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Differentiation question

2023 · 29 Jan · Shift 2 · Q38
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  5. /2023 · 29 Jan · Shift 2 · Q38

Differentiation question

2023 · 29 Jan · Shift 2 · Q38

JEE MainMathematicsDifferentiationMCQ+4 / −1
Let fff and ggg be the twice differentiable functions on R\mathbb{R}R such that f′′(x)=g′′(x)+6xf′(1)=4g′(1)−3=9f(2)=3g(2)=12f''(x)=g''(x)+6xf'(1)=4g'(1)-3=9f(2)=3g(2)=12f′′(x)=g′′(x)+6xf′(1)=4g′(1)−3=9f(2)=3g(2)=12. Then which of the following is NOT true?
  1. A
    g(−2)−f(−2)=20g(-2)-f(-2)=20g(−2)−f(−2)=20
  2. B
    There exists x0∈(1,3/2)x_0\in(1,3/2)x0​∈(1,3/2) such that f(x0)=g(x0)f(x_0)=g(x_0)f(x0​)=g(x0​)
  3. C
    ∣f′(x)−g′(x)∣<6⇒−1<x<1|f'(x)-g'(x)| \lt 6\Rightarrow -1 \lt x \lt 1∣f′(x)−g′(x)∣<6⇒−1<x<1
  4. D
    If −1<x<2-1 \lt x \lt 2−1<x<2, then ∣f(x)−g(x)∣<8|f(x)-g(x)| \lt 8∣f(x)−g(x)∣<8
View written solutionFree

Correct answer: A, B AND C ARE NOT TRUE; D IS TRUE. THE STORED ANSWER D APPEARS INCORRECT.

Let h(x)=f(x)−g(x).h(x)=f(x)-g(x).h(x)=f(x)−g(x). Then the given condition f′′(x)=g′′(x)+6xf''(x)=g''(x)+6xf′′(x)=g′′(x)+6x becomes h′′(x)=6x.h''(x)=6x.h′′(x)=6x.

Also, from

\qquad 4g'(1)-3=9 \,\Rightarrow\, g'(1)=3,$$ we get $$h'(1)=f'(1)-g'(1)=1.$$ And from $$9f(2)=3g(2)=12,$$ we obtain $$f(2)=\frac{12}{9}=\frac43, \qquad g(2)=\frac{12}{3}=4,$$ so $$h(2)=f(2)-g(2)=\frac43-4=-\frac83.$$ Now solve for $h$. --- ### 1. Find $h(x)$ Since $$h''(x)=6x,$$ integrating, $$h'(x)=3x^2+C_1.$$ Using $h'(1)=1$, $$3+C_1=1 \Rightarrow C_1=-2.$$ Hence $$h'(x)=3x^2-2.$$ Integrating again, $$h(x)=x^3-2x+C_2.$$ Using $h(2)=-\frac83$, $$8-4+C_2=-\frac83 \Rightarrow 4+C_2=-\frac83 \Rightarrow C_2=-\frac{20}{3}.$$ Thus $$h(x)=f(x)-g(x)=x^3-2x-\frac{20}{3}.$$ So also, $$g(x)-f(x)=-h(x)=-x^3+2x+\frac{20}{3}.$$ --- ### 2. Check option A We need $g(-2)-f(-2)$. Compute: $$h(-2)=(-2)^3-2(-2)-\frac{20}{3}=-8+4-\frac{20}{3}=-4-\frac{20}{3}=-\frac{32}{3}.$$ Hence $$g(-2)-f(-2)=-h(-2)=\frac{32}{3}.$$ This is **not** equal to $20$. So **A is false**. --- ### 3. Check option B We need to see whether there exists $x_0\in(1,3/2)$ such that $f(x_0)=g(x_0)$, i.e. $h(x_0)=0$. Now $$h(1)=1-2-\frac{20}{3}=-\frac{23}{3}<0,$$ and $$h\left(\frac32\right)=\left(\frac32\right)^3-2\left(\frac32\right)-\frac{20}{3} =\frac{27}{8}-3-\frac{20}{3}.

Compute:

so h\left(\frac32\right)=\frac38-\frac{20}{3}= rac{9-160}{24}=-\frac{151}{24}<0. Thus hhh is negative at both endpoints, so this alone does not prove anything.

Now check monotonicity on (1,3/2)(1,3/2)(1,3/2): h′(x)=3x2−2.h'(x)=3x^2-2.h′(x)=3x2−2. For x>1x>1x>1, h′(x)>3(1)−2=1>0,h'(x)>3(1)-2=1>0,h′(x)>3(1)−2=1>0, so hhh is strictly increasing on (1,3/2)(1,3/2)(1,3/2). But since both endpoint values are negative, h(x)<0h(x)<0h(x)<0 throughout (1,3/2)(1,3/2)(1,3/2). Therefore there is no x0∈(1,3/2)x_0\in(1,3/2)x0​∈(1,3/2) such that h(x0)=0h(x_0)=0h(x0​)=0.

