Let f(x) = x4 - 4x3 + 4x2 +c, c ∈ mathbbR. Then

Question

Let f(x) = x4 - 4x3 + 4x2 +c, c ∈ mathbbR. Then (A) f(x) has infinitely many zeros in (1, 2) for all c (B) f(x) has exactly one zero in (1, 2) if -1 < c < 0 (C) f(x) has double zeros in (1, 2) if -1 < c < 0 (D) Whatever be the value of c, f(x) has no zero in (1, 2)

Tardigrade Admin · Accepted Answer

f(x) = x4 - 4x3 + 4x2 +c, c ∈ R. Then F'(x) = 4x3 - 12x2 + 8x = 4x (x2 - 3x + 2) = 4x(x - 1) (x - 2) if -1 < c < 0 f(1) = 1 - 4 + 4 + c = 1 + c > 0 f(2) = 16 - 32 + 16 + c = c < 0 f(x) has exactly are zero in (1, 2) if -1 < c < 0

Q. Let $f (x) = x^{4} - 4 x^{3} + 4 x^{2} + c, c \in R .$ Then

f(x) has infinitely many zeros in (1, 2) for all c

f(x) has exactly one zero in (1, 2) if -1 < c < 0

f(x) has double zeros in (1, 2) if -1 < c < 0

Whatever be the value of c, f(x) has no zero in (1, 2)

$f (x) = x^{4} - 4 x^{3} + 4 x^{2} + c, c \in R$ . Then $F^{'} (x) = 4 x^{3} - 12 x^{2} + 8 x = 4 x (x^{2} - 3 x + 2) = 4 x (x - 1) (x - 2)$
if $- 1 < c < 0$
$f (1) = 1 - 4 + 4 + c$
$= 1 + c > 0$
$f (2) = 16 - 32 + 16 + c$
$= c < 0$
$f (x)$ has exactly are zero in $(1, 2)$ if $- 1 < c < 0$

Q. Let f(x)=x4−4x3+4x2+c,c∈R. Then