Mthemtis 7. f (x) g (x) (g) If (x) f (x) g (x) then f (x) g f (x) (x) g (x) (x) + or f (x) g (x) f (x) g (x) f (x) g (x) f (x) g (x) + f (x) g (x) f (x) g (x) (h) If f(x) g(x) h(x) (x) α β γ then f(x)dx g(x)dx h(x)dx (x)dx α β γ Solved Exmples JEE Min/Bords Exmple : Prove tht R.H.S. pqr( + + ). (ii) From eq. (i) nd (ii), we get p q r q r p pqr. Use p + q + r 0. r p q Sol: By using the expnsion formul of determinnts we n prove this. p q r L.H.S. q r p r p q r p q p q r p q + r p q r q r p p( qr p ) q(q pr ) + r(pq r ) pqr p q + pqr r pqr( + + ) (p + q + r ) p+ q+ r 0 (given) (p + q + r) 0 p + q + r pqr 0 p + q + r pqr L.H.S. pqr( + + ) (pqr) L.H.S. pqr( + + ). (i) R.H.S. pqr + pqr[( ) ( ) + ( )] pqr[ + + ] L.H.S. R.H.S. Exmple : Prove tht the determinnt x sinθ osθ sinθ x osθ x is independent of θ. Sol: Simply y expnding the given determinnt we n prove it. x sinθ osθ We hve, sinθ x osθ x x sinθ sinθ x x sinθ + osθ x osθ x osθ x( x ) sin θ( xsinθ os θ ) + os θ( sinθ+ xos θ) x x + xsin θ+ sinθ osθ sinθ osθ+ xos θ x x + x(sin θ+ os θ ) x x+ x Thus, the determinnt is independent of θ. x+ x x Exmple : Solve the eqution x x+ x 0, 0. x x x+ Sol: We n expnd the ove determinnt y pplying the invrine nd slr multiple properties, nd hene we n esily solve this prolem.
7. Determinnts We hve, x+ x x x x+ x 0 x x x+ Here, Opertion: C C + C + C x + x x x x x+ x+ x 0 (x+ ) x+ x 0 x+ x x+ x x+ Operting R R R, x x We get (x + ) 0 0 0 0 0 0 (x + ) 0 0 x + 0, [ 0] Hene Proved. R R R (x + ) 0 x Exmple 4: Solve, using Crmer s rule x y + 4z 5; x + y + z ; x + y z Sol: By defining D, D, D, D nd y using Crmer s Rule we will get required result. 4 D 5 5 4 D 5 D 5 4 D, D By Crmer s Rule, x, D 5 5 D y ; D 5 5 D z D 5 5 Exmple 5: Solve the following system of equtions y Crmer s Rule x y+ z 9; x+ y+ z 6; x y+ z Sol: By defining, x, y, z nd y using Crmer s Rule we will get the required result. ( + ) + ( ) + ( ), 9 x 6 9( + ) + (6 ) + ( 6 ) 9 y 6 (6 ) 9( ) + ( 6) 4 9 z 6 ( + 6) + ( 6) + 9( ) 6 By Crmer s Rule x y z x, y, z Exmple 6: Show tht + + + + ( + + ) + + Sol: By using invrine nd slr multiple property we n expnd given determinnt nd n prove it. ( + + ) (+ + ) + + ( + + ) + + [C C + C + C ] (+ + ) + + + + (+ + )0 + + 0 0 + + [y R R R nd R R R ] ( + + )[{( + + ) 0}] ( + + )( + + ) ( + + ) Exmple 7: Using determinnts, show tht the points (, 7), (5, 5) nd (, ) re olliner.
Mthemtis 7. Sol: If these points re olliner then the re of tringle mde y joining these points will e zero. The re of the tringle formed y the given points 7 5 5 Operte: R R R; R R R 6 0 6 0 0 0 ( R nd R re identil) Hene, the given points re olliner. Exmple 8: If A nd B re two mtries suh tht AB B nd BA A, then A + B. Sol: By using the multiplition property of mtries we n solve given prolem. A + B AA + BB A(BA) + B(AB) [Given AB B nd BA A] (AB)A + (BA)B [Mtrix multiplition is ssoitively]ba + AB [Given AB B nd BA A] [Given AB B nd BA A] Exmple 9: Find the vlue of A+B 4 4 5 4 5 6 4 5 6 7 Sol: By pplying the invrine property we n find the vlue of the given determinnt. 4 4 9 6 4 9 6 4 5 4 9 6 5 5 7 9 4 5 6 9 6 5 6 5 7 9 4 5 6 7 6 5 6 49 7 9 [Applying R4 R4 R, R R R, R R R] 4 9 6 5 7 9 [Applying 0 R4 R4 R, R R R ] JEE Advned/Bords Exmple : Without expnding, evlute the determinnt sinα osα sin( α+δ) sinβ osβ sin( β+δ) sinγ os γ sin( γ+δ) Sol: By using the formul sin(a+ B) sin A os B + os A sin B nd invrine property of determinnts we n expnd the given determinnt. sinα osα sin( α+δ) Let sinβ osβ sin( β+δ) sinγ os γ sin( γ+δ) sinα osα sinαosδ+ osαsinδ sinβ osβ sinβosδ+ osβsinδ sinγ os γ sinγosδ+ os γsinδ sinα osα 0 sinβ osβ 0 sinγ os γ 0 [Applying C C osδ. C sin δ. C ] 0 [ C onsists of ll zeroes] Exmple : By using properties of determinnts prove tht x x x x ( x ) x x Here in this prolem y using invrine nd slr multiple properties we will expnd the given determinnt nd we will prove it. Sol: L.H.S. x x x x x x [Applying C C + C + C] x x ( + x + x ) x x x x ( + x + x ) 0 x x x 0 x x x + x+ x x x + x+ x x + x+ x x
7.4 Determinnts [Applying R R R nd R R R ] ( + x + x )(){( x)( x ) (x x)(x x )} ( + x + x )( x) { + x + x } {( x)( + x + x )} ( x ) R.H.S. Exmple : Show tht x ( + + ) is one root of the eqution: eqution ompletely. x+ x+ 0 x+ nd solve the Sol: We n expnd given determinnt using the invrine nd slr multiple properties nd y solving we will find out required result. By C C + C + C, we get x+ + + x+ + + x+ 0 x+ + + x+ (x+ + + ) x+ 0 x+ (x+ + + )0 x + 0 0 x+ R R R ; R R R On expnding y first olumn, we get (x + + + )[(x + )(x + ) ( )( )] 0 (x + + + )[x ( ) ( + )] 0 (x + + + )(x + + + ] 0 (x + + + )(x + + + ) 0 Either x + + + 0 x ( + + ) or x + + + 0 x ± + + Exmple 4: If the re of tringle is 5 sq. units with verties (, 6), (5, 4) nd (k, 4), then find k. Sol: As we know tht the re of the tringle d e f where (, ) (, d) (e, f) re the verties of tringle. Therefore y sustituting the vlue of verties we will get required result. Let the verties of tringle e A(, 6), B(5, 4) nd C(k, 4). Sine the re of the tringle ABC is 5 sq. units, we hve, 6 5 4 ± 5 k 4 6 0 0 ± 5 k 0 0 [Applying R R R nd R R R ] 0 k 0 ±5 [Expnding long C ] {0 0(k )} ± 5 0 0k + 0 ± 70 0k 50 70 k + or k Exmple 5: Solve the following system of equtions y using determinnts: x+ y+ z, x + y + z k ; x+ y+ z k Sol: Here in this prolem first define D, D, D nd D. then y using Crmer s rule we n solve it. We hve, D 0 0 [Applying C C C nd 0 0 ( )( ) + + ( )( ).. + + [Expnding long R ] ( )( )( + ) C C C ] ( )( )( ) (i)
Mthemtis 7.5 D k ( )( k)(k ) k [Repling y k in (i)] D k (k )( )( k) k [Repling y k in (i)] nd D k ( )( k)(k ) k [Repling y k in (i)] D ( )( k)(k ) ( k)(k ) x, D ( )( )( ) ( )( ) D (k )( )( k) (k )( k) y D ( )( )( ) ( )( ) D ( )( k)(k ) (k )( k) z D ( )( )( ) ( )( ) Exmple 6: Show tht + + + + + + + + + + + + + ( + ) + ( ) ( ) i i i j j i i < i j Sol: By putting α, β, then α + β u, v, then u + v. Using the invrine property expnd the given determinnt, nd then ompring it to the R.H.S. of the given prolem we n prove it. Let Now R.H.S. i i i j j i i < i j + ( + ) + ( ) ( ) + ( + + + + + ) + ( )( ) + ( )( + ) + ( )( ) + ( + + + + + ) αu βv ( α+β )(u + v) + ( + + + + + ) αu βv βu α v.... (i) +α α α Now L.H.S. β +β β + + + + [R R R, R R R ] α β u v + + [ C C C,C C C] 0 α β u + v v + + [( + + ) (u + v )( β )] [ C C + C] + [(v ) (u + v )] + ( +β ) uβ vβ+β+ u +v +α ( v + u) + ( + ) + β+ u + v +α uβ vβ αv uα + ( +β ) + ( ) + + ( β ) + αu βv uβ vα + ( + + + + + ) αu βv uβ vα RHS [From (i)] Exmple 7: Find vlues of for whih the equtions x + y ; ( + )x + ( + 4)y + 6 ( + ) x + ( + 4) y ( + 6) re onsistent nd hene solve the eqution. Sol: Here in this prolem first define given equtions s nd solve it s 0 y using the invrine method. The eqution will e onsistent, if + + 4 + 6 0 ( + ) ( + 4) ( + 6) Applying C C C, we get 0 + + 4 0 ( + ) ( + 4) 4( + 5) Solving, we get + 0 0 or 0, 0 (i) If 0, the system of equtions eomes x + y x + 4y 6 x, y. (ii)
