Home
Class 12
MATHS
Let f(x)=x+2|x+1|+x-1|dotIff(x)=k has ex...

Let `f(x)=x+2|x+1|+x-1|dotIff(x)=k` has exactly one real solution, then the value of `k` is 3 (b) 0 (c) 1 (d) 2

Text Solution

Verified by Experts

The correct Answer is:
3
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • GRAPHICAL TRANSFORMATIONS

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|10 Videos

  • GRAPHICAL TRANSFORMATIONS

    ARIHANT MATHS|Exercise Exercise (Passage Based Questions)|4 Videos

  • FUNCTIONS

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|24 Videos

  • HYPERBOLA

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|17 Videos

Similar Questions

Explore conceptually related problems

Let f(x)=x+2|x+1|+2|x-1|* If f(x)=k has exactly one real solution,then the value of k is (a) 3 (b) 0(c)1(d)2

Let f(x)=|x-1|+a|-4,quad if f(x)=0 has three real solution,then the values of lies in

Let f(x)=[2+x,,xgt=0 , 4-x,,xlt0, If f(f(x))=k has at least one solution,then smallest value of k is

Let f(x)=2+x, x>=0 and f(x)=4-x , x < 0 if f(f(x)) =k has at least one solution, then smallest value of k is

Let f(x)=(e^(x))/(x^(2))x in R-{0}. If f(x)=k has two distinct real roots,then the range of k, is

Let f(x)=x + 2|x+1|+2| x-1| . Find the values of k if f(x)=k (i) has exactly one real solution, (ii) has two negative solutions, (iii) has two solutions of opposite sign.

Let |x-2|+|x|+|x+3|=k ,have real solution then possible value of k is are

Let f(x)=x^(3)-3x^(2)+2x. If the equation f(x)=k has exactly one positive and one negative solution then the value of k equals.-(2sqrt(3))/(9)(b)-(2)/(9)(2)/(3sqrt(3)) (d) (1)/(3sqrt(3))

Let f(x)=min{e^(x), (3)/(2), 1+e^(-x)} if f(x)=k has at least 2 solutions then k can not be

If the equation |2-x|-|x+1|=k has exactly one solution,then number of integral values of k is (A)7(B)5(C)4(D)3

Home
Profile