Q103 / 155 · WBJEE 2019 · Mathematics
Let $f(x)=x^{4}-4 x^{3}+4 x^{2}+c, c \in R .$ Then,
- A$f(x)$ has infinitely many zeros in (1,2) for all $c$
- Correct answerB$f(x)$ has exactly one zero in (1,2) if $-1 < c < 0$
- C$f(x)$ has double zeros in (1,2) if $-1 < c < 0$
- Dwhatever be the value of $c, f(x)$ has no zero in (1,2)
Solution
Given,
$$
\begin{array}{ll}
f(x) & =x^{4}-4 x^{3}+4 x^{2}+c \\
f(1) & =1+c \\
\text { and } \quad f(2) & =2^{4}-4(2)^{3}+4(2)^{2}+c
\end{array}
$$
$$
\begin{array}{l}
\begin{aligned}
=& 16-32+16+c \\
&=c
\end{aligned} \\
\text { Now, } f(1) \cdot f(2)=c(1+c) \\
\text { Let } f(1) \cdot f(2) < 0 \\
\Rightarrow \quad c(1+c) < 0 \Rightarrow c \in(-1,0)
\end{array}
$$
For $c \in(-1,0), f(x)$ has exactly one zero is (1,2)
Hence, the correct answer is option: 'B'.