 5

(a) 3 (b) 5 (c) 4 (d) 3 3 Sol : (a) Usig C C + C + C3, we get y z z y z z y y y z y z z 0 z = = yz = yz y y 0 0 y z 0 z y 0 0 0 0 0 0 Takig commo ad applyig C C C, C3 C3 C, we get 0 0 4yz 0 0 = 4 yz Now, 4yz = yz = 3 Usig C C C, C3C3 C, we get [usig C C +C + C3] [usig C C + C + C3] The umber 3 ca be assiged to ay of, y, z. Therefore, the umber of positive itegral solutio of = is 3. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 6

. If (a) a iteger 6 i 3 6 3 8 i 3 6 i (c) a irratioal umber Sol : (a) Takig 8 i 7 i 6 commo formc, we get i 3 6 6 3 i 3 6i 3 3i 3 3 i the is (b) a ratioal umber (d) a imagiary umber Applyig R R - R, ad R3 R3 3 R, we i 3 6 6 0 3 6i 3 0 i 3 = 6 3 3 3 = 6( 3 6 6) = 6, which is a iteger. [usig C C - i C] PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 7

. Let f() = π si π ( )! si cos cos π π π the value of 3 (a) 0 (b) (c) Sol : (a, d) we kow that d d π = ( )!π ; d π [si π] π si π d, ad d [cos π ] d Thus f () = f () = π cos π π ( )!π π π π si π π cos π π π 3 ( )! si cos π π ( )!π π si π π cos π π π ( )! si cos 3 = 0 [ R ad R3 are proportioal] This also show that f () is idepedet of a. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 8 d d [f ()]at = is (d) idepedet of

3. The determiat a ab ac ab b bc ac bc c is divisible by (a) (b) (c) 3 (d) oe of these Sol : (a, b) We ca write as 3 a a ab ac a b b bc a a c bc c Applyig C C + bc + cc3 ad takig a + b + c + commo, we get a ab ac (a b c ) b b bc a c bc c Applyig C C + bc ad C3 C3 cc, we get a 0 0 (a b c ) b 0 a (a b c ) (a ) = a = (a + b + c + ) Thus ad. 4. If c 0 z z ( y)/ z (y z)/ / / y(y z)/ z ( y z)/ z ( y)/ z (a) is idepedet of (c) depedets oly o z (d) = 0 Sol : (a, b, d) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 9, the (b) is idepedet of y Multiplyig C by, C by y ad C3 by z, we obtai

y y z z z y z y z yz z y(y z) y( y z) y( y) z z z Applyig C C + C + C3 we get 0 y / z ( y)/ z 0 y / z / yz 0 y( y z)/ z y( y)/ z = 0 This shows that is idepedet of, y ad z. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 0

5. If p 0, solutio set of the equatio p p p 3 = 0 is (a) {, } (b) {, 3} (c) {, p, } (d) {,, -p} Sol : (a) Applyig C C C, we get 0 p 0 p ( ) p p 3 = ( ) ( )p ( ) 0 p p Thus = 0 =,. 6. If a b c a a b c a a b c 3 3 3 bb c b c bb3 c3 c3 c c3 c c3 ad is equal to a b c a b c a b c 3 3 3 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH, the (a) a b c3 (b) a a a3 (c) a3 b c (d) a b c+ abc+ a3b3c3 Sol : (a) Takig c3 commo form R3 ad applyig RR R3 ad R R3. We obtai a b a a b a a b 3 3 c3 bb b bb3 c c c 3 Takig b commo from R ad applyig R R R we get

a a a a a 3 b c b b b 3 3 = a b c3 c c c 3 a a a 3 b b b 3 c c c 3 = abc3 = abc3 0 if is a it eger 7. Let f () = [] ad g() the otherwise (a) (g o f) () = (c) f o g is differetiable o R~I Sol : (c) g o f () = 0 for all R ad f o g () = 0 if is a it eger if R ad or R ad So, f o g is differetiable o R I. (b) g of is ot cotiuous (d) f o g is a differetiable fuctio f( ) 8. Let f : R R be such that f () = 3 ad f () =6. The lim equals. 0 f() (a) (b) e / (c)e (c) e 3 Sol : (b) f( ) lim 0 f() 0 lim u / f() f (). u f() PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH /

