Straight lines JEE 2
If x = 2 , 4 are the solutions of the equation ∣ m x + α ∣ + ∣ m x + β ∣ = c where m > 0 and α , β , c are non zero constants , then ∣ m α + β ∣ is equal to
Also check this Optics Problem
The answer is 6.00.
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2 solutions
W h e n x = 2 , w e h a v e , ( 2 m + α ) + ( 2 m + β ) = c , a n d ( 2 m + α ) − ( 2 m + β ) = c . S u b t r a c t i n g t h e f i r s t e q u a t i o n f r o m t h e s e c o n d w e g e t 2 m = − β . . . . . . . . . . . . ( A ) S i m i l a r l y f o r x = 4 , w e g e t 4 m = − α . . . . . ( B ) S o l v i n g ( A ) a n d ( B ) , w e g e t 6 m = − ( α + b e t a ) . ⟹ ∣ ∣ ∣ m α + β ∣ ∣ ∣ = 6 . N o t e t h a t s o l v i n g a b s v a l u e s , w e g e t f o u r e q u a t i o n s b u t p a i r s o f t w o s a r e i d e n t i c a l . e . g . ∣ a + b ∣ + ∣ a − b ∣ = 0 , g i v e s 2 a = 0 , . . . . 2 b = 0 , . . . . − 2 a = 0 , . . . . − 2 b = 0 S i n c e t h e e q u a t i o n i s e q u a l t o 0 , a n d 0 = − 0 , f i r s t a n d T h i r d a n d s e c o n d a n d f o u r t h e q u a t i o n s a r e i d e n t i c a l .
|x+alpha ÷m|+|x+beta÷m|=c÷m (since m>0) draw graph. x coordinate of Midpt of (2,c÷m) and (4,c÷m) is same as that of (-alpha÷m,0) and (-beta÷m,0) . so answer =6