7.6 Determinnts If 0, then system of equtions eomes x + y x 8x 6y 4 6x + 9y 4, 4 y. (iii) Hene the solutions re given y (ii) nd (iii). Exmple 8: If ( r, r ), r,, e the verties of tringle, prove tht ( ) + ( ) ( ) + ( ) 0 (i) ( ) + ( ) nd hene show tht the ltitudes of tringle re onurrent. Sol: Using the invrine method we n expnd the given determinnt nd using the equtions of ltitude we n prove it ( ) + ( ) ( ) + ( ) 0 ( ) + ( ) [Applying R R + R + R ] 0 0 0 ( ) + ( ) ( ) + ( ) F B(, ) A(, ) D Eqution of ltitude AD is: y (x ) E C(, ) or x( ) + y( ) ( ) + ( ) (ii) Similrly eqution of ltitudes BE nd CF re x( ) + y( ) ( ) + ( ). (iii) x( ) + y( ) ( ) + ( ). (iv) Altitudes (ii), (iii), (iv) re onurrent, sine the determinnt given y L.H.S. of (i) is zero. Exmple 9: Let λ nd α e rel. Find the set of ll vlues of λ nd α for whih the system of liner equtions λ x + (sin α )y + (os α )z 0 x + (os α )y + (sin α )z 0 x + (sin α)y (os α )z 0 hs non-trivil solution. If λ, find ll vlues of α. Sol: Here in this prolem first define the given equtions s nd s we know tht for non-trivil solution 0. For non-trivil solution, ondition is 0. or λ sinα osα osα sinα 0 sinα osα λ [ os α sin α] sin α[ osα+ sin α] + os α[sinα+ os α ] 0 or λ sinα+ osα α R ; λ If λ, then sinα+ osα π π os α os 4 4 π π π π α nπ± : n I α nπ± + : n I 4 4 8 8 Exmple 0: For fixed positive integer n, if n! (n + )! (n + )! (n + )! (n +!) (n + )! (n + )! (n + )! (n + 4)! then show tht (n!) 4 is divisile y n. Sol: By using the slr multiple property of determinnts we n tke (n!),(n+) nd (n + ) ommon nd using the invrine property we n solve the given prolem. n + (n + )(n + ) (n!) n + (n + )(n + ) (n + )(n + )(n + ) (n + )(n + ) (n + )(n + )(n + ) (n + 4)(n + )(n + )(n + ) Tking (n+) nd (n + ) (n + ) ommon from C nd C respetively, we get (n!) (n + )(n + ) n+ n+ n+ (n + )(n + ) (n + )(n + ) (n + 4)(n + ) [Apply C C C nd C C C then
Mthemtis 7.7 (n!) (n + ) (n + ) 0 0 n+ (n + )(n + ) (n + ) 4n + 0 (n!) (n + ) (n + )[4n + 0 4(n + )] (n!) (n + ) (n + ) (n!) (n + n + )(n + 4) (n!) 4 n 8n 0n, + + n(n + 4n + 5), whih is divisile y n. (n!) (n + 8n + 0n + 4) JEE Min/Bords Exerise Q. Find x, if Q. It mtrix A Q. Given x 4 8 x 4. 4 0 4 0, find [A]. 0, find (i) M (ii) C. Q.4 Are of tringle with verties (k, 0), (, ) nd (0, ) is 5 sq. units. Find the vlue(s) of k. Q. Find the djoint of mtrix A Q. Find the inverse of mtrix. 5 4, if possile. 6 8 8 Q. Without expnding, find the vlue of 4 6. 5 4 x 0 Q.4 If 4 is singulr mtrix, find x. 0 Q.5 Find the re of the tringle whose verties re (, ) (4, ) nd ( 5, 4). Q.5 Find the re of tringle, whose verties re (0, ) (, 4), (, 6). Q.6 Given determinnt Find the vlue of C + C + C.. Q.6 Find the vlue of x, if re of tringle is 5 squre ms with verties (x, 4), (, 6) nd (5, 4). Q.7 Show tht the following determinnt vnishes: 5 5 5 7 0. 8 4 4 Q.7 Find the vlue of p, suh tht the mtrix is singulr. Q.8 Given I. Find I. Also find I. 4 4 p Q.8 Using properties of determinnts, prove tht : + + 0. + Q.9 Find the vlue of x, suh tht the points (0,), (, x), (, ) re olliner. Q.0 For two given squre mtries A nd B of the sme order, suh tht A 0 nd B 0, find AB. Q.9 If points (, 0), (0, 5) nd (x, y) re olliner, then x y show tht +. 5 Q.0 If for mtrix A, A find 5A, where mtrix A is of order.
7.8 Determinnts Q. Given A C + C., suh tht A 0. Find Q. Without expnding prove tht, the vlue of determinnt + + is zero. + Q. A is non-singulr mtrix of order nd A 4. Find dj.a. Q.4 Is it possile to find the inverse of mtrix 5? Given resons. 0 Q.5 Given squre mtrix A of order, suh tht A, find the vlue of A.dj. A. Q.6 Compute tht A A. 9 Q.7 Let A 5 A for the mtrix Verify tht (i) (dj A) - (ii) (A ) A. nd show 5 Q.8 Using mtrix method, exmine the system of equtions: x + 5y 7, 6x + 5y for onsisteny. Q.9 Find the inverse of mtrix A show tht A ( + + )I A. Q.0 If A show tht tnx, tnx osx sinx AA. sinx osx + nd 0 Q. If A 5 0, prove tht A A 6A + I. 0 Q. If A nd B, verify tht (AB) B A. (-8) Using properties of determinnt, prove tht Q. ( + + ) ( + + ). Q.4 Q.5 y+ z x y z+ x y x (x + y + z) (x z). x+ y z z + + + + (++). + + ( + ) Q.6 ( + ) ( ) ( ) ( ) (+ +) Q.7 Q.8 ( + ) + ( + + ). + (++) ( + + ). + x + y x + y x + y x + y 0 ( )(x +xy+y ) Q.9 Write the minors nd oftors of the elements of seond row of the following determinnt: 4 6. 7 9 Q.40 Find the qudrti funtion defined y the eqution f(x) x + x +, if f(0) 6() nd f( ) 6, using determinnts. Q.4 Exmine whether the system of equtions: x y 5, 4x y 0 is onsistent or inonsistent. Q.4 Verify, whether the system of equtions: x y z, y z, x 5y is onsistent or inonsistent.
Mthemtis 7.9 Without expnding the determinnts, show tht Q.4. Q.5 Show tht x 6x + 7 0. Hene find A. A 4 stisfies the eqution Q.44 Q.45 0 p q p r q p 0 q r 0 r p r q 0 Q.46 Solve for x, x 0 x 4. 0 Q.5 Find mtrix A if, Q.54 Given (AB). Q.55 For the mtrix 4 A 0 6. 5 0 4 A nd B 4. Compute 4 A 6A + 9A 4I O, hene find A, verify tht A. Q.47 If, verify tht A 4A I O 0 0 Exerise 0 0 nd O nd hene find 0 0 Q.48 Evlute : / / /. A. Q.49 Show tht { ( + + )} is root of the following eqution: x+ x+ 0. x+ Q.50 Using properties of determinnts, prove tht : x+ 4 x x x x+ 4 x 6(x+ 4). x x x+ 4 Q.5 Using properties of determinnts, prove tht: + + +. Single Corret Choie Type Q. If,, re ll different from zero nd + + 0, then the vlue of + + + is: (A) (B) (C) (D) Q. If,, re ll different nd then (A) ( + + ) + + (B) ( + + ) ( + + ) (C) ( + + ) + + (D) None of these 4 4 0, 4 Q. If (sin x + sin w) (sin y + sin z)p, then N N x y ; (N N N,N,N,N4 N) z w 4
7.0 Determinnts (A) Hs mximum vlue. (B) Hs minimum vlue. (C) In independent of N,N,N,N 4 (D) None of these n n Q.4 If ( + x + x ) 0 + x + x +... + nx then n n n+ n 6 n n+ is n 4 n 7 n+ 7 (A) (B) (C) 0 (D) Q.5 The solute vlue of the determinnt + + is (A) 6 (B) 8 (C) 8 (D) None of these Q.6 D, D + + + + + + D + + + + + + then + + + + + + (A) D 0, if ++>0 (B) (C) D D D (D) D D D D D Q.7, where,, re distint positive rels, then is lwys less thn (A) 4 (B) 79 (C) 7 (D) 8 Q.8 The vlue of for whih the system of equtions, ( + ) x + ( + ) y ( + ), logx xyz logx y logx z logy xyz logy z nd x + y re onsistent is log xyz log y z z (A) (B) (C) 0 (D) None, Q.9 The following system of equtions x 7y + 5z ; x + y + 5z 7 nd x + y + 5z 5 re (A) Consistent with trivil solution (B) Consistent with unique non-trivil solution (C) Consistent with infinite solution (D) Inonsistent with no solution Q.0 The system of equtions (sin θ )x + z 0, (os θ )x + (sin θ )y 0, (os θ )y + z hs (A) Non unique solution (B) A unique solution whih is funtion of nd θ (C) A unique solution whih is independent of nd θ (D) A unique solution whih is independent of θ only Q. The eqution ( + x) ( x) ( + x ) x + x 5x + x + x x (+ x) x+ x+ ( x) x x 0 x x x (A) Hs no rel solution (B) Hs 4 rel solutions (C) Hs two rel nd two non-rel solutions (D) Hs infinite numer of solutions, rel or non-rel Q. The system of eqution : xos θ+ y sinθ sinθ 0 ; xsinθ+ y sin θ osθ; xsinθ yosθ 0, for ll vlues of θ, n (A) Hve unique nontrivil solution (B) Not hve solution (C) Hve infinite solutions (D) Hve trivil solution Q. If x, y, z re not ll simultneously equl to zero, stisfying the system of equtions (sin θ)x y + z 0; (os θ )x + 4y + z 0; x + 7y + 7z 0, then the numer of prinipl vlues of θ is (A) (B) 4 (C) 5 (D) 6
