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1 + {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 =

2 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 = 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 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

4 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

5 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 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

6 (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 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

7 . If (a) a iteger 6 i i 3 6 i (c) a irratioal umber Sol : (a) Takig 8 i 7 i 6 commo formc, we get i 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 i 3 0 i 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

8 . 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

9 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

10 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

11 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 bb c b c bb3 c3 c3 c c3 c c3 ad is equal to a b c a b c a b c 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

12 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 /

13 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

14 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

15 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! ! 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

16 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 (+ ) = = - y = e ee e ! 3 y e + 3 e e 0() 0()... y e e lim e e is PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 6 (c) e 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

17 for all. But lim 5 = 0 = lim ( 5 8 ) so lim 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

18 (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 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

19 = 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

20 =. Hece, lim g() lim ( + ) = 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 = 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

21 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) = = 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

22 (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

23 Now ED = FD = 0 ad EF = 7 so that h = ad k = (08) (5 6) = = = 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 = ( - 8) 5 9 3y + 35 = 0 () ad equatio of BC is y 5 = (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

24 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 m + 4m - 3 = 0. m = 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

25 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

26 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

27 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 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

28 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 + = 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 y - 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 = = 7. g f = 6. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 8

29 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 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

30 Sol : (c) We have 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

31 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

32 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 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 ad ta - (/5) = π 5π 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

33 α + β = - 5 π/, α - β = 35π/ α β = - 5π / ad 3 α + 4 β = 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 ()( ) = 0 (4 - ) = 0 = 0, = / oe of which satisfies < <. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 33

34 Aswers Good Luck PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 34

JEE(Advanced) 2018 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 20 th MAY, 2018)

JEE(Advanced) 2018 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 20 th MAY, 2018) JEE(Advaced) 08 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 0 th MAY, 08) PART- : JEE(Advaced) 08/Paper- SECTION. For ay positive iteger, defie ƒ : (0, ) as ƒ () j ta j j for all (0, ). (Here, the iverse