JEE MAIN - Mathematics (2022 - 24th June Morning Shift - No. 8)
For the function
$$f(x) = 4{\log _e}(x - 1) - 2{x^2} + 4x + 5,\,x > 1$$, which one of the following is NOT correct?
$$f(x) = 4{\log _e}(x - 1) - 2{x^2} + 4x + 5,\,x > 1$$, which one of the following is NOT correct?
f is increasing in (1, 2) and decreasing in (2, $$\infty$$)
f(x) = $$-$$1 has exactly two solutions
$$f'(e) - f''(2) < 0$$
f(x) = 0 has a root in the interval (e, e + 1)
Explanation
Lets draw the curve $y=f(x)=4 \log _e(x-1)-2 x^2$ $+4 x+5, x>1$
_24th_June_Morning_Shift_en_8_1.png)
$$ \begin{aligned} &f(x)=4 \log _e(x-1)-2 x^2+4 x+5, x > 1 \\\\ &f^{\prime}(x)=\frac{4}{x-1}-4(x-1) \\\\ &f^{\prime \prime}(x)=\frac{-4}{(x-1)^2}-4\\\\ &\text {For} 1 < x< 2 \Rightarrow f^{\prime}(x) > 0 \\\\ &\text {For}~ x > 2 \Rightarrow f^{\prime}(x)<0 \text { (option } A \text { is correct) } \\\\ &f(x)=-1 \text { has two solution (option } B \text { is correct) } \\\\ &f(e) > 0 \\\\ &f(e+1) < 0 \\\\ &f(e) \cdot f(e+1)<0(\text { option } D \text { is correct) } \\\\ &f^{\prime}(e)-f^{\prime \prime}(2)=\frac{4}{e-1}-4(e-1)+8 > 0 \end{aligned} $$
(option C is incorrect)
$$ \begin{aligned} &f(x)=4 \log _e(x-1)-2 x^2+4 x+5, x > 1 \\\\ &f^{\prime}(x)=\frac{4}{x-1}-4(x-1) \\\\ &f^{\prime \prime}(x)=\frac{-4}{(x-1)^2}-4\\\\ &\text {For} 1 < x< 2 \Rightarrow f^{\prime}(x) > 0 \\\\ &\text {For}~ x > 2 \Rightarrow f^{\prime}(x)<0 \text { (option } A \text { is correct) } \\\\ &f(x)=-1 \text { has two solution (option } B \text { is correct) } \\\\ &f(e) > 0 \\\\ &f(e+1) < 0 \\\\ &f(e) \cdot f(e+1)<0(\text { option } D \text { is correct) } \\\\ &f^{\prime}(e)-f^{\prime \prime}(2)=\frac{4}{e-1}-4(e-1)+8 > 0 \end{aligned} $$
(option C is incorrect)
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