For the function f(x) = 4\log _e(x - 1) - 2x^2 + 4x + 5,\,x \gt 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(x) = 0 has a root in the interval (e, e + 1)

Application of Derivatives PYQ — JEE Main 2022 · 24 Jun · Shift 1 · Q29 with Solution

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Correct answer: C

Given function

$f(x)=4\ln(x-1)-2x^2+4x+5, \qquad x>1$

We need to find which statement is NOT correct.

First derivative and monotonicity

Differentiate:

$f'(x)=\frac{4}{x-1}-4x+4$

Simplify:

$f'(x)=\frac{4}{x-1}-4(x-1)$

Let $t=x-1>0$ . Then

$f'(x)=4\left(\frac{1}{t}-t\right)=4\frac{1-t^2}{t}$

So,

$f'(x)>0 \iff 1-t^2>0 \iff t<1 \iff x<2$
$f'(x)=0$ at $x=2$
$f'(x)<0 \iff x>2$

Hence $f$ is increasing on $(1,2)$ and decreasing on $(2,\infty)$ .

So Option A is correct.

Second derivative

Differentiate again:

$f''(x)=-\frac{4}{(x-1)^2}-4$

In particular,

$f''(2)=-\frac{4}{1}-4=-8$

Also,

$f'(e)=\frac{4}{e-1}-4e+4$

Thus,

=\frac{4}{e-1}-4e+12$$ Now check its sign. Using $e\approx 2.718$, $$\frac{4}{e-1}\approx \frac{4}{1.718}\approx 2.33$$ So, $$f'(e)-f''(2)\approx 2.33-10.87+12=3.46>0$$ Therefore, $$f'(e)-f''(2)<0$$ is **false**. So **Option C is NOT correct**. --- 4. **Check Option B: equation $f(x)=-1$** We solve $$4\ln(x-1)-2x^2+4x+5=-1$$ or equivalently, $$f(x)+1=0$$ Let $$g(x)=f(x)+1=4\ln(x-1)-2x^2+4x+6$$ Since $f$ increases on $(1,2)$ and decreases on $(2,\infty)$, the same is true for $g$. Now evaluate: - As $x\to 1^+$, $\ln(x-1)\to -\infty$, so $g(x)\to -\infty$ - At $x=2$, $$g(2)=4\ln 1-8+8+6=6>0$$ - As $x\to \infty$, $g(x)\to -\infty$ because of the term $-2x^2$ So by continuity: - one root lies in $(1,2)$ - one root lies in $(2,\infty)$ And because $g$ is strictly increasing then strictly decreasing, there can be **exactly two** roots. So **Option B is correct**. --- 5. **Check Option D: equation $f(x)=0$ has a root in $(e,e+1)$** Evaluate at the endpoints: At $x=e$, $$f(e)=4\ln(e-1)-2e^2+4e+5$$ Using $\ln(e-1)\approx \ln(1.718)\approx 0.541$, $$f(e)\approx 4(0.541)-2(2.718)^2+4(2.718)+5$$ $$\approx 2.164-14.778+10.872+5$$ $$\approx 3.258>0$$ At $x=e+1$, $$f(e+1)=4\ln e-2(e+1)^2+4(e+1)+5$$ Since $\ln e=1$, $$f(e+1)=4-2(e^2+2e+1)+4e+4+5$$ $$=4-2e^2-4e-2+4e+9$$ $$=11-2e^2$$ Now, $$11-2e^2\approx 11-2(7.389)=11-14.778=-3.778<0$$ Since $f(e)>0$ and $f(e+1)<0$, by the Intermediate Value Theorem there is a root in $(e,e+1)$. So **Option D is correct**. --- 6. **Conclusion** The only statement that is NOT correct is: $$\boxed{\text{C}}$$

More from Application of Derivatives

View written solutionFree

Correct answer: C

Given function

$f(x)=4\ln(x-1)-2x^2+4x+5, \qquad x>1$

We need to find which statement is NOT correct.

First derivative and monotonicity

Differentiate:

$f'(x)=\frac{4}{x-1}-4x+4$

Simplify:

$f'(x)=\frac{4}{x-1}-4(x-1)$

Let $t=x-1>0$ . Then

$f'(x)=4\left(\frac{1}{t}-t\right)=4\frac{1-t^2}{t}$

So,

$f'(x)>0 \iff 1-t^2>0 \iff t<1 \iff x<2$
$f'(x)=0$ at $x=2$
$f'(x)<0 \iff x>2$

Hence $f$ is increasing on $(1,2)$ and decreasing on $(2,\infty)$ .

So Option A is correct.

Second derivative

Differentiate again:

$f''(x)=-\frac{4}{(x-1)^2}-4$

In particular,

$f''(2)=-\frac{4}{1}-4=-8$

Also,

$f'(e)=\frac{4}{e-1}-4e+4$

Thus,

=\frac{4}{e-1}-4e+12$$ Now check its sign. Using $e\approx 2.718$, $$\frac{4}{e-1}\approx \frac{4}{1.718}\approx 2.33$$ So, $$f'(e)-f''(2)\approx 2.33-10.87+12=3.46>0$$ Therefore, $$f'(e)-f''(2)<0$$ is **false**. So **Option C is NOT correct**. --- 4. **Check Option B: equation $f(x)=-1$** We solve $$4\ln(x-1)-2x^2+4x+5=-1$$ or equivalently, $$f(x)+1=0$$ Let $$g(x)=f(x)+1=4\ln(x-1)-2x^2+4x+6$$ Since $f$ increases on $(1,2)$ and decreases on $(2,\infty)$, the same is true for $g$. Now evaluate: - As $x\to 1^+$, $\ln(x-1)\to -\infty$, so $g(x)\to -\infty$ - At $x=2$, $$g(2)=4\ln 1-8+8+6=6>0$$ - As $x\to \infty$, $g(x)\to -\infty$ because of the term $-2x^2$ So by continuity: - one root lies in $(1,2)$ - one root lies in $(2,\infty)$ And because $g$ is strictly increasing then strictly decreasing, there can be **exactly two** roots. So **Option B is correct**. --- 5. **Check Option D: equation $f(x)=0$ has a root in $(e,e+1)$** Evaluate at the endpoints: At $x=e$, $$f(e)=4\ln(e-1)-2e^2+4e+5$$ Using $\ln(e-1)\approx \ln(1.718)\approx 0.541$, $$f(e)\approx 4(0.541)-2(2.718)^2+4(2.718)+5$$ $$\approx 2.164-14.778+10.872+5$$ $$\approx 3.258>0$$ At $x=e+1$, $$f(e+1)=4\ln e-2(e+1)^2+4(e+1)+5$$ Since $\ln e=1$, $$f(e+1)=4-2(e^2+2e+1)+4e+4+5$$ $$=4-2e^2-4e-2+4e+9$$ $$=11-2e^2$$ Now, $$11-2e^2\approx 11-2(7.389)=11-14.778=-3.778<0$$ Since $f(e)>0$ and $f(e+1)<0$, by the Intermediate Value Theorem there is a root in $(e,e+1)$. So **Option D is correct**. --- 6. **Conclusion** The only statement that is NOT correct is: $$\boxed{\text{C}}$$

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