where u = f( ) f() f( ) f() r f '() 0 0 as 0. f() 3 3 f ( ) f (). u f () = lim ( u) u0 = e e e f'()/f () 6/3 Note that we caot take the logarithm as the fuctio may take egative values ad also the L Hospital rule is ot applicable. e 9. Let f: R R be a fuctio defied by f() e (a) f is both oe-oe ad oto (c) f is oto but ot oe-oe e e. The (b) f is oe-oe but ot oto (d) f is either oe-oe or oto. Sol : (d) f is ot oe-oe as f (0) = 0 ad f( ) = 0. f is also ot oto as for y = there is o R such that f () =. If there is such a R the e - e - = e + e -. Cleary 0. For >0, this equatio gives e - = e - which is ot possible. For < 0, the above equatio gives e = e - which is also ot possible. 0. The graph of the fuctio f() = loga is symmetric about (a) ais (b) origi (c) y ais (d) the lie y = Sol : (b) f () = loga loga log a = - f(). Hece f is a odd fuctio, therefore, the graph of f is symmetric about origi. Sice f - () = a a, so graph of f, caot by symmetric about the lie y =. No-symmetry about -ais ad y-ais is clear.. Let f() = log ( si + cos ). The rage of f() is (a) [, 0] (b) [0, /] (c) [ /, 0] (d) [0, ] Sol : (b) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 3

Let g() = si + cos is defied for all real ad is periodic with period π/. For [0, π/] g() = si + cos = si ( + π/4). Sice g(0) = ad g(π/4) = ad g icrease o [0, π/4] ad decrease o { π/4, π/], so the rage of g is [, ]. Hece the rage of f is [0, /].. The rage of the fuctio f () = cos [], for π/< < π/ cotais (a) {-,, 0} (b) {cos,, cos } (c) {cos, - cos, } (d) [-, ] Sol : (b) For the give rage of, we have [] = - for - π/ < < - f() = cos(-) = cos, [] = - for - < 0 f () = cos (-) = cos, [] = 0 for 0 < f () = cos 0 =, ad [] = for π/ f () = cos, 3. If [] deotes the greatest iteger less tha or equal to the lim [k,] equals (a) / (b) /3 (c) (d) 0 Sol : (a) For ay iteger k, k < [k] < k + (k ) [k] (k ) k k k ( ) ( ) [k] k ( ) [k] k < ( ) Takig limit as, we have PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 4 k

lim [k] k Hece lim [k]. k 4. The iteger of for which lim 0 (cos )(cos e ) is a fiite ozero umber is (a) (b) (c) 3 (d) 4 Sol : (c) (cos )(cos e ) 4 4 3........0! 4!! 4!! 3! 4 3......! 4! 3!... 3! 4! 3! For lim 0 (cos ) (cos e ) 5. If f() = (/), the o the iterval [0, π] (a) ta (f ()) ad are both cotiuous. f() (b) ta (f()) ad are both discotiuous f() (c) ta (f()) ad f - () are both cotiuous (d) either ta f() or f - () is cotiuous to eist we must have 3 = 0 = 3. Sol : (c) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 5

f () = (/) is cotiuous o [0, π] ad y = ta is cotiuous o [-, π/ - ] (=rage of f), therefore, ta (f()) is cotiuous o [0, π] as /f() is ot defied at =, so /f is ot defied at =, thus /f() is ot cotiuous o [0, π]. Now f - () = ( + ) which is clearly cotiuous o [0, π]. Hece (c) is true ad (d) is ot. 6. The value of (a) e 4 Sol : (a) Let y = ( + ) / lim 0 / ( ) e e (b) - e 4 log y = 3 4 log (+ ) =... 3 4 = - y = e 3... 3 4...... 3 3 ee e......... 3! 3 y e + 3 e e 0() 0()... y e e lim e e 0 3 8 4 is PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 6 (c) e 4 8 7. The lim 0 3 (where [] is greatest iteger fuctio) is (a) a ozero real umber (b) a ratioal umber(c) a iteger Sol : (b, c, d) Sice [] for all R so (d) oe of these (0() i terms cotaiig ) (d) zero

3 3 3 8 5 5 3 3 for all. But lim 5 = 0 = lim ( 5 8 ) so lim 0 0 8 3 8. If A = 0 0 I Q. ta π lim lim the (a) A > 3 (b) A > 4 (c) A < 4 (d) A is a trascedetal umber Sol : (a, b, d) ta π ta π( ) lim lim π ta y lim π [y π( )] y0 y ad (/) lim 0 lim lim e ta π A lim lim π 4 ad is a trascedetal umber. 9. Let f() = at = is (log( ) log)(3.4 3) /3 /,. The value of f() so that f is cotiuous {(7 ) ( 3) }si π (a) a algebraic umber (b) a ratioal umber (c) a trascedetal umber (d) - 9 log e π 4 Sol : (c, d) we have PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 7

