MHT CET20226 Aug 2022Morning ShiftMathematicsApplication of DerivativesActual
If the tangent to the curve y= x x^2-3 , x R,(x 3 ) at a point ( , ) (0,0) on it, is parallel to the line 2 x+6 y-11=0 , then
Options
- A|2 +6 |=11
- B|6 +2 |=9
- C|6 +2 |=19
- D|2 +6 |=19
Correct answer
C. |6 +2 |=19
Step-by-step solution
y= x x^2-3 d y d x = - (3+x^2 ) (x^2-3 )^2 Now slope of 2 x+6 y-11=0 is -1 3 aligned & A Q - 1 3 = - (3+x^2 ) (x^2-3 )^2 & x^4-9 x^2=0 & x^2 (x^2-9 )=0 & x=0 or x= 3 aligned But x 0 so x= 3 y= 1 2 Hence, = 3, = 1 2 |6 +2 |= | (6 3+2 1 2 ) |=19