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

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.

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

If the equation x^(log_(a)x^(2))=(x^(k-2))/a^(k),a ne 0 has exactly one solution for x, then the value of k is/are

Let f(x)=sin^(-1)x+|sin^(-1)x|+sin^(-1)|x| If the equation f(x) = k has two solutions, then true set of values of k is

Let h(x)=|k x+5| ,then domainof f(x) be [-5,7], the domain of f(h(x)) be [-6,1] ,and the range of h(x) be the same as the domain of f(x). Then the value of k is. (a)1 (b) 2 (c) 3 (d) 4

Let f:""R vec R be defined by f(x)={k-2x , if""xlt=-1 (-2x+3),x >-1} . If f has a local minimum at x=-1 , then a possible value of k is (1) 0 (2) -1/2 (3) -1 (4) 1

Let f(x)= 3/(x-2)+4/(x-3)+5/(x-4) . Then f(x)=0 has (A) exactly one real root in (2,3) (B) exactly one real root in (3,4) (C) at least one real root in (2,3) (D) none of these

Let f:""RrarrR be defined by f(x)={(k-2x , if xle-1),( 2x+3, if x gt-1):} . If f has a local minimum at x""=""1 , then a possible value of k is (1) 0 (2) -1/2 (3) -1 (4) 1

Consider the polynomial f(x)=1 + 2x + 3x^2 +4x^3 for all x in R So f(x) has exactly one real root in the interval