(log( ) log)(3.4 3) /3 / lim {(7 ) ( 3) }si π (log( y) log)(3.4 3) y0 /3 / lim {(7 ) ( 3) }si [y = - ] y y log.(3(4 ) 3y) lim y0 /3 / y 3 y si π y 8 4 y log( y / ) 3(4 ) lim. 3 y0 y y π y y y 3 4 8 si π y y O(y ) 9 4 = {3log 4 3}( 3) log π π e which is trascedetal umber. 30. Let f() = lim, the (a) f() = for > (b) f () = - for < (c) f () is ot defied for ay value of (d) f () = for = Sol : (a, b) If >, the lim. Thus for >, f () = lim. If <, the If = the lim = 0, therefore for <, f() = -. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 8

= for ay ad therefore f() = 0 3. Let f: R R be give by f() = 5, if Q ad f() = + 6 if R Q the (a) f is cotiuous at = ad = (b) f is ot cotiuous at = ad = (c) f is cotiuous at = but ot at = (d) f is cotiuous at = but ot at = Sol : (d) f is cotiuous at = as lim ' as ad f () = 5 f() = 0 ad f() = 5 = 0. Let = + ad ' 5 as. Also ad ad f ( ) = 6 7 as. Therefore f is ot cotiuous at =. 3. If g() is a polyomial satisfyig g() g(y) = g() + g(y) + g(y) for all real ad y ad g() = 5 the lim 3 g() is (a) 9 (b) 5 (c) 0 (d) oe of these Sol : (c) Puttig = ad y =, we have g () g() = g() + g() + g() 5 g () = 8 + g() g () =. Puttig y = / i the give equatio, we have g() g(/) = g() + g(/) + g() = g() + g(/). Hece, g() = + or g() = - +. But 5 = g() = - + so - = 4 which is ot possible. Thus 5 = g() = +. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 9

=. Hece, lim g() lim ( + ) = 0. 3 3 33. Let (a) = 4 ad (b) = 6. The the umber of oe-oe fuctio from A to B is (a) 0 (b) 360 (c) 4 (d) oe of these Sol : (b) Let f : A B be ay fuctio. Suppose that A = {,, 3, 4} ad B = {y, y,, y6}. Now f() ca be chose to be as ay oe of y, y y6 say f() = y, f () ca be chose as ay oe of y, y6 so those are five choices for f(). This meas that we ca have 6.5.4.3 = 360 such fuctios from A to B. Remarks: If (a) = m ad (b) = m. The the umber of oe-oe fuctios from A to B is! ( m)! PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 0

34. If f() is a polyomial satisfyig f() f(/) = f() + f(/), ad f(3) = 8, the f(4) is give by (a) 63 (b) 65 (c) 67 (d) 68 Sol : (b) By cosiderig a geeral th degree polyomial ad writig the epressio f (). f(/) = f()+f(/) i terms of it, it ca be proved by comparig the coefficiets of, -, ad the costat term, that the polyomial satisfyig the above equatio is either of the form + or +. Now, from f(3) = 3 + = 8, we get 3 = 7, or = 3. But f(3) = -3 + = 8 is ot possible, as -3 = 7 is ot true for ay value of. Hece f(4) = 4 3 + = 65. 35. The equatio y = 4 ad + 4y + y = 0 represet the sides of (a) a equilateral triagle (c) a isosceles triagle Sol : (a) (b) a right agled triagle (d) oe of these Acute agle betwee the lies + 4y + y = 0 is ta - 4 ta 3 π /3 Agle bisector of + 4y + y = 0 are give by y y = ± y y 0 As + y = 0 is perpedicular to y = 4, the give triagle is isosceles with vertical agle equal to π/3 ad hece it is equilateral. 36. The lie + y = meets -ais at A ad y-ais at B. P is the mid-poit of AB. P is the foot of the perpedicular form P to OA; M is that from P to OP; P is that from M to OA; M is that from P to OP; P3 is that from M to OA ad so o. If P deotes the th foot o the perpedicular o OA form M-, the OP = PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH

(a) / (b) / (c) / / (d)/ Sol : (b) + y = meets -ais at A(, 0) ad y-ais at B(0, ) The coordiates of P are (/, /) ad PP is perpedicular t OA. OP - P P = / Equatio of lie OP is y =. we have (OM-) + (OP) + (P m - ) = (O P) = P (say) Also, (OP ) = (OM-) + (P- M ) = P P p p p p 4 OP p p p... p 37. The coordiates of the feet of the perpediculars from the vertices of a triagle o the opposite sides are (0, 5), (8, 6) ad (8, 9). The coordiates of a verte of the triagle are (a) (5, 0) (b) (50, -5) (c) (5, 30) (d) 0, 5 Sol : (a, b, c) we use the fact that the orthoceter O of the triagle ABC is the icetre of the pedal triagle DEF. Let (h, k) be the coordiates of O. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH

Now ED = FD = 0 ad EF = 7 so that h = ad k = (08) (5 6) = 5 700858 705 = 0 750659 = 5 705 thus the coordiates of 0 are (0, 5) 0 8 Sice AC is perpedicular to OE, equatio of AC is y 6= ( - 8) 56 y = 0 () Similarly equatio of AB is y 9 = - 0 8 ( - 8) 5 9 3y + 35 = 0 () ad equatio of BC is y 5 = - 0 0 (0) 5 5 y + 45 = 0 (3) Solvig () ad () we get a (5, 0) From () ad (3) we get B (50, -5) ad from (3) ad () we get C (5, 30) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 3

38. If y = m bisect the agle betwee the lies (ta θ + cos θ) + y ta θ y si θ = 0 whe θ = π/3 the value of m is (a) 7 3 Sol : (a, b) (b) 7 3 (c) 7 (d) 3 Equatio of the bisector of the agles betwee the give lies is y y ta θ cos θ si θ taθ y y ta θ taθ y y wheθ π /3 3 3 which is satisfied by y = m if m m 4 3 3 m + 4m - 3 = 0. m = 7 3 39. The lie L has itercepts a ad b o the coordiate aes. The coordiate aes are rotated through a fied agle, keepig the origi fied. If p ad q are the itercepts of the lie L o the ew aes, the is equal to a p b q (a) (b) 0 (c) (d) oe of these Sol : (b) Equatio of the lie L i the two coordiate systems is y, X y a b p q where (X, Y) are the ew coordiates of a poit (, y) whe the aes are rotated through a PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 4

fied agle, keepig the origi fied. As the legth of the perpedicular from the origi has ot chaged. a b p q a b p q 0. a p b q or 40. A circle C of radius b touches the circle + y = a eterally ad has its cetre o the positive -ais; aother circle C of radius c touches the circle C eterally ad has its cetre o the positive -ais. Give a<b<c, the the three circles have a commo taget if a, b, c are i (a) A. P. (b) G.P. (c) H.P. (d) oe of these Sol : The cetre of C is (a + b, 0) ad the cetre of C is (a + b + c, 0) Let y = m + k be a taget commo to the three circles. Sice it touches + y = a, C ad C K m(a b) k a, b m m ad m(ab c) k c m As the cetre of the three circles lie o the same side of the lie y = m + k, takig the same sig, say positive, i the three relatios we get, PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 5

k a b m (a b) c m(a b c) m a b a b c b a c a (a b)(c a) (b a)(a b c) (elimiatig m) ac a bc ba ba a b ab bc ac ac b ac b a, b,c are i G.P. 4. If the lie cos α + y si α = p represet the commo chord APQB of the circles + y = a ad + y = b (a>b) as show i the fig. the AP is equal to (a) Sol : (c) a p b p (b) a p b p (c) a p b p (d) a p b p The give circles are cocetric with cetre at (0, 0) ad the legth of the perpedicular from (0, 0) o the give lie is p. Let OL = p the Al = ad PL = AP = (OA) (OL) a p (OP) (OL) b p a p b p 4. C ad C are circles of uit radius with cetres at (0, 0) ad (, 0) respectively. C3 is a circle of uit radius, passes through the cetres of the circles C ad C ad have its cetres PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 6

above -ais. Equatio of the commo taget to C ad C3 which does ot pass through C is (a) - 3 y + = 0 (b) 3 y + = 0 (c) 3 y = 0 (d) + 3 y + = 0 Sol : (b) Equatio of ay circle through (0, 0) ad (, 0) is ( - 0) ( - ) + (y - 0) (y - 0) + y λ 0 0 0 0 + y + y = 0 If it represet C3, its radius = = (/4) + ( λ /4) = 3 As the cetre of C3, lies above the -ais, we take λ = 3 ad thus a equatio of C3 is + y - 3 y = 0 Sice C ad C3 itersect ad are of uit radius, their commo tagets are parallel to the lie joiig their cetres (0, 0) ad (/, 3 /). So, let the equatio of a commo taget be 3 -y + k = 0 It will touch C, if PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 7

