Q12 Solve for x a 17 x 10 8 x 5 5 x 3 5 x 4 4 x 3 3 x 2 4 x 3 3 x 2 2 x 0 b 64

# Q12 solve for x a 17 x 10 8 x 5 5 x 3 5 x 4 4 x 3 3 x

This preview shows page 6 - 8 out of 11 pages.

Q.12 Solve for x (a) 17 x 10 8 x 5 5 x 3 5 x 4 4 x 3 3 x 2 4 x 3 3 x 2 2 x + + + + + + + + + = 0. (b) 64 x 3 27 x 2 8 x 16 x 3 9 x 2 4 x 4 x 3 3 x 2 2 x = 0. Q.13 If a + b + c = 0 , solve for x : x c a b a x b c b c x a = 0. Q.14 If a 2 + b 2 + c 2 = 1 then show that the value of the determinant θ + + θ θ θ θ + + θ θ θ θ + + cos ) b a ( c ) cos 1 ( bc ) cos 1 ( ac ) cos 1 ( cb cos ) a c ( b ) cos 1 ( ab ) cos 1 ( ca ) cos 1 ( ba cos ) c b ( a 2 2 2 2 2 2 2 2 2 simplifies to cos 2 θ . Q.15 If p + q + r = 0 , prove that qa pc rb pb ra qc rc qb pa = pqr a c b b a c c b a . Q.16 If a , b , c are all different & 1 c c c 1 b b b 1 a a a 4 3 4 3 4 3 = 0, then prove that, abc(ab + bc + ca) = a + b + c. Q.17 Show that, λ + λ + λ + 2 2 2 c bc ac bc b ab ac ab a is divisible by λ 2 and find the other factor. Q.18 Prove that : 1 1 1 c b a c b a 4 ) 1 c ( ) 1 b ( ) 1 a ( ) 1 c ( ) 1 b ( ) 1 a ( c b a 2 2 2 2 2 2 2 2 2 2 2 2 = + + + . Q.19 In a Δ ABC, determine condition under which 1 1 1 tan tan tan tan tan tan cot cot cot 2 B 2 A 2 A 2 C 2 C 2 B 2 C 2 B 2 A + + + = 0 Q.20 Prove that : ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) a p a q a r b p b q b r c p c q c r ap aq ar bp bq br cp cq cr = + + + + + + + + + 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1
ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005) 7 Q.21 Prove that 2 3 3 2 2 3 2 1 3 2 3 2 2 2 2 2 1 2 2 3 1 2 2 1 2 1 1 ) b a ( ) b a ( ) b a ( ) b a ( ) b a ( ) b a ( ) b a ( ) b a ( ) b a ( = 2(a 1 a 2 )(a 2 a 3 )(a 3 a 1 )(b 1 b 2 )(b 2 b 3 )(b 3 b 1 ) Q.22 Prove that αβγδ β + α γδ + δ + γ αβ γδ + αβ β + α γδ + δ + γ αβ δ + γ β + α δ + γ + β + α γδ + αβ δ + γ + β + α 2 ) ( ) ( ) ( ) ( ) ( ) ( 2 2 = 0. Q.23 If ax 1 ² + by 1 ² + cz 1 2 = ax 2 2 + by 2 2 + cz 2 2 = ax 3 2 + by 3 2 + cz 3 2 = d and ax 2 x 3 + by 2 y 3 + cz 2 z 3 = ax 3 x 1 + by 3 y 1 + cz 3 z 1 = ax 1 x 2 + by 1 y 2 + cz 1 z 2 = f, then prove that 3 3 3 2 2 2 1 1 1 z y x z y x z y x = (d f) 2 / 1 abc f 2 d + (a , b , c 0) Q.24 If S r = α r + β r + γ r then show that 4 3 2 3 2 1 2 1 0 S S S S S S S S S = ( α β ) 2 ( β − γ ) 2 ( γ α ) 2 . Q.25 If u = ax 2 + 2 bxy + cy 2 , u = a x 2 + 2 b xy + c y 2 . Prove that y b x a by ax u u y 1 y c x b y b x a cy bx by ax c b a c b a x xy y 2 2 + + = + + + + = . EXERCISE–II Q.1 Solve the following using Cramer’s rule and state whether consistent or not. (a) 0 3 z 2 y x 0 1 z y x 2 0 6 z y x = + + = + = + + (b) 0 y 2 x 6 z y x 3 1 z y 2 x = + = + + = + + (c) 5 z 5 y 3 x 2 7 z 5 y x 3 3 z 5 y 7 x 7 = + + = + + = + Q.2 For what value of K do the following system of equations possess a non trivial (i.e. not all zero) solution over the set of rationals Q? x + K y + 3 z = 0 , 3 x + K y 2 z = 0 , 2 x + 3 y 4 z = 0. For that value of K , find all the solutions of the system. Q.3 The system of equations α x + y + z = α – 1 x + α y + z = α – 1 x + y + α z = α – 1 has no solution. Find α .

#### You've reached the end of your free preview.

Want to read all 11 pages?

• Winter '16
• SFZ
• Math, Determinant, Elementary algebra, Indraprastha Industrial Area, Mahindra Showroom, Evergreen Motors