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

Let `f(x)=x+2|x+1|+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

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 Matching Type Questions|1 Videos

  • FUNCTIONS

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

  • HYPERBOLA

    ARIHANT MATHS|Exercise Hyperbola Exercise 11 : Questions Asked in Previous 13 Years Exams|3 Videos

Similar Questions

Explore conceptually related problems

If x=2k-1 and y=k is a solution of the 3x-5y-9=0 find the value of k.

Let f(x) be a polynominal with real coefficients such that xf(x)=f'(x) times f''(x). If f(x)=0 is satisfied x=1,2,3 only, then the value of f'(1)f'(2)f'(3) is

Knowledge Check

  • If f(x)= {(1,-,kx, x,le3),(2x,+,3,x,>3):} is a continuous function, then the value of k is

    A
    `8/3`
    B
    `-8/3`
    C
    `3/8`
    D
    None of these

  • If x ne 1 and f(x) = ( x + 1)/( x - 1) is a real function, then f(f(f(2))) is

    A
    1
    B
    2
    C
    3
    D
    4

    • Similar Questions

    Explore conceptually related problems

    Let f(x)=(alphax)/((x+1)),x!=-1. The for what value of alpha is f(f(x))=x ? (a) sqrt(2) " " (b) -sqrt(2) " " (c) 1 " " (d) -1

    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 a. 0 b. 0.5 c. 1.25 d. 1.5

    Let f(x)=(1+b^2)x^2+2b x+1 and let m(b) the minimum value of f(x)dot As b varies, the range of m(b) is (a) [0,1] (b) (0,1/2] (c) [1/2,1] (d) (0,1]

    If f(x) is continuous and derivable, AA x in R and f'(c)=0 for exactly 2 real value of 'c'. Then the number of real and distinct value of 'd' for which f(d)=0 can be

    Let f(x) = Max. {x^2, (1 - x)^2, 2x(1 - x)} where x in [0, 1] If Rolle's theorem is applicable for f(x) on largest possible interval [a, b] then the value of 2(a+b+c) when c in [a, b] such that f'(c) = 0, is

    Let f_(1) (x) and f_(2) (x) be twice differentiable functions where F(x)= f_(1) (x) + f_(2) (x) and G(x) = f_(1)(x) - f_(2)(x), AA x in R, f_(1) (0) = 2 and f_(2)(0)=1. "If" f'_(1)(x) = f_(2) (x) and f'_(2) (x) = f_(1) (x) , AA x in R then the number of solutions of the equation (F(x))^(2) =(9x^(4))/(G(x)) is...... .

    Home
    Profile