k k 3 From the figure, we observe that the required taget makes positive itercept o the y-ais ad egative o the-ais ad hece its equatio is 3 -y + = 0. 43. If the circle + y + g + fy + c = 0 cuts each of the circles + y 4 = 0, + y 6 8y + 0 = 0 ad + y + 4y =0 at the etremities of a diameter, the (a) c = -4 (b) g + f = c (c) g + f - c = 7 (d) gf = 6 Sol : (a, b, c ad d) Sice the circle + y + g + fy + c - 0 cuts the three give circles at the etremities of a diameter, the commo chords will pass through the cetre of the respective circles, so that g + fy + c + 4 = 0 passes through (0, 0) c = - 4 () Net g + fy + c + 6 + 8y - 0 = 0 passes through (3, 4) (g + 6)3 + (f + 8)4 4 = 0 3g + 4f+ 8 = 0 (ii) ad g + fy + c - + 4y + = 0 passes through (-, ) (g - ) (-) + (f + 4) - = 0 g f - 4 = 0 (iii) From (ii) ad (iii) we get g = - ad f = - 3. g + f = c = - 5 g + f - c = 4 + 9 + 4 = 7. g f = 6. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 8

44. A straight lie through the verte P of a triagle PQR itersects the side QR at the poit S ad the circumcircle of the triagle PQR at the poit T. If S is ot the cetre of the circumcircle the (a) PS ST QS SR Sol : (b, d) (b) (c) 4 (d) PS ST QS SR PS ST QR Sice S is ot the cetre of the circum-circle PS ST,QS SR sice A.M > G.M. PS ST PSST PS ST QS SR [ PSST QS SR ] We kow that if +y = k, a costat, the y is maimum if = y = k/. Thus QSSR < 4 (QR) 4 QS SR (QR) 4 PS ST QS SR QR 45. The value of si - 3 4 4 cot si cos sec is equal to 4 PS ST QR (a) π/4 (b) π/6 (c) 0 (d) π/ PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 9

Sol : (c) We have 3 4 3 3 4 8 3 3 π 4 si si Also cos - 3 π cos 4 6 ad sec - π / 4 so the give epressio is equal to si - π π π 6 4 cot si cot 0 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 30

46. If ta - (a) Sol : (c) ta - ta ta... ta = ta - θ, the θ = ()(3) (3)(4) ( ) (b) ta ( ) ( ) = ta - ( +) ta - () so that L.H.S. of the give equatio is PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 3 (c) ta - ta - + ta - 3 ta - +. + ta - ( + ) ta -. = ta - ( + ) ta - = ta - =ta - ( ) so that ta - = ta - θ θ. 47. If a si - b cos - = c, the α si - + b cos - is equal to (a) 0 (b) Sol : (d) πab c(b a) a b we have b si - + b cos - = bπ ad a si - b cos - = c (give) (a + b) si - = bπ c si - = (bπ)/ c a b Similarly cos - = (aπ)/ c a b (c) π / (d) (d) πab c(a b) a b

πab c(a b) so that a si - + b cos - = a b 48. cos - is equal to (a) si - Sol : (a, c) (b) cos - Let cos - = y, so that = cos y. The si - Also cos - cos y si (y /) y si Y si cos y cos (y / ) y cos y cos y cos si cos y y (c) cos - (d) si - 49. If α ad β are the roots of the equatio (ta - (/5)) + ( 3 -) ta - (/5) - 3 = 0, α > β the (a) α β 5π / (b) α β 35π / (c) Sol : (a, b, c, d) (ta - (/5) + 3 ) (ta- (/5) -) = 0 ta - (/5) = - 3 π 5π 5 3 3 ad ta - (/5) = π 5π 5 4 4 Let α = - 5π, β 5π 3 3 αβ 5π / PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 3 (d) 3α 4β 0

α + β = - 5 π/, α - β = 35π/ α β = - 5π / ad 3 α + 4 β = 0. 50. If < <, the umber of solutios of the equatio ta - ( - ) + ta - + ta - ( + ) = ta - 3 is (a) 0 (b) (c) (d) 3 Sol : (a) The give equatio ca be writte as ta - ( - ) + ta - ( + ) = ta - 3 ta- ta - 3 ta ()( ) 3 3 3 3 3 4 3 = 0 (4 - ) = 0 = 0, = / oe of which satisfies < <. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 33

Aswers 3 4 5 6 7 8 9 0 3 4 5 6 7 8 9 0 3 4 5 6 7 8 9 30 3 3 33 34 35 36 37 38 39 40 4 4 43 44 45 46 47 48 49 50 Good Luck PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 34