Mthemtis 7. Q.4 For non-zero, rel, nd + + α + then the vlue of α is (A) 4 (B) 0 (C) (D) 4 Q.5 Numer of vlue of for whih the system of 5 equtions, x + ( )y 4 + ; x + ( )y possess no solution is (A) 0 (B) (C) (D) Infinite Previous Yers Questions Q. The determinnt xp + y x y yp + z y z 0 0 xp+ y yp+ z (A) x, y, z re in AP (C) x, y, z re in HP (B) x, y, z re in GP (D) xy, yz, zx re in AP (997) x x+ Q. If f(x) x x(x ) (x + )x, x(x ) x(x )(x ) (x + )x(x ) then f(00) is equl to (999) (A) 0 (B) (C) 00 (D) 00 Q. If the system of equtions x ky z 0, kx y z 0, x + y z 0 hs non-zero solution, then possile vlues of k re (000) (A), (B), (C) 0, (D), Q.4 The numer of distint rel roots of (00) sinx osx osx π π osx sinx osx 0 in the intervl x is 4 4 osx osx sinx (A) 0 (B) (C) (D) Q.5 If the system of equtions x + y 0, z + y 0 nd x + z 0 hs infinite solutions, then the vlue of is (00) (A) (B) (C) 0 (D) No rel vlues Q.6 The numer of vlues of k for whih the system of equtions (k + )x + 8y 4k, kx+ (k + )y k hs infinite solutions. Assertion Resoning Type (A) Both ssertion nd reson re true nd reson is the orret explntion of Assertion. (B) Both ssertion nd reson re true nd reson is not the orret explntion of ssertion. (C) Assertion is true ut reson is flse (D) Assertion is flse ut reson is true. Q.7 Consider the system of equtions x y + z, x y + 4z nd x + y z k Sttement-I: The system of eqution hs no solution for k. nd Sttement-II: The determinnt k 0, for 4 k 0. (008) Q.8 Given, x y + z, y z + x, z x + y, where x, y, z re not ll zero, prove tht + + + (978) Q.9 If α e repeted root of qudrti eqution f(x) 0 nd A(x), B(x) nd C(x) e polynomils of degree, 4 nd 5 respetively, then show tht A(x) B(x) C(x) A( α) B( α) C( α) A( α) B( α) C( α) is divisile y f(x), where prime denotes the derivtives. (984) Q.0 If mtrix A, where,, re rel T positive numer, nd AA I, then find the vlue of + +. (00) Q. The numer of vlues ofk, for whih the system of equtions: ( k + ) x + 8y 4k kx + ( k + ) k Hs no solution, is: (0) (A) Infinite (B) (C) (D) n n Q. If αβ, 0, nd f( n ) α +β
7. Determinnts + ( ) + ( ) ( ) ( ) ( ) ( ) ( ) ( ) f f ( ) ( )( ) + g + f + f K α β α β + f + g + f 4 then K is equl to: (04) (A) αβ (B) α β (C) (D) - Q. The set of ll vlues of λ for whih the system of liner equtions: x x + x λx x x + x λx x + x λx Hs non trivil solution. (05) (A) Is n empty set, (B) Is singleton (C) Contins two elements (D) Contins more thn two elements Q.4 The system of liner equtions (06) x+λy z 0 λx y z 0 x+ y λ z 0 hs non-trivil solution for: (A) Infinitely mny vlues of λ (B) Extly one vlue of λ (C) Extly two vlues of λ (D) Extly three vlues of λ JEE Advned/Bords Exerise Q. Solve the following using Crmer s rule nd stte whether onsistent or not. () x + y + z x + y + z 6 x + y 0 () 7x 7y + 5z x + y + 5z 7 x + y + z 5 () x+ y+ z 6 0 x + y z 0 x + y z + 0 Q.5 Given x y + z; y z + x; z x + y where x, y, z re not ll zero, prove tht + + +. x y Q.6 Given ; ; y z z z x x y where x, y, z re not ll zero, prove tht + + + 0. Q.7 If sinq osq nd x, y, z stisfy the equtions xosp y sinp + z osq + xsinp + y osp + z sinq x os(p + q) y sin(p + q) + z Then find the vlue of x + y + z. Q. For wht vlue of K do the following system of equtions possess non-trivil (i.e. not ll zero) solution over the set of rtionl Q? x + Ky + z 0, x + Ky z 0, x + y 4z 0. For tht vlue of K, find ll the solutions of the system. Q. The system of equtions α x+ y+ z α x+α y+ z α ; x+ y+α z α hs no solution. Find α. Q.4 If the equtions (y + z) x, (z + x) y, (x + y) z hve non-trivil solutions, then find the vlue of + + + + +. Q.8 Investigte for wht vlues of λµ, the simultneous equtions x + y + z 6; x+ y + z 0 nd x + y +λ z µ hve; () A unique solution () An infinite numer of solutions () No solution Q.9 For wht vlues of p, the equtions: x + y + z ; x + y + 4z p nd x + 4y + 0z p hve solution? Solve them ompletely in eh se. Q.0 Solve the equtions : Kx + y z ; 4x + Ky z ; 6x + 6y + Kz onsidering speilly the se
Mthemtis 7. when K. Q. () Let,,, d re distint numers to e hosen from the set {,,, 4, 5}. If the lest possile positive x + y solution for x to the system of eqution x + dy n e expressed in the form p where p nd q re q reltively prime, then find the vlue of (p + q). () Find the sum of ll positive integrl vlues of for whih every solution to the system of equtions x + y nd x + 4y 6 stisfy the inequlities x >, y > 0. Q. If the following system of equtions ( t)x + y + z 0, x + ( t)y + z 0 nd x + y + ( t)z 0 hs non-trivil solutions for different vlues of t, then show tht we n express produt of these vlues of t in the form of determinnt. Q. Show tht the system of equtionsx y + 4z, x + y z nd 6x + 5y + λ z hs tlest one solution for ny rel numer λ. Find the set of solutions of λ 5. Q.4 Solve the system of equtions: z + y + x + 0 z + y + x + 0 z + y + x + 0 Q.5 () Consider the system of equtions αx y+ z α x α y+ z x y+α z If L, M nd N denotes the numer of integrl vlues of α in intervl [ 0, 0] for whih the system of the equtions hs unique solution, no solution nd infinite solutions respetively, then find the vlue of (L M + N). () If the system of equtions is x + y z 0 x + y + kz 0 4x + y + z 0 hve set of non-zero integrl solutions then, find the smllest positive vlue of z. () Given, {0,,,, 4,, 9, 0}. Consider the system of equtions x+ y+ z 4 x + y + z 6 x + y + z Let L: denotes numer of ordered pirs (, ) so tht the system of equtions hs unique solution, M: denotes numer of ordered pirs (, ) so tht the system of equtions hs no solution nd N: denotes numer of ordered pirs (, ) so tht the system of equtions hs infinite solutions. Find (L + M N). Q.6 () Prove tht the vlue of the determinnt 7 5 + i 4i 5 i 8 4 + 5i is rel. + 4i 4 5i 9 () On whih one of the prmeter out of, p, d or x vlue of the determinnt does not depend. x + x x z + z z os(p d)x ospx os(p + d)x sin(p d)x sinpx sin(p + d)x () If y + y y 0 nd x, y, z re ll different then, prove tht xyz. Q.7 Prove tht () () + + + + ( ) x y z [(x y)(y z)(z x)(x + y + z)] x y z Q.8 () Let vlue of f(x) (given x > ). x f(x) Find the minimum 0 x () If + + + + + 0,, R, then find the vlue of the determinnt
