Consider the determinant `f(x)=|[0,x^2-a, x^3-b], [x^2+a,0,x^2+c], [x^4+b,x-c,0]|dot` Statement 1: `f(x)=0` has one root `x=0.` Statement 2: The value of skew symmetric determinant of odd order is always zero.

If a ,b ,c are different, then the value of x satisfying |[0,x^2-a, x^3-b],[ x^2+a,0,x^2+c],[ x^4+b, x-c,0]|=0 is a. c b. a c. b d. 0

If |(0,x-a,x-b),(x+a,0,x-c),(x+b,x+c,0)|=0 then the value of x is -

Consider f(x)={(x^2/|x|, x ne 0),(0, x=0):}

Consider the function f(x)=|x-2|+|x-5|,x in RR . Statement-I : f'(4)=0 . Statement-II : f is continuous is [2, 5], differentiable in (2, 5) and f(2)=f(5) .

Statement 1: If 27 a+9b+3c+d=0, then the equation f(x)=4a x^3+3b x^2+2c x+d=0 has at least one real root lying between (0,3)dot Statement 2: If f(x) is continuous in [a,b], derivable in (a , b) such that f(a)=f(b), then there exists at least one point c in (a , b) such that f^(prime)(c)=0.

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

If f(x)=x^3+b x^2+c x+d and f(0),f(-1) are odd integers, prove that f(x)=0 cannot have all integral roots.

The domain of f(x)=ln(a x^3+(a+b)x^2+(b+c)x+c), where a >0, b^2-4ac=0 , is

The polynomial f(x)=x^4+a x^3+b x^3+c x+d has real coefficients and f(2i)=f(2+i)=0. Find the value of (a+b+c+d)dot

Consider the determinant Delta = |[a_1+b_1x^2,a_1x^2+b_1,c_1],[a_2+b_2x^2,a_2x^2+b_2,c_2],[a_3+b_3x^2,a_3x^2+b_3,c_3]| = 0 , \ w h e r e \ a_i ,b_i , c_i in R \ (i = 1,2,3) \ a n d \ x in R . Statement 1: The value of x satisfying Delta=0 are x=1,-1. Statement 2: If |[a_1,b_1,c_1],[a_2,b_2,c_2],[a_3,b_3,c_3]|=0,t h e n \ Delta=0.