So B is false.

4. Check option C

We have f′(x)−g′(x)=h′(x)=3x2−2.f'(x)-g'(x)=h'(x)=3x^2-2.f′(x)−g′(x)=h′(x)=3x2−2. So ∣f′(x)−g′(x)∣<6  ⟺  ∣3x2−2∣<6.|f'(x)-g'(x)|<6 \iff |3x^2-2|<6.∣f′(x)−g′(x)∣<6⟺∣3x2−2∣<6. This means −6<3x2−2<6.-6<3x^2-2<6.−6<3x2−2<6. Add 222: −4<3x2<8.-4<3x^2<8.−4<3x2<8. Since 3x2≥03x^2\ge 03x2≥0, the left inequality is automatic. Thus 3x2<8  ⟺  x2<83  ⟺  ∣x∣<83.3x^2<8 \iff x^2<\frac83 \iff |x|<\sqrt{\frac83}.3x2<8⟺x2<38​⟺∣x∣<38​​. Now 83>1.\sqrt{\frac83}>1.38​​>1. So the implication ∣f′(x)−g′(x)∣<6⇒−1<x<1|f'(x)-g'(x)|<6 \Rightarrow -1<x<1∣f′(x)−g′(x)∣<6⇒−1<x<1 is false, because values such as x=1.2x=1.2x=1.2 satisfy ∣x∣<8/3|x|<\sqrt{8/3}∣x∣<8/3​ but not ∣x∣<1|x|<1∣x∣<1.

Hence C is false.

5. Check option D

We have ∣f(x)−g(x)∣=∣h(x)∣=∣x3−2x−203∣.|f(x)-g(x)|=|h(x)|=\left|x^3-2x-\frac{20}{3}\right|.∣f(x)−g(x)∣=∣h(x)∣=​x3−2x−320​​. The statement is: −1<x<2⇒∣h(x)∣<8.-1<x<2 \Rightarrow |h(x)|<8.−1<x<2⇒∣h(x)∣<8. Check a point in this interval, say x=0x=0x=0: ∣h(0)∣=∣−203∣=203<8.|h(0)|=\left|-\frac{20}{3}\right|=\frac{20}{3}<8.∣h(0)∣=​−320​​=320​<8. But need it for all x∈(−1,2)x\in(-1,2)x∈(−1,2).

Since h′(x)=3x2−2,h'(x)=3x^2-2,h′(x)=3x2−2, critical points occur at x=±23.x=\pm\sqrt{\frac23}.x=±32​​. Evaluate relevant values:

At x=−1x=-1x=−1, h(−1)=−1+2−203=1−203=−173,h(-1)=-1+2-\frac{20}{3}=1-\frac{20}{3}=-\frac{17}{3},h(−1)=−1+2−320​=1−320​=−317​, so ∣h(−1)∣=173<8.|h(-1)|=\frac{17}{3}<8.∣h(−1)∣=317​<8.

As x→2−x\to 2^{-}x→2−, h(2)=−83,h(2)=-\frac83,h(2)=−38​, so ∣h(2)∣=83<8.|h(2)|=\frac83<8.∣h(2)∣=38​<8.

At x=−2/3x=-\sqrt{2/3}x=−2/3​, using x2=23x^2=\frac23x2=32​, we get x3=x⋅23.x^3=x\cdot\frac23.x3=x⋅32​. Thus

For x=−2/3x=-\sqrt{2/3}x=−2/3​, h(−23)=4323−203,h\left(-\sqrt{\frac23}\right)=\frac43\sqrt{\frac23}-\frac{20}{3},h(−32​​)=34​32​​−320​, which is still negative and clearly greater than −8-8−8.

At x=2/3x=\sqrt{2/3}x=2/3​, h(23)=−4323−203,h\left(\sqrt{\frac23}\right)=-\frac43\sqrt{\frac23}-\frac{20}{3},h(32​​)=−34​32​​−320​, whose magnitude is 203+4323<203+43<8.\frac{20}{3}+\frac43\sqrt{\frac23}<\frac{20}{3}+\frac43<8.320​+34​32​​<320​+34​<8. So indeed ∣h(x)∣<8|h(x)|<8∣h(x)∣<8 on (−1,2)(-1,2)(−1,2).

Therefore D is true.

6. Final conclusion

The false statements are:

  • A
  • B
  • C

So there is more than one statement which is NOT true. Hence the given problem, as a single-correct MCQ, is inconsistent.

The stored answer says D is NOT true, but our work shows D is actually true.

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