7.4 Determinnts ( + + ) + ( + + ) + + ( + + ) () x x x 4 x 4 x 9 x 6 0 x 8 x 7 x 64 Q.9 If D nd + + + D + + + then prove tht D D. + + + Q.0 Prove tht + + ( + + ) Q. Let sinx sin(x + h) sin(x + h) f(x) sin(x + h) sinx sin(x + h). sin(x + h) sin(x + h) sinx Q.5 If + + 0, solve for x : x x 0 x Q.6 Let,, re the solutions of the ui x 5x + x 0, then find the vlue of the determinnt Q.7 Show tht, + + + nd find the other ftor. Q.8 Prove tht +λ λ +λ is divisile y λ f(x) If Lim h 0 h find k N. hs the vlue eqution to k(sinx + sin x) ( + ) ( + ) ( + ) 4 ( ) ( ) ( ) Q. Prove tht 4 ( β+γ α δ) ( β+γ α δ) 4 ( γ+α+β δ) ( γ+α β δ) 4 ( α+β γ δ) ( α+β γ δ) 64( α β)( α γ)( α δ)( β γ)( β δ)( γ δ) Q. If, nd re the roots of the ui x x + 0 then find the vlue of the determinnt. ( + ) ( + ) ( + ) Q.4 Solve for x () x + x + x + 4 x + x + 4 4x + 5 0 x + 5 5x + 8 0x + 7 Q.9 In ABC, determine ondition under whih A B C os ot ot B C C A A B tn + tn tn + tn tn + tn 0 Exerise Single Corret Choie Type Q. Let m e positive integer & D r m r m r C m m+ sin (m ) sin (m) sin (m + ) then the vlue of (A) 0 m r 0 D (B) m (C) r is given y m (D) (0 t m), m m sin ( )
Mthemtis 7.5 Q. If α,β nd γ re rel numers, then os( β α) os( γ α) D os( α β) os( β γ) os( γ β ) os( α γ) (A) (B) osαosβos γ (C) osα+ osβ+ os γ (D) Zero Q. If, nd re non-zero rel numers, then + D + 0 (A) + (C) + + (B) (D) Zero mx mx p mx + p Q.4 If f (x) n n+ p n p then yf(x) represents mx + n mx + n + p mx + n p (A) A stright line prllel to x-xis (B) A stright line prllel to y-xis (C) Prol (D) A stright line with negtive slope x (x ) x Q.5 If D(x) x x (x + ) x (x + ) (x + ) then the oeffiient of x in D(x) is (A) 5 (B) (C) 6 (D) 0 Q.6 The numer of integrl solutions of D 8, where y+ z z y D z z+ x x is y x x+ y (A) (B) 8 (C) 6 (D) 4 + sin x os x 4sinx Q.7 Let f(x) sin x + os x 4sinx, sin x os x + 4sinx,, then the mximum vlue of f(x) is equl to (A) (B) 4 (C) 6 (D) 8 Q.8 If 4 px + qx + rx + sx + t x + x x x + x+ x x then t is equl to x x + 4 x (A) (B) 0 (C) (D) None Q.9 If D (A) (C) + +, then D is equl to + + + + (B) + + ( + + ) (D) None of these Q.0 If α+β+γ π, then the vlue of sin( α+β+γ) sinβ os γ sinβ 0 tnα os( α+β) tnα 0 (A) 0 (B) (C) (D).sinβ. os γ.tnα Q. If the entries of determinnt re zero or one then the vlue of the determinnt (A) Cnnot e (B) Cnnot e (C) Cn e is (D) Is essentilly zero Q. In third order determinnt, eh element of the first olumn onsists of sum of two terms, eh element of the seond olumn onsists of sum of three terms nd eh element of the third olumn onsists of sum of four terms. Then it n e deomposed into n determinnts, where n hs the vlue (A) (B) 9 (C) 6 (D) 4 Q. If the system of equtions x + y + z 4, x + py + z, µ x + 4 y + z hs n infinite numer of solutions, then (A) p, µ (B) p, µ 4 (C) p µ (D) None of these
7.6 Determinnts Q.4 Numer of triplets of, nd for whih the system of equtions, x y nd ( + )x + y 0 + hs infinitely mny solutions nd x, y is one of the solutions, is (A) Extly one (C) Extly three (B) Extly two (D) Infinitely mny Q.5 If the system of equtions x + y + z 0, x + y + z 0 & x + y + z 0 (,, ) hs nontrivl solution, then the vlue of + + is (A) (B) 0 (C) (D) None of these Q.6 The determinnt os( θ+φ) sin( θ+φ) osφ sinθ osθ sinφ osθ sinθ osφ (A) 0 (B) Independent of θ (C) Independent of φ (D) Independent of θ nd φ oth Q.7 The vlues θ, λ for whih the following equtions xsinθ yos θ+ ( λ+ )z 0 ; xosθ+ y sinθ λ z 0 ; λ x + ( λ+ )y + zosθ 0 re onsistent with infinite solution, re (A) θ n π, λ,λ R {0} (B) θ n π, λ is ny rtionl numer + (C) θ (n + ) π, λ R, n I π (D) θ (n + ), λ R, n I Q.8 If the system of equtions, x y nd x + ( )y + possess unique solution x, y then (A) ; (B), (C) 0, 0 Q.9 Let is (D) None of these n+ n+ n+ 4 Cn Cn+ Cn+ n+ n+ 4 n+ 5 n+ n+ n+ n+ 4 n+ 5 n+ 6 Cn+ Cn+ Cn+ 6 D C C C nd n N then the vlue of D is equl to (A) (B) 0 (C) (D) (n + ) (n + ) (n + 4) (n + 5) (n + 6) Q.0 The set of equtions λx y + (os θ )z 0 ; x + y + z 0 ; (os θ )x + y + z 0, 0 θ< π, hs nontrivil solution(s) (A) For no vlue of λ nd θ (B) For ll vlues of λ nd θ (C) For ll vlues of λ nd only two vlues of θ (D) For only one vlue of λ nd ll vlues of θ Multiple Corret Choie Type Q. The determinnt os(x y) os(y z) os(z x) os(x + y) os(y + z) os(z + x) sin(x + y) sin(y + z) sin(z + x) (A) sin(x y)sin(y z)sin(z x) (B) sin(x y)sin(y z)sin(z x) (C) os(x y)os(y z)os(z x) (D) os(x y)os(y z)os(z x) π π Q. The vlue of θ lying etween nd nd 4 π 0 A nd stisfying the eqution + sin A os A sin4θ sin A + os A sin4θ re sin A os A + sin4θ π π π (A) A, θ (B) A θ 4 8 8 π π (C) A 9 θ (D) A π π θ 5 8 6 8 Q. If x x 0 (A) x (B) x (C) Q.4 The determinnt is equl to zero, if (A),, re in AP (B),, re in GP (C) α is root of the eqution (D) (x α ) is ftor of x (D) x α+ α+ α+ α+ 0 x + x + x + x + 0
Mthemtis 7.7 Q.5 The set of equtions x y + z, x y + z 4, x y +α z hs (A) Unique solution only for α 0 (B) Unique solution for α 8 (C) Infinite numer of solution of α 8 (D) No solution for α 8 Q.6 Whih of the following determinnt(s) vnish(es)? (A) (C) ( + ) ( + ) ( + ) 0 0 0 (B) (D) + + + log xyz log y log z x x x log xyz log z y log xyz log y Q.7 If the system of eqution x y nd x y + 4 possess n infinite numer of solutions then the possile vlues of nd re (A), (B), (C), (D), z z Q.8 If p, q, r, s re in A.P, nd p + sinx q + sinx p r + sinx f(x) q + sinx r + sinx + sinx suh tht r + sinx s + sinx s q + sinx 0 f(x)dx 4 n e, then the ommon differene of the A.P. (A) (B) ½ (C) (D) None of these Q.9 If the system of equtions x+ y 0, ( + K)x + ( + K)y 8 0 nd x ( + K) y + ( + K) re onsistent then the vlue of K is (A) (B) /5 (C) 5/ (D) (x + y) z z z (y + z) Q.0 If D x x x y(y + z) x + y + z y(x + y) xz x z xz y then (A) D is independent of x (B) D is independent of y (C) D is independent of z (D) D is dependent of x, y, z Previous Yers Questions Q. The prmeter, on whih the vlue of the determinnt os(p d)x ospx os(p + d)x sin(p d)x sinpx sin(p + d)x does not depend upon, is (997) (A) (B) p (C) d (D) x Q. Let λ nd α e rel. Find the set of ll vlues of λ for whih the system of liner equtions λ x + (sin α )y + (os α )z 0 x + (os α )y + (sin α )z 0 nd x + (sin α)y (os α )z 0 hs non-trivil solution. For λ, find ll vlue of α. (99) Q. Let,, e rel numers with + +. Show tht the eqution x y x + y x + x + y x + y y + 0 x + y + x y + represents stright line. Q.4 For wht vlue of k does the following system of equtions possess non-trivil solution over the set of rtionls x + y z 0; x y + z 0 nd x 5y + 4z k. Find ll the solution. (979) Q.5 For wht vlue of m does the system of equtions x + my m nd x 5y 0 hs solution stisfying the onditions x > 0, y > 0. (979) Q.6 Prove tht for ll vlues of θ sinθ osθ sinθ π π 4π sin θ+ os θ+ sin θ+ 0 π π 4π sin θ os θ sin θ (000)
