Home
Class 12
MATHS
y=x^(2)+1, …. x = 0, x = 3...

`y=x^(2)+1, …. x = 0, x = 3`

A

12

B

21

C

18

D

9

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Topper's Solved these Questions

  • APPLICATIONS OF DEFINITE INTEGRALS

    MARVEL PUBLICATION|Exercise MULTIPLE CHOICE QUESTIONS (PART - B : Mastering The BEST)|8 Videos
  • APPLICATIONS OF DEFINITE INTEGRALS

    MARVEL PUBLICATION|Exercise MULTIPLE CHOICE QUESTIONS (PREVIOUS YEARS (MHT-CET EXAM QUESTIONS))|2 Videos
  • APLICATIONS OF DERIVATIVES

    MARVEL PUBLICATION|Exercise MULTIPLE CHOICE QUESTIONS (TEST YOUR GRASP - II : CHAPTER 12)|19 Videos
  • CONTINUITY F FUNCTIONS

    MARVEL PUBLICATION|Exercise MULTIPLE CHOICE QUESTIONS|131 Videos

Similar Questions

Explore conceptually related problems

Solve for x and y: 2x + 3y + 1 = 0, 3x + 2y - 11 = 0

The tangent from the point of intersection of the lines 2x – 3y + 1 = 0 and 3x – 2y –1 = 0 to the circle x^(2) + y^(2) + 2x – 4y = 0 is

{:((2a - 1)x + 3y - 5 = 0),(3x + (b - 1) y - 2 = 0):}

Find area of region bounded by y=x^(2)-3x+2,x=1,x=2 and y=0

( 5)/(x)- ( 3) /( y ) = 1 , ( 3) /( 2 x ) + ( 2) /( 3y ) = 5 ( x ne 0 , y ne 0 )

(3) / (x) - (1) / (y) + 9 = 0 (2) / (x) + (3) / (y) = 5, (! = 0, y! = 0)

(x+2y +1) dx - (2x + 4y +3)dy = 0, x+2y = u

y-1=m_1(x-3) and y - 3 = m_2(x - 1) are two family of straight lines, at right angled to each other. The locus of their point of intersection is: (A) x^2 + y^2 - 2x - 6y + 10 = 0 (B) x^2 + y^2 - 4x - 4y +6 = 0 (C) x^2 + y^2 - 2x - 6y + 6 = 0 (D) x^2 + y^2 - 4x - by - 6 = 0