Let x=2 be a local minima of the function f(x)=2x^4-18x^2+8x+12,x\in(-4,4) . If M is local maximum value of the function f in ( -4,4) , then M =

Application of Derivatives PYQ — JEE Main 2023 · 25 Jan · Shift 1 · Q26 with Solution

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Correct answer: C

Given function

We need the local maximum value of

\quad x\in(-4,4).$$ It is given that $x=2$ is a local minimum. We will find all critical points and then identify the local maximum value. --- 2. **Find critical points** Differentiate: $$f'(x)=8x^3-36x+8=4(2x^3-9x+2).$$ Since $x=2$ is a critical point, $$2(2)^3-9(2)+2=16-18+2=0,$$ so $(x-2)$ is a factor of $2x^3-9x+2$. Now factor it: $$2x^3-9x+2=(x-2)(2x^2+4x-1).$$ So, $$f'(x)=4(x-2)(2x^2+4x-1).$$ Solve $$2x^2+4x-1=0.$$ Using quadratic formula: $$x=\frac{-4\pm\sqrt{16+8}}{4}=\frac{-4\pm 2\sqrt6}{4}=\frac{-2\pm\sqrt6}{2}.$$ That is, $$x=-1\pm\frac{\sqrt6}{2}.$$ Hence the critical points are $$x=2,\quad x=-1+\frac{\sqrt6}{2},\quad x=-1-\frac{\sqrt6}{2}.$$ All lie in $(-4,4)$. --- 3. **Use second derivative test** $$f''(x)=24x^2-36=12(2x^2-3).$$ At $x=2$: $$f''(2)=24(4)-36=60>0,$$ so $x=2$ is indeed a local minimum. Now test the other two critical points. Let $$\alpha=-1+\frac{\sqrt6}{2}, \qquad \beta=-1-\frac{\sqrt6}{2}.$$ For these points, note that $$|\alpha|<\sqrt{\frac32}, \quad |\beta|>\sqrt{\frac32}.$$ Actually it is easier to check directly: - Near the smaller root $\beta$, $f''(\beta)>0$, so it is a local minimum. - Near the larger root $\alpha$, $f''(\alpha)<0$, so it is a local maximum. Thus the local maximum occurs at $$x=-1+\frac{\sqrt6}{2}.$$ --- 4. **Compute the local maximum value** We need $$M=f\left(-1+\frac{\sqrt6}{2}\right).$$ Let $$x=-1+\frac{\sqrt6}{2}.$$ Since $x$ satisfies $$2x^2+4x-1=0,$$ we can simplify powers of $x$. From $$2x^2+4x-1=0 \implies x^2=\frac{1-4x}{2}.$$ Then $$x^4=(x^2)^2.$$ But a cleaner method is to reduce $f(x)$ using the relation $2x^2+4x-1=0$. Given $$f(x)=2x^4-18x^2+8x+12.$$ Using $$x^2=\frac{1-4x}{2},$$ first compute $x^4$: $$x^4=x^2\cdot x^2=\left(\frac{1-4x}{2}\right)^2=\frac{1-8x+16x^2}{4}.$$ Now substitute $x^2=\frac{1-4x}{2}$ again: $$x^4=\frac{1-8x+16\cdot \frac{1-4x}{2}}{4} =\frac{1-8x+8-32x}{4} =\frac{9-40x}{4}.$$ Therefore, $$2x^4=\frac{9-40x}{2}, \qquad -18x^2=-18\cdot \frac{1-4x}{2}=-9+36x.$$ So $$f(x)=\frac{9-40x}{2}-9+36x+8x+12.$$ Combine terms: $$-9+12=3,$$ and $$36x+8x=44x.$$ Thus $$f(x)=\frac{9-40x}{2}+44x+3 =\frac{9-40x+88x+6}{2} =\frac{15+48x}{2}.$$ Now substitute $$x=-1+\frac{\sqrt6}{2}:$$ $$15+48x=15+48\left(-1+\frac{\sqrt6}{2}\right)=15-48+24\sqrt6=24\sqrt6-33.$$ Hence, $$M=f(x)=\frac{24\sqrt6-33}{2}=12\sqrt6-\frac{33}{2}.$$ --- 5. **Match with options** $$M=12\sqrt6-\frac{33}{2}$$ which corresponds to **Option C**. --- 6. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** So the answer agrees with the stored answer.