7.8 Determinnts Q.7 The totl numer of wys in whih 5 lls of different olours n e distriuted mong 5 persons so tht eh person gets t lest one ll is (0) (A) 75 (B) 50 (C) 0 (D) 4 Q.9 The totl numer of distint x R for whih x x + x x 4x + 8x 0 is (06) x 9x 7x Q.8 Whih of the following vlues of α stisfy the eqution ( +α ) ( + α ) ( + α) ( ) ( ) ( ) ( +α ) ( + α ) ( + α) +α + α + α 648 α? (A) 4 (B) 9 (C) 9 (D) 4 (05) MASTERJEE Essentil Questions JEE Min/Bords Exerise Q.5 Q.9 Q.8 Q.48 Q.5 Q.54 Q.55 Exerise Q. Q.6 Q. Previous Yers Questions Q. Q.6 Q.7 Q.0 JEE Advned/Bords Exerise Q. Q. Q.8 Q. Q.5 Q. Exerise Q. Q. Q.4 Q.0 Q.5 Q.0 Previous Yers Questions Q. Q.5 Q.7 Q.0
Mthemtis 7.9 Answer Key JEE Min/Bords Exerise Single Corret Choie Type Q. x ± Q. Q. (i) (ii) 6 Q.4 7, Q.5 5 sq. units Q.6 0 Q.7 96 Q.8 ; 9 Q.9 5 Q.0 400 4 5 Q. Q. Not possile Q. 0 Q.4 x Q.5 9 5 8 Q.6 - Q.0 75 Q. 0 Q. 6 Q.4 No sq. units Q.5 78 Q.6 + Q.9 A Q.8 Inonsistent 9 5 Q.9 M 9, M, M, C 9, C, C Q.40 f(x) x + x + 6 Q.4 Consistent Q.4 Inonsistent Q.46 7 x or Q.47 A 4 Q.48 0 Q.5 4 A 7 Q.5 49 8 4 0 Q.54 9 4 8 5 4 9 7 Q. 55 4 Exerise Single Corret Choie Type Q. D Q. A Q. A Q.4 C Q.5 A Q.6 C Q.7 C Q.8 A Q.9 B Q.0 B Q. D Q. B Q. C Q.4 D Q.5 C
7.40 Determinnts Previous Yers Questions Q. B Q. A Q. D Q.4 C Q.5 A Q.6 Q.7 A Q.0 4 Q. B Q. C Q. C Q.4 D JEE Advned/Bords Exerise Q. () x, y, z ; onsistent () x, y, z ; onsistent () Inonsistent 5 Q. K,x:y:z :: Q. Q.4 Q.7 Q.8 () λ () λ, µ 0 () λ, µ 0 Q.9 x + K, y K, z K, when p ; x K, y K, z K when p ; where K R Q.0 If K x y z, (K + 6) K + 6(K ) (K + K + 5), If K, then λ x λ,y nd z 0 where λ R Q. () 9 () 4 Q. Q. If λ 5, then 4 9 x ;y nd z 0; If λ 5 then 7 7 Q.4 x ( + + ), y + +, z 4 5K K 9 x ;y nd z K where K R. 7 7 Q.5 () () 5 () 9 Q.6 () p Q.8 () 4 () 65 Q. Q. 08 Q.4 () x or x ; () x 4 Q.5 X 0 or x ± ( + + ) Q.6 80 Q.7 λ ( + + +λ ) Q.9 Tringle ABC is isoseles. Exerise Single Corret Choie Type Q. A Q. D Q. A Q.4 A Q.5 A Q.6 D Q.7 C Q.8 C Q.9 A Q.0 A Q. C Q. D Q. D Q.4 B Q.5 C Q.6 B Q.7 D Q.8 A Q.9 A Q.0 A
Mthemtis 7.4 Multiple Corret Choie Type Q. A,D Q. A, B, C, D Q. A, D Q.4 B, D Q.5 B, D Q.6 A, B, C, D Q.7 A, B, C, D Q.8 A, C Q.9 A, C Q.0 A, B, C Previous Yers Questions Q. B Q. Zero Q.4 k 0, the given system hs infinitely mny solutions 5 Q.5 m < or m > 0 Q.7 B Q.8 B C Q.9 Solutions JEE Min/Bords Exerise k + 0 k + 0 or k 0 7 k or k Sol : 4 8 x x 4 Sol 5: Verties of tringle (0, ) (, 4) (, 6) 8 8 ( 4) x 8 x x ± 8 ± 0 Are 4 (6) 8 [ ] + + 6 Sol : A 0, A [] (0) 4 9 + 5 5 Sq. Unit Sol : 4 0 4 0, Sol 6: D (i) M 4 0 C ( ) + (ii) C ( ) + 0 ( ) 4 0 4 6 Sol 4: Are of tringle, [(k, 0), (, ), (0, )] 5 unit C + C + C + + ( ) + [ ] + [ ] [ ] 0 It n e diretly sid s it is property k 0 0 [ 0] + k( ) 5 Sol 7: 4 4 P A (ssume)
7.4 Determinnts A 4 4 P P 4(4) (96 + P) 0 So 0 x[ 8] + {() ( )()}] 8x + 4 + 0 P 96 8x 5 x 5 8 0 Sol 8: I, I 0 I 0 0 0 I 9 0 0, 0 0 0 Sol 5: Verties (, ) (4, ) nd ( 5, 4) Are 4 5 4 [6 + 5 + [ 4] + [ 5 4]] [ 9] 9 sq. unit Sol 9: (0, ), (, x) nd (, ) points re olliner 0 So x 0 [ x] + [ ] 0 x + 4 0 x 5 x 5 Sol 0: A 0, B 0 AB A B 0( 0) 400 Sol : A 5 4 C 4, C 5, C, C dja C C C C T 4 5 Sol : A 6 8 A 8 [] [ 6] 8 + 8 0 So, A does not exist 8 Sol : 4 6 (8) 5 4 4 5 Two olumns re sme, so determinnt is 0 x 0 Sol 4: 4 is singulr 0 Sol 6: Verties (x, y) (, 6), (5, 4) x 4 Are 6 [8 + 0 + x[ 6 4] + 4[5 ]] 5 4 8 0x + 70 0x 50 70 0 x 0 0 Sol 7: 5 5 5 7 0 8 4 4 () 5 5 5 7 7 0 8 8 4 Two olumn re sme so Determinnts is 0 Sol8: + + + + + + + + + + + + (++) ( + + ) + + + 0 + 0 + 0 + C C + C C C C 0 Sol 9: (, 0), (0, 5) nd (x, y) re olliner 0 0 5 x y 0
Mthemtis 7.4 [ 5x] + [5 y] 0 5x + 0 y 0 5x + y 0 x + y - 5 Sol 0: A, A s order 5A (5) A 5 75 Sol : A, A 0 A C + C (long first row) A 0 + + R R,R R Sol : + 0 + 0 + ( )( ) 0 0 Sine the olumns re linerly dependent, hene the vlue of determinnt is zero. Sol : A 4 Order of A dja A ( 4) 6 5 Sol 4: 0 It is not squre mtrix, so inverse not exist Sol 5: A, A s order A. dja A A 78 Sol 6: A 5, A 9 A C, C 5, C, C A dja A Sol 7: A ( 4 5) 5 5 9 5 9 A C 5 4 C + 0 C 5, C + 0 C 5 4 C C 5, C C 4 A [4] [] + [ 5] 4 5 4 5 AdjA 4, 5 A dja A 4 5 A 4, 5 A For dja, C A 4 C 5 + 6 C + 0, C 5 + 6, C 4 5 9 C C, C C, C 55 + 4, C 8 56 65 4 9, 6 So A 6 9 A 65 6 6 9 A 65 5
7.44 Determinnts Sol 8: x + 5y 7, 6x + 5y 5 D 0 0 0 6 5 D 0. So system is inonsistent tnx os x sinxosx A A tnx sinxosx os x os x sin x sinxosx sinxosx sinxosx + sinxosx sin x + os x Sol 9: A ( + ) / osx sinx sinx osx R.H.S. A + + ( + ) / dja dja ( + ) / A A ( + + )I A + + 0 0 + + ( + ) / + + + + ( + ) ( + ) / A R.H.L. L.H.S. tnx Sol 0: A, tnx tnx A tnx A + tn x os x tnx dja tnx sinx dja A os x osx A sinx osx os x sinxosx sinxosx os x 0 Sol : A 5 0, 0 Assume A xi 0 A xi 0 x 0 So 5 x 0 0 0 x ( x) [x + 4x] [5] 0 x x + 4x + x + 6 8x 5 x 6x + x x 6x + x x (A xi) 0 A 6A + I A Sol : A + + 6+ 4+ AB + + 4 6 + 6 4 + + 4 6 6 5 5 5 5 For A C + 4 6, C + 4 6, C 4, C 4 6, C 4, C + 4 6,.. (i)
Mthemtis 7.45 C 4, C 4 6, C + 4 6 A (6) + (6) + () 7 6 6 dja A 6 6 A 7 6 6 9 For B C +, C +, C 5, C 6 +, C 4 5, C 6 9, C, C, C B [] + [] + [ 5] 0 djb B 5 B 5 9 B A 4 6+ 4 + + 6+ 6 0 6 5 0 9 + + + + 0 8 6 4 5 + 8 4 5 0 9 5 9 0 9 6 5 5 AB 5, 5 AB 6( ) + 5( 9) + 5(8) 8 45 + 90 +7 C 5 C 9 C 5 + 8 C 5 + 0 5 C 5 7 C 0 + 5 45 C 5 + 55 0 C 0 + 5 5, C 66 5 8 5 0 So (AB) 9 7 5 8 45 8 5 0 9 5 9 0 9 B A Sol : (++) ( + + ) C C + C + C + + + + (++) + + R R R, R R R 0 (++) 0 (++)[( ) ( ) ( ) ( )] (++) ( + + ) +... + y+ z x y Sol 4: z+ x z x (x + y + z) (x z) x+ y y z R R + R + R (x+ y+ z) x+ y+ z x+ y+ z z+ x z x x+ y y z (x+y+z) z+ x z x x+ y y z
7.46 Determinnts C C C C (x+y+z) (x+y+z) z+ x z x z x x+ y y z y z 0 0 z x x z y z (x + y + z) (x z) (x z) R R + R ( )( ) + + ( + ) + + (+ + ) + ( + ) ( )( ) + ( + ) Sol 5: + + + + + + ( + + ) 0 0 + + + C C + C, C C + C ( + ) + ( + ) + [ ( + )] + (+) [ ( +) + (+) ] ( + ) [ ( + ) ] [ 4 4 + (+) ( + + + + ] + ( + )[ + + ] This on simplifition omes out to e equl to ( + + ) C C + C C ( )( )( ) z z + + + + + + + ( )( )( ) z z + + + + + + + ( )( ) ( ) ( + + )( ) ( )( )( )( + + )(++) Sol 6: ( + ) ( + ) ( + ) ( ) ( ) ( ) Sol 7: + + + R R R, R R R ( ) ( + ) ( + ) ( + ) ( + ) ( )[ + + ] ( )( ) ( ) ( )[ ] ( )( + ) ( + ) ( + ) + + ( ) ( ) (+ + ) + ( + ) ( + ) ( + + ) + 0 + + + + + 0 + + + 0 + + 0
Mthemtis 7.47 + + + 0 0 0 + + 0 + 0 0 Using, C C C nd C C -C in (i), (ii), (iii) + + 0 0 + 0 0 + 0 + 0 + 0 0 [ + + ] + [ + + ] + [ + + ] ( + + ) ( + + ) Sol 8: x + y x + y x + y x + y 0 R R xr yr 0 x + y x + y 0 0 (x + yx + yx + y ) 0 (x + xy + y ) ( ) 0 ( ) (x + xy + y ) + 0 0 + + Sol 9: M 0 0 + 0 4 6 7 9 7 9 + 0 + 0 0 8 + 9 M M 9 9 6 7 7 4 C 9, C, C Sol 40: f(x) x + x +, f(0) 6 f(), f( ) 6 0 0 4 9 D 8 0 D 6 0 6 6 6 6[+]+[ ] 0 5 D 5 D 0 D 0 6 4 9 6 6[9 4] + [4 99] 0 + 4 99 45 45 0 0 0 6 D 4 7 6[ 8] 0 6 9 6 D C 0 6 C 6 D 0 x Eqution x +x+ + x +6 Sol 4: x y 5 4x y 0 D 0 4 5 D x 0, D 0 y 5 4 0 0 So system hs infinite solution (onsistent).
