Home
Class 12
MATHS
Let k and K be the minimum and the maxim...

Let k and K be the minimum and the maximum values of the function `f(x) = ((1+x)^(0.6))/(1+x^(0.6)`, and `x in [0,1]` respectively,then the ordered pair (k, K) is equal to

A

`(1, 2^(0.6))`

B

`(2^(-0.4), 2^(0.6))`

C

`(2^(-0.6), 1)`

D

`(2^(-0.4), 1)`

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Topper's Solved these Questions

  • APPLICATIONS OF DERIVATIVES

    MCGROW HILL PUBLICATION|Exercise Question for Previous Year.s B-Architecture Entrance Examination Papers|27 Videos
  • APPLICATIONS OF DERIVATIVES

    MCGROW HILL PUBLICATION|Exercise Exercise ( NUMERICAL ANSWER TYPE QUESTIONS)|17 Videos
  • AREA BY INTEGRATION

    MCGROW HILL PUBLICATION|Exercise Question from Previous Years. B-Architecture Entrance Examination Papers|12 Videos

Similar Questions

Explore conceptually related problems

If the function f defined as f(x)=(1)/(x)-(k-1)/(e^(2x)-1),x!=0, is continuous at x=0, then the ordered pair (k,f(0)) is equal to:

For what value of k is the function f(x)={k(|x|)/(x),x =0 continuous at x=0?

Find the value of k, such that the function f(x) = {(k(x^2-2x), if x = 0):} at x=0 is continuous.

If the function f(x)={[(cos x)^((1)/(x)),x!=0k,x=0k,x=0]} is continuos at x=0 then the value of k]}

If the range of function f(x) = (x + 1)/(k+x^(2)) contains the interval [-0,1] , then values of k can be equal to

If the function f(X)={{:(,(cos x)^(1//x),x ne 0),(,k,x=0):} is continuous at x=0, then the value of k, is

If the function f(x) = {((cosx)^(1/x), ",", x != 1),(k, ",", x = 1):} is continuous at x = 1, then the value of k is

For what value of k , the function defined by f(x) =(log (1+2x) sin x^(0))/(x^(2)) for x ne 0 =K for x=0 is continuous at x=0 ?

If the function f(x)={((1-cosx)/(x^(2))",","for "x ne 0),(k,"for " x =0):} continuous at x = 0, then the value of k is

MCGROW HILL PUBLICATION-APPLICATIONS OF DERIVATIVES-Question for Previous Year.s AIEEE/JEE Main Paper
  1. Let f(x) be a polynomial of degree four having extreme values at x"...

    Text Solution

    |

  2. The equation of a normal to the curve, sin y = x (sinpi/3+y)at x=0 is

    Text Solution

    |

  3. Let k and K be the minimum and the maximum values of the function f(x)...

    Text Solution

    |

  4. From the top of a 64 metres high tower, a stone is thrown upward verti...

    Text Solution

    |

  5. If x= 2cost+cos 2t,y=2sin t-sin 2t ,then " at " t= ( pi)/(4) ,(dy)/(d...

    Text Solution

    |

  6. Tangents are drawn to x^2+y^2=16 from the point P(0, h)dot These tange...

    Text Solution

    |

  7. Consider f(x)=tan^(-1)(sqrt((1+sinx)/(1-sinx))), x in (0,pi/2)dot A n...

    Text Solution

    |

  8. If m and M are the minimum and the maximum values of 4+1/2 sin^2 2x-2c...

    Text Solution

    |

  9. If the tangent at a point P with parameter t, on the curve x=4t^2+3, y...

    Text Solution

    |

  10. Let f(x) =sin^4x+cos^4x. Then f is increasing function in the interval

    Text Solution

    |

  11. Let C be a curve given by y=1 + sqrt(4x-3), x gt (3)/(4) . If P is a p...

    Text Solution

    |

  12. The normal to the curve y(x-2)(x-3)=x+6 at the point where the curve i...

    Text Solution

    |

  13. The tangent at the point (2,-2) to the curve, x^2y^2 - 2x= 4(1-y) does...

    Text Solution

    |

  14. A tangent drawn to the curve y = f(x) at P(x, y) cuts the x and y axes...

    Text Solution

    |

  15. If a point P has co-ordinates (0, -2) and Q is any point on the circle...

    Text Solution

    |

  16. The function f defined by f(x) = x^(3) - 6x^(2) - 36 x + 7 is incr...

    Text Solution

    |

  17. Twenty metres of wire is available for fencing off a flower-bed in ...

    Text Solution

    |

  18. Let f(x) be a polynomial of degree four having extreme values at x=1 a...

    Text Solution

    |

  19. If the curves y^2=6x, 9x^2+by^2=16 intersect each other at right angle...

    Text Solution

    |

  20. If a right circular cone, having maximum volume is inscribed in a sphe...

    Text Solution

    |