Home
Class 12
MATHS
Let f(t)=|t-1|-|t|+|t+1|, AA t in R. Fin...

Let `f(t)=|t-1|-|t|+|t+1|, AA t in R`. Find `g(x) = max {f(t):x+1letlex+2}, AA x in R`. Find `g(x)` and the area bounded by the curve `y=g(x)`, the X-axis and the lines `x=-3//2 and x=5`.

Text Solution

Verified by Experts

The correct Answer is:
`g(x)={(-x-1,x <=-5//2),(4+x,-5//2 -1//2):}`
and area = `101/4` sq units.
Promotional Banner

Topper's Solved these Questions

  • AREA OF BOUNDED REGIONS

    ARIHANT MATHS ENGLISH|Exercise Area of bounded Regions Exercise 7: Subjective Type Questions|1 Videos
  • AREA OF BOUNDED REGIONS

    ARIHANT MATHS ENGLISH|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|23 Videos
  • AREA OF BOUNDED REGIONS

    ARIHANT MATHS ENGLISH|Exercise Exercise (Single Integer Answer Type Questions)|8 Videos
  • BIONOMIAL THEOREM

    ARIHANT MATHS ENGLISH|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|21 Videos

Similar Questions

Explore conceptually related problems

The area bounded by the curve y=f(x) the coordinate axes and the line x = t is given by te^(t) then f(x) =

Let g(t)=abs(t-1)-abs(t)+abs(t+1),forall " " t in R. Find f(x)=max{g(t):-3/2 le t le x},forall x in ((-3)/2,infty) .]

Let f(x) = 1 + 4x - x^(2), AA x in R g(x) = max {f(t), x le t le (x + 1), 0 le x lt 3min {(x + 3), 3 le x le 5} Verify conntinuity of g(x), for all x in [0, 5]

Find the value of t for which the area bounded by the lines y = 0, x = 0, x = 1 and the curve y=t^2x^2+tx+1 is minimum.

Let f (x) = [{:(1-x,,, 0 le x le 1),(0,,, 1 lt x le 2 and g (x) = int_(0)^(x) f (t)dt.),( (2-x)^(2),,, 2 lt x le 3):} Let the tangent to the curve y = g( x) at point P whose abscissa is 5/2 cuts x-axis in point Q. Let the prependicular from point Q on x-axis meets the curve y =g (x) in point R .Find equation of tangent at to y=g(x) at P .Also the value of g (1) =

If the tangent to the curve x = t^(2) - 1, y = t^(2) - t is parallel to x-axis , then

Let f (x) = x^3-x^2-x+1 and g(x) = {max{f(t); 0<=t<=x}, 0<=x<=1, 3-x, 1<=x<=2 Discuss the continuity and differentiability of the function g (x) in the interval (0, 2).

If f: R^+ to R , f(x) + 3x f(1/x)= 2(x+1),t h e n find f(x)

If x=f(t) and y=g(t) , then find (dy)/(dx)

A function g(x) is defined g(x)=2f(x^2/2)+f(6-x^2),AA x in R such f''(x)> 0, AA x in R , then m g(x) has

ARIHANT MATHS ENGLISH-AREA OF BOUNDED REGIONS-Exercise (Subjective Type Questions)
  1. Find the continuous function f where (x^4-4x^2)lt=f(x)lt=(2x^2-x^3) su...

    Text Solution

    |

  2. Let f(t)=|t-1|-|t|+|t+1|, AA t in R. Find g(x) = max {f(t):x+1letlex+2...

    Text Solution

    |

  3. Let f(x)= minimum {e^(x),3//2,1+e^(-x)},0lexle1. Find the area bounded...

    Text Solution

    |

  4. Find the area bounded by y=f(x) and the curve y=2/(1+x^2) satisfying ...

    Text Solution

    |

  5. The value of overset(sin^(2)x)underset(0)int sin^(-1)sqrt(t)dt+overs...

    Text Solution

    |

  6. Let T be an acute triangle Inscribe a pair R,S of rectangle in T as sh...

    Text Solution

    |

  7. Find the maximum area of the ellipse that can be inscribed in an isoce...

    Text Solution

    |

  8. Find the area of the region bounded by curve y=25^(x)+16 and the curve...

    Text Solution

    |

  9. If the circles of the maximum area inscriabed in the region bounded by...

    Text Solution

    |

  10. Find limit of the ratio of the area of the triangle formed by the orig...

    Text Solution

    |

  11. Find the area of curve enclosed by |x+y|+|x-y|le4,|x|le1, y ge sqrt(x^...

    Text Solution

    |

  12. Calculate the area enclosed by the curve 4lex^(2)+y^(2)le2(|x|+|y|).

    Text Solution

    |

  13. Find the area enclosed by the curve [x]+[y]-4 in 1st quadrant (where [...

    Text Solution

    |

  14. Sketch the region and find the area bounded by the curves |y+x|le1,|y-...

    Text Solution

    |

  15. Find the area of the region bounded by the curve 2^(|x|)|y|+2^(|x|-1)l...

    Text Solution

    |

  16. The value of the parameter a(a>=1) for which the area of the figure bo...

    Text Solution

    |