7.48 Determinnts Sol 4: x y z y z x 5y R.H.S. D 0 5 0 [ 5] [ ] [ 6] 5 + + 0 D x 5 0 C C nd then C C 0 p q p r Sol 45: q p 0 q r r p r q 0 (p q) q p q r r p 0 + (p r) q p 0 r p r q [ + 4] 5[ + ] 5 0 5 0 So system is inonsistent. Sol 4: L.H.S. C C ( ) Sol 44: L.H.S. C C R.H.S. +(p q (q r) (r p) (p q) (q r) (r p)0 Sol 46: x 0 x 4 0 x [8] + [x ] 4x + x 7 0 (x ) (4x + 7) 0 (x )(4x + 7) 0 x or 7 4 Sol 47: A, A 4A I 0 Assume A xi 0 x 0 x ( x)( x) 4 0 + x x x 4 0 x 4x 0 A xi 0 A 4A I 0 Hene proved. A [A 4A I] 0 A 4I A 0 A 4I 4 A 0 0 4 4
Mthemtis 7.49 Sol 48: () / / / 0 () / / / 0 + ( + + ) 0 + ( + + ) [( )( + ) ( )( + )] + +.. + x+ Sol 49: x+ 0 x+ Hve to show tht x ( + + ) R R + R + R Sol 5: A 4 Assume A xi 0 0 x 4 0 0 x+ + + x+ + + x+ + + x+ x+ (x+++) x+ x+ 0 x + + + 0 x ( + + ) Sol 50 : x+ 4 x x+ 4 x x x x+ 4 C C C,C C C 4 0 x 0 4 x 4 4 x+ 4 ( 4x + 6 + 4x) + 6x 48x+64 0 x 0 0 4 x ( x) (4 x) + 9 0 8 + x 4x x + 9 0 x 6x + 7 0 nd A xi 0 So, A stisfied this eqution A 6A + 7 I 0 A [A 6A + 7I] 0 A 6I + 7A 0 7A (A 6I) A 0 (A 6I) 6 7 7 4 0 A 6 4 7 4 6 7 A 4 7 + Sol 5: + + C C + C + + + + (++) + + R R R, R R R Sol 5: 4 A 0 6 Assume BAC D B Adj B 4 6 4 4 B 4
7.50 Determinnts B BAC B D AC B D 4 6 + + 8 + 0 + 45 9 + 4 + 8 + 4 0 8 + 45 4 8 5 + + 4 8 + 40 + 60 4 9 7 AC 8 + 4 + 6 6 9 5 8 0 C djc, C 0 ACC B DC A 9 5 8 0 0 7 + 8 49 8 A + 4 5 8 0 4 0 5 0 4 Sol 54: A, B (AB) B A C 6, C, C 4, C 8, C 5 4, C 0, C, C 8 5 7, C 5 A 5( ) + 4() 5 + 4 A dja A B A 4 4 4 4 8 7 0 5 8 7 0 5 Sol 55: A, Assume (A XI) 0 x x 0 x ( x) [( x) [ + x] + [ + x] 0 ( x)[4 + x 4x ] + x + x 0 6 x + x 8x x + 4x + x 0 x + 6x 9x + 4 0 x 6x + 9x 4 0 A xi 0, so this eqution stisfied A A 6A + 9A 4I 0 A [A 6A + 9A 4I] A 0 0 A 6A + 9I 4A 0 A 4+ + + + + + 4+ + + + + 4 6 5 5 A 5 6 5, 5 5 6 4A A 6A + 9I 6 5 5 0 0 4A 5 6 5 6 + 9 0 0 5 5 6 0 0 4A A 6 + 9 5+ 6 5 6 5+ 6 6 + 9 5+ 6 5 6 5+ 6 6 + 9 4
Mthemtis 7.5 Exerise Single Corret Choie Type + Sol : (D) + 0 + + + ` 0 + R R + R + R + + + + + + + + + + + + + + + + C, C 0 0 + + + 0 0 0 0 + + + 0 + + Sol : (A) 4 4 4 4 4 4 + 0 0 0 R R R, R R R 0 0 0 0 [( ) ( ) ( ) ( ) [( ) ( ) ( ) ( ) ( ) ( )[( + ) ( + + ) ( + ) ( + + ) ( ) ( ) [ + + ( + + )] [ + + + + + ] + + ( ) ( + + ) ( ) [ + ( + )] ( ) ( + + ) [ + + ] [ + + ] Sol : (A) (sin x + sin w) (sin y + sin z) p D x z N N y N N4 w (x, y, w, z) N N4 x w z N N4 y If x y 7 w x z N N y N N4 w ( ) N+ N4 ( ) For mx vlue N + N 4 n. N + N m + n,m N Vlue ( )n ( ) n ( ) m+ ( ) Min vlue Dependent of N, N, N, N 4 N+ N
7.5 Determinnts Sol 4: (C) (+ x + x ) n 0 + x + x + + n x n n n n+ n 6 n n+ n 4 n 7 n+ 7 ( + x + x ) n (x + x + ) n n n+ 0 n n r n + r O r n So determinte C C C 0 n n 0 n 6 n 0 n 4 n 7 Sol 5: (A) ( ) 0 n n n n 6 n n n 4 n 7 n 7 + + C C C C + + + 4 0 + + 4 [ + ( )] 0 + 4 [ 4 ] 6 Sol 6: (C) D D D, + + + + + + + + + + + + + + + + + + D (given) in D C C C, C C C nd ssume T + + T 0 + + + 0 T + + T T + + 0 + + T 0 + + + + T [ + + ++++++()] T [ + + + ( + + )] D C C + C + C D D T T T T R R R, R R R D T 0 + + 0 + + D T [( + ) ( + ) ( + ) ( + )] D ( T) ( T) [T + ( + + )] D T [ + + + ( + + )] D D D Sol 7: (C) L.H. S. ( ) ( ) ( ) in L.H.S. C C C, C C C
Mthemtis 7.5 0 0 ( ) ( ) 0 0 + + + + ( ) ( )[ + + ] ( ) ( ) ( ) ( + + ) + + ( ) ( ) ( ) + + A.M. G.M + + () / ; ()/ is lwys less thn /7 7 Sol 8: (A) ( + ) x + ( + ) y ( + ) (4 + )x + ( + )y ( + ) x + y Here for two vrile thus eqution So D D x D y0 for onsistent D ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) [ + + 4 4] ( + ) ( + ) ( ) (i) D x ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + 9 + 6 4 4) ( + ) ( + ) (5 + ) (ii) D y ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) [ + + 9 6] ( + ) ( + ) ( 4 8) 4 ( + ) ( + ) ( + ) D D x D y 0 (ommon solution in ll) Sol 9: (B) x 7y + 5z, x + y + 5z 7 x + y + 5z 5 D 7 5 5 5 [5 5] 7 [0 5] + 5 [9 ] 0 + 5 + 5 40 0 So system is onsistent with unique non trivil solution. Sol 0: (B) (sin θ)x + 7 0 (os θ)x + sin θ y 0 (os θ)y + z 0 D sinθ 0 osθ sinθ 0 0 osθ D sin θ (sin θ.) + (os θ) (sin θ + os θ) Constnt 0 0 0 0 C 0, D x 0 sinθ 0 osθ So system hs unique solution whih is funtion of nd θ Sol : (D) ( + x) ( x) ( + x ) x + x 5x x + x x (+ x) x+ x+ + ( x) x x 0 x x x In nd determinte R + R + R (+ x) x+ x+ ( x) x x - B x + ( + x) + ( + x ) ( 5x) x + x + x x + x
7.54 Determinnts R R A ( + x) (x + ) x + ( x) x x B (+ x ) 5x x Now ll rows of A is equl to olumns of B B A A A 0 (lwys) For every vlued of x A + B is zero Therefore infinite solutions Sol : (B) x os θ + y sin θ sin θ 0 x sin θ + y sin θ os θ (ii) x sin θ y os θ 0 (iii) for (i) & (ii) os θ sinθ D sinθ sin θ 4 sin θ os θ 4 sin θ os θ 0 (sin θ sin θ os θ) D x sinθ osθ sinθ sin θ 4 sin θ + 4 sin θ os θ 4 sin θ(sin θ + os θ) 4 sin x for onsistent D x 0 4 sin θ 0 θ n π, n I D y sin θ sin θ sinθ sinθ 4 os θ 4 sin θ os θ (i) 4 os θ [sin θ + os θ) 4 os θ D y 0 θ (n + )π/, n I Sin θ nd os θ oth re not zero for sme θ, so for every vlue of θ system hs not solution Sol : (C) (sin θ)x y + z 0 (os θ)x + 4y + z 0 x + 7y + 7z 0 ( 0 0 0 So D x D y D z 0 D sin θ osθ 4 7 7 x, y, z re not ll simultneously equl to zero so for solution (not-trivil), D 0 sin θ [8 ] [6 7os q] + [7 osθ 8] 0 7 sin θ 6 + 7 os θ + 7 osθ 8 0 7 sin θ + 4 os θ 4 Sin θ + os θ Sin θ + os θ sin θ 4 sin θ + [ sin q] sin θ 4 sin θ + 4 sin θ ssume sin θ x 4x + 4x x 0 x[4x + 4x ] 0 x[4x + 6x x ] 0 x 0 or x (x + ) (x + ) 0 (x + ) (x ) 0 x / or / x 0, ½, / ut sin θ sin θ / sin θ {0, ½}, x 0, π/6, 5π/6, π, π etween [0, p] No. of priniple vlue 5 Sol 4: (D) + + + + + + + ( + )( + ) α