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View written solutionFree

Correct answer: C

Given function

We need the local maximum value of

\quad x\in(-4,4).$$ It is given that $x=2$ is a local minimum. We will find all critical points and then identify the local maximum value. --- 2. **Find critical points** Differentiate: $$f'(x)=8x^3-36x+8=4(2x^3-9x+2).$$ Since $x=2$ is a critical point, $$2(2)^3-9(2)+2=16-18+2=0,$$ so $(x-2)$ is a factor of $2x^3-9x+2$. Now factor it: $$2x^3-9x+2=(x-2)(2x^2+4x-1).$$ So, $$f'(x)=4(x-2)(2x^2+4x-1).$$ Solve $$2x^2+4x-1=0.$$ Using quadratic formula: $$x=\frac{-4\pm\sqrt{16+8}}{4}=\frac{-4\pm 2\sqrt6}{4}=\frac{-2\pm\sqrt6}{2}.$$ That is, $$x=-1\pm\frac{\sqrt6}{2}.$$ Hence the critical points are $$x=2,\quad x=-1+\frac{\sqrt6}{2},\quad x=-1-\frac{\sqrt6}{2}.$$ All lie in $(-4,4)$. --- 3. **Use second derivative test** $$f''(x)=24x^2-36=12(2x^2-3).$$ At $x=2$: $$f''(2)=24(4)-36=60>0,$$ so $x=2$ is indeed a local minimum. Now test the other two critical points. Let $$\alpha=-1+\frac{\sqrt6}{2}, \qquad \beta=-1-\frac{\sqrt6}{2}.$$ For these points, note that $$|\alpha|<\sqrt{\frac32}, \quad |\beta|>\sqrt{\frac32}.$$ Actually it is easier to check directly: - Near the smaller root $\beta$, $f''(\beta)>0$, so it is a local minimum. - Near the larger root $\alpha$, $f''(\alpha)<0$, so it is a local maximum. Thus the local maximum occurs at $$x=-1+\frac{\sqrt6}{2}.$$ --- 4. **Compute the local maximum value** We need $$M=f\left(-1+\frac{\sqrt6}{2}\right).$$ Let $$x=-1+\frac{\sqrt6}{2}.$$ Since $x$ satisfies $$2x^2+4x-1=0,$$ we can simplify powers of $x$. From $$2x^2+4x-1=0 \implies x^2=\frac{1-4x}{2}.$$ Then $$x^4=(x^2)^2.$$ But a cleaner method is to reduce $f(x)$ using the relation $2x^2+4x-1=0$. Given $$f(x)=2x^4-18x^2+8x+12.$$ Using $$x^2=\frac{1-4x}{2},$$ first compute $x^4$: $$x^4=x^2\cdot x^2=\left(\frac{1-4x}{2}\right)^2=\frac{1-8x+16x^2}{4}.$$ Now substitute $x^2=\frac{1-4x}{2}$ again: $$x^4=\frac{1-8x+16\cdot \frac{1-4x}{2}}{4} =\frac{1-8x+8-32x}{4} =\frac{9-40x}{4}.$$ Therefore, $$2x^4=\frac{9-40x}{2}, \qquad -18x^2=-18\cdot \frac{1-4x}{2}=-9+36x.$$ So $$f(x)=\frac{9-40x}{2}-9+36x+8x+12.$$ Combine terms: $$-9+12=3,$$ and $$36x+8x=44x.$$ Thus $$f(x)=\frac{9-40x}{2}+44x+3 =\frac{9-40x+88x+6}{2} =\frac{15+48x}{2}.$$ Now substitute $$x=-1+\frac{\sqrt6}{2}:$$ $$15+48x=15+48\left(-1+\frac{\sqrt6}{2}\right)=15-48+24\sqrt6=24\sqrt6-33.$$ Hence, $$M=f(x)=\frac{24\sqrt6-33}{2}=12\sqrt6-\frac{33}{2}.$$ --- 5. **Match with options** $$M=12\sqrt6-\frac{33}{2}$$ which corresponds to **Option C**. --- 6. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** So the answer agrees with the stored answer.

Application of Derivatives question

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Application of Derivatives question

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