Mthemtis 7.55 + + + ( + + + )( + ) 4 4 4 + + 4 4 4 4 + + + + + 4 4 4 4 + + + 4 4 4 4 + 4 4 4 + + 4 α 4 Sol 5: (C) x + ( )y 4 + x + ( )y 5 D ( ) + ( ) + For D 0 ( ) +,, 0 4+ D x 5 (4 + ) ( ) + ( ) ( 5 ) 8 4 + + 6 5 + 4 t 0 D x 0 4+ So D y 0 4 + 0 0 0 So t 0, system hs infinite solution At, +, D 0, nd D x, D y 0 No solution, no. of vlues Previous Yers Questions Sol : (B) Given xp + y x y yp + z y z 0 0 xp+ y yp+ z Applying C C (pc+ C ) 0 x y 0 y z 0 (xp + yp + yp + z) xp + y yp + z (xp + yp + z)(xz y ) 0 Either xp + yp + z 0 or x, y, z re in GP. y xz Sol : (A) Given x x+ f(x) x x(x ) (x + )x x(x ) x(x )(x ) (x + )x(x ) Applying C C (C + C ) x 0 x x(x ) 0 0 x(x ) x(x )(x ) 0 f(x) 0 f(00) 0 Sol : (D) Sine, the given system hs non-zero solution. k k 0 Applying C C C, C C + C + k k + k 0 0 0 (k + ) (k + ) 0 (k + )( k ) 0 k ± Note: There is golden rule in determinnt tht n one s (n ) zero s or n(onstnt) (n ) zero s for ll onstnt should e in single row or single olumn. Sol 4: (C) Given sinx osx osx osx sinx osx 0 osx osx sinx Applying C C + C + C sinx + osx osx osx sinx + osx sinx osx sinx + osx osx sinx
7.56 Determinnts osx osx (osx + sinx) sinx osx 0 osx sinx Applying R C R,R C R osx osx (osx + sinx) 0 sinx osx 0 0 0 0 sinx osx (osx + sinx)(sinx osx) 0 osx + sinx 0 or sinx osx 0 osx sinx or sinx osx π π ot x gives no solution in x nd sin 4 4 x os x tn x x π 4 Sol 5: (A) Given equtions x + y 0, z + y 0 nd x + z 0 hs infinite solutions. 0 0 0 0 + 0 or Sol 6: () For infinitely mny solution, we must hve k + 8 4k k k + k k Sol 7: (A) The given system of eqution n e expressed s x 4y z k Applying R R R, R R + R x ~ 0 y 0 z k x ~ 0 y 0 0 0z k R R R When k, the given system of eqution hs no solution. Sttement I is true. Clerly, Sttement II is lso true s it is rerrngement of rows nd olumns of 4 Sol 8: Given systems of eqution n e rewritten s x + y + z 0 x y + z 0 nd x + y z 0 Aove system of equtions re homogeneous eqution. Sine, x, y nd z re not ll zero, so it hs non-trivil solution. Therefore, the oeffiient of determinnt must e zero 0 ( ) ( ) + ( + ) 0 + + + 0 + + + Sol 9: Sine α is repeted root of f(x) 0. f(x) (x α ), onstnt ( 0) A(x) B(x) C(x) Let φ (x) A( α) B( α) C( α) A( α) B( α) C( α) [To show φ (x) is divisile y (x α ), it is suffiient to show tht φα ( ) nd φ ( α ) 0]. A( α) B( α) C( α) φα ( ) A( α) B( α) C( α) 0 A( α) B( α) C( α) [ R nd R re identil] A (x) B (x) C (x) Agin, φ (x) A( α) B( α) C( α) A( α) B( α) C( α) A( α) B( α) C( α) φ ( α ) A( α) B( α) C( α) 0 A( α) B( α) C( α) [ R nd R re identil] Thus, α is repeted root of φ (x) 0 Hene, φ (x) is divisile y f(x).
Mthemtis 7.57 Sol 0: (4) Given A, T nd AA (i) Now, T AA 0 0 0 0 0 0 + + + + + + + + + + + + + + + + + + 0 0 0 0 0 0 + + nd + + 0 (ii) We know, + + ( + + )( + + ) + + ( + + )( 0) + [from eqution (i) nd (ii)] + + ( + + ) + (iii) Now, ( + + ) + + + ( + + ) (iv) From eqution (iii), + + 4 Sol : (B) + + + k + 8 k + 4k + 8k k k+ k 4k + ( k )( k ) 4k 8 4k + k 4k + 8 k k + 4k k + 8 4 ( k k ) 4k k + ( )( ) k + 4k k + k 4k k k k + k ( k ) As given no solution & 0 0 k Sol : (C) + + +α+β +α +β +α+β +α +β +α +β +α +β +α +β +α +β α α α α α β β β ( α) ( α β) ( β ) K 4 α α β β Sol : (C) ( λ) x x + x 0 x ( +λ ) x + x 0 x + x λ x 0 Non-trivil solution 0 λ λ 0 λ ( ){ } { } ( ) ( ) λ λ+λ 4 +. λ+ + 4 λ 0 6λ+ λ 8 λ λ + 4λ 4λ+ 4+ λ 0 λ λ 5λ+ 0 x λ + λ λ λ+ 0 ( ) ( ) ( ) ( λ )( λ + λ ) 0 λ λ + λ λ λ 0 ( λ )( λ+ )( λ ) 0 λ,, Sol 4: (D) x+λy z 0 λx y z 0 x+ y λ z 0 For non-trivil solution 0 λ λ 0 λ { } ( ) λ( λ ) 0 λ+ λ λ + λ+ 0 λ 0, ±
7.58 Determinnts JEE Advned/Bords Exerise Sol : () x + y + z 6 x + y z x + y z D D x D y + 0 + 0 6 + 6 0 6 C C C [+ + ] C C + C [6 ] + ( + ) 5 + 8 [ ] + 6[ + 4] + [ 6 ] 5 + 8 7 6 D z 6 [ ] + [ + 6] + 6[ ] 4 + 7 + 6 9 x Dx, y Dy 6, z Dz 9 D D D Here, it is onsistent () x + y + z x + y + z 6 x + y 0 D 0 [6 ] + [0] 5 Dz 0 y D 5 D y 6 [ 6] 5, 0 0 D y 5 y D 5 D z 6 [6 ] + 6[0] 5m, 0 Dz 5 Z D 5 () 7x 7y + 5z x + y + 5z 7 x + y + 5z 5 7 7 5 D 5 R R R ; R R R 5 5 0 0 0 5[ 0 + 0] 0 5 7 5 D x 7 5 R R R ; R R R 5 5 0 0 0 5[4 + 0] 0 0 5 5 D 0 ut Dx 0, so, system is inonsistent Sol : x + ky + z 0 x + ky z 0 x + y 4z 0 Eqution hs non-trivil solution. So, D D x D y D z 0 k D k 4 (i) (ii) (iii) D x 6 0 0 [6 ] 0, [ 4k + 6] + k[ 4 + ] + [9 k] 4k + 6 + 8k + 7 6k k 0
Mthemtis 7.59 K, ssuming x t From eqution (ii) (i) x 5z 0 z x t, (x t) 5 5 In (iii) t + y 4z 0 z 4z t t 8t 0t y 4 t t 5 5 5 y t 5 t t (x, y, z) t,, 5 5 t R Sol : x + y + z α x + y + z α x + y + z α D α α α D [ ] + [ ] + [ ] α + α α + α +, At α + 0 So (α ) is ftor of α + Now, α + n e witten s + α α + (α ) + α(α ) (α ) (α ) ( + α ) D (α ) ( + α ) D (α ) ( + α α ) (α ) D (α )[α(α + ) (α + )] D (α ) (α + ) (α ) For D 0, α or For α, D x 0 0 0 0, so onsistent So on D y nd D t 0 \α α Sol 4: (y + z) x x y z 0 (z + x) y x y + z 0 (x + y) z x + y z 0 0 0 0, so D x D y D z 0 So for non-trivil solution, D 0 D C C C ; C C C + 0 D 0 ( + ) + + + + + + + or from eqution x y+ z y x+ z z x+ y + x + y + z ; + x + y + z y+ z x+ z ; + x + y + z x+ y + + x + y + y + z + z + x + + + x+ y+ z (x + y + z) (x + y + z) Sol 5: x y + z x y z 0 y z + x x y + z 0 z x + y x + y z 0 0 0 0 D x D y D z 0, But system hs solution. So D 0 D [ ] + [ ) [ +] 0 0 + + +
7.60 Determinnts x Sol 6: x y + z 0 y z y x + y z 0 z x z x y z 0 x y 0 0, so D x D y D z 0, 0 For solution D 0 D [ ] [ + ] + [ ] + ( + + + ) 0 + + + 0 Sol 7: sinq osq Xosp ysinp + z osq +... (i) xsinp + yosp + z sinq... (ii) xos(p + q) ysin(p + q) + z... (iii) os(a + B) osa osb sinasinb sin(p + q) sinposq + osp sinq eqution (i) + eqution (ii) x (sin p+os p) + y (os p+sin p) xy ospsinp + xospz yzsinp + xysinposp + xzsinp + z + yzosp z + sinq + osq x + y + z + xyosp yzsinp + xzsinp + z + yzosp + sinq + osq From eqution (iii) nd (i) x + y + z + z ( + osq z) + q( sinq z) z (sinq osq) x +y +z + z( + osq sinq z) + (osq sinq) For eqution (iii) z( + osq sinq z) + osq sinq \x + y + z Sol 8: x + y + z 6 x + y + z 0 x + y + lz µ () A unique solution, D 0 D λ [λ 6] + [ λ + ] + 0 λ 6 + λ λ 0 λ () Infinite solution So D 0, Dx Dy Dz 0 D 0 λ 6 6 D x 0 0 µ λ µ 6[0] + [µ 0] + [0 µ] (µ 0)5 0 µ 0 () No solution D 0, D x 0 λ, µ 0 Sol 9: x + y + z x + y + 4z p x + 4y + 0z p D 4 4 0 D [0 6] + [4 0] + [4 ] 4 6 + 0 So for solution, D x D y D z 0 D x p 4 p 4 0 [0 6] + [4p 0p] + [4p p ] 0 4 + 4p 0p + 4p p 0 p 6p + 4 0 p p + 0 p p p + 0
Mthemtis 7.6 (p ) (p ) 0 p or For p x + y + z x + y + 4z x + 4y + 0z Assume tht x k Eqution (ii) ii(i) x + z z x z k (k ) So y z x k y k k + k k k (x,y,z) k,, At p... (i) (ii) (iii) x + y + z () x + y + 4z () x + 4y + 0z 4 () Assume x k Eqution () () x + z 0 x z k z k y x z k k k k Sol 0: Kx + y z 4x + Ky z 6x + 6y + Kz K D 4 K t K (given) 6 6 K 4 4 0 6 6 x + y z (i) 4x + 4y z 6x + 6y + z Assume x l Eqution (iii), (ii) (iii).(ii) 7z 0 z 0 y + z x λ (x, y, z) (λ, λ, λ) If K K D 4 K 6 6 K K[K + 6] + [ 6 4K] [4 K] K + 6K 8K 48 + 4K K + K 60 (K + K 0) At K (8 + () 0) 0 So (K ) is ftor k + k 0 K + K + 5 k D (K ) (K + K + 5) D x K 6 K K + 6 + [ K] [ 6K] K + 6 6 4K 4 + K K + 8K 4 [K + 4K ] [K + 6K K ] [K(K+6) (K + 6)] (K ) (K + 6) Similrly, D y (K ) (K + ) nd D z 6(K ) if K, x y z (K+ 6) K+ 6(K ) (K + K+ 5) Sol : (),,, d re distint no.,,, d {,,, 4, 5} x + y x + dy (ii) (iii)
7.6 Determinnts D D x x d d (i) d d, Dx d D d for lest possile +ve vlue of x d (lest nturl numer) (d, ) (, ) or (5, ) d should e mximum for lest x (, ) (, ) (d ) ( ), {7, 4, 5} Mx. (5) 5 x If, (5, ), d 5,, {,, 4} Mx. 5 5(4) () 8 x 8 p q (min.) p + q + 8 9 () x + y nd x + 4y 6 x >, y > 0 D D x D y 4 4, 6 4 6 6 6, D x x > 0, D > 6( ) > ( )( + ) 6 ( ) >, + < 6, y D y ( ) D 6( ) So is nd + 4 + ve 6 Sol : ( t)x + y + z 0 x + ( t)y + z 0 x + y + ( t)z 0 Hs non-trivil solution, So D 0 D t t t 0 Assume D 0 t + 0 t + 0 t + d 0 0 So t t t d 0 0 At t 0, D d 0 So d 0 And 0 is oeffiient of t ( )( )( ) t t t d 0 d 0 Sol : x y + 4z X + y z 6x + 5y + lz, D 4 6 5 λ (λ + 5) [ 8 λ) + 4[5 ] 6λ + 45 + 8 + λ 8 7λ + 5 7(λ + 5) D 7(λ + 5) D x 4 5 λ [λ + 5] + [ λ 9] + 4[ 0 + 6] 6λ + 45 λ 9 6 4λ + 0 4(λ + 5); Dx 4( λ+ 5) 4 x D 7( λ+ 5) 7 D y 4 6 λ
Mthemtis 7.6 [ λ 9] + [ 8 l] + 4[ + ] 6λ 7 54 λ + 6 9λ 45 (λ + 5) D x y D z D y 9( λ+ 5) D 7( λ+ 5) 6 5 9 7 [ 6 + 0] + [ + ] + [5 ] + 9 0, D z z D 0 So x, y, z is not dependent on λ (if λ 5) At λ 5 x y + 4z x + y z 6x + 5y 5z Assume z k, (iii) (ii)(i) 7y z 9 y k 9 7 So, x z y z k 4 (k 9) z 5k 7 7 4 5k k 9 (x, y, z),,k 7 7 Sol 4: z + y + x + 0 z + y + x + 0 z + y + x + 0 Now, D x D x D, y Dy D, z ( ) ( ) ( ) Dz D (i) (ii) (iii) ( + + ) ( ) ( ) ( ); \x ( + + ) D y ( + + ) ( )( ) ( ) \y [ + + ] D z ( ) ( ) ( ) \ z Sol 5: () x y + z α x y + z x y + z D [ + ] [ ] + [ +] + α + ( + α ) ( α + ) At α D ( + ) 0 So (α ) is ftor α α+ + α α So D (α ) ( + α ) (α )( + α α ) (α )[α(α + ) (α + )] (α ) (α ) (α + ) α [ 0, 0] So, α hs n integrl vlue α D x α α So x, D x (α ) (α + )
7.64 Determinnts α α (), {0,,,., 0} D y 0, x + y + z 4 α x + y + z 6 D z α α α 0 () Unique solution, So D 0 α, Numer of vlues for α in [ 0, 0] 9 L () Numer solution is not possile for every vlue of α, system hs tlest one solution. So M 0 () Infinite solution D 0 α, N L M + N 9 + () x + y z 0 x + y + kz 0 4x + y + z 0 Hs non-trivil solution So, D 0 k 4 [ k] + [4k ] [ 8] 0 4 k + k 9 + 5 0k 0 k 0 x + y z 0 x + 7y 0 4x + y + z 0 (iii) (ii) (i) 5y + 0 z 5y x y x y, (i) (ii) (iii) z 5 y, y y x, y, z re integer, so t for x nd z to e integer x n y y n (lso n integer) So t n, 7y, z 5 (minimum +ve vlue) x + y + z D ( 6) + ( ) + (4 ) 6 + + 9 4 D x 6 4( 6) + ( 6) + ( ) 4 4 + 6 + + 4 D y 6 6 + 4[ ] + 6 6 + 8 + 6 + 6 4 D z 6 + 6 + 4[4 ] + 6 + + 6 (i) Unique solution so D 0 0 {,,., 0}, {0,,., 0} L 0 0 (ii) Numer solution D 0, 0 D x 0 6, nd D y 0 M ( ) 9 (iii) Infinite solution D0 0, D x D y D z 0
Mthemtis 7.65 But D x nd D z n t e zero t sme times, so no possile ommon solution N 0 L + M N 0 + 9 0 9 Sol 6: 7 5 + i 4i 5 i 8 4 + 5i + 4i 4 5i 9 () Assume z 5 + i, z + 4i z 4 + 5i (z ) 4 + 5 4 7 z z z 8 z z z 9 7[7 z z ]+z [z z 9 z ]+ z [ z z 8z ] [sin x (p+d p) + [sinx (p d p d)] + [sin x (p p + d)] sin x d + sin ( d) + sin dx It dose not depend upon p () xyz x + x x y + y y z + z z x x y y z z + x x x y y y z z z x x y y z z (xyz + ) (x y) (y z) (z x) 0 (given) x, y, z re ll different So (xyz + ) 0 xyz + x x y y z z 7[7 4] + (5 + i) + 4i ( 4 + 5i ) 9 4i Sol 7: () + + + + ( ) + 4i 7 () + (5 + i) ( 5 i )(4 5i) 8 + 4i 8 0 + 6i + i 0 6 + 6i + 4i 7 + (5 + i) 6 0 5 i 5i i Coeffiient of i 70 + 800 46 + 4 () 70 60i + + 4i 06 + 804 os(p d)x ospx os(p + d)x sin(p d)x sinpx sin(p + d)x [os px sin (p+d)x os(p+d)x sin px] 69i R R R, R R R + + 0 0 ( ) ( + ) 4( ) ( ) [( + ) 4 ( )] ( ) [( ) ( + ) 4( )] ( ) [ + 4] ( ) 0 0 () x y z x y x y z x z y z z 0 0 (x z) (y z) x + y + xz y + z + yz z ( ) ( + + ) (x z) (y z) (y + z + yz x z xz) + [os(p+d)x sin (p d)x ws(p d)x sin (p + d)x] + [os (p d)x sin px os px sin (p d)x] (x z) (y z) [ z (y x) + (y x ) (x y) (y z) (z x) (x + y + z)