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Definite Integration question

2015 · Shift 2 · Q36
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Definite Integration question

2015 · Shift 2 · Q36

JEE AdvancedMathematicsDefinite IntegrationMCQ+4 / −1
Let f′(x)=192x32+sin⁡4 πxf'\left( x \right) = {{192{x^3}} \over {2 + {{\sin }^4}\,\pi x}}f′(x)=2+sin4πx192x3​ for all x∈R  x \in R\,\,x∈R with f(12)=0f\left( {{1 \over 2}} \right) = 0f(21​)=0. If m≤∫1/21f(x)dx≤M,m \le \int\limits_{1/2}^1 {f\left( x \right)dx \le M,}m≤1/2∫1​f(x)dx≤M, then the possible values of mmm and MMM are
  1. A
    m=13,M=24m=13,M=24m=13,M=24
  2. B
     m=14,M=12\,m = {1 \over 4},M = {1 \over 2}m=41​,M=21​
  3. C
    m=−11,M=0m=-11,M=0m=−11,M=0
  4. D
    m=1,M=12m=1,M=12m=1,M=12
View written solutionFree

Correct answer: D

  1. Given data

We have

f′(x)=192x32+sin⁡4(πx)f'(x)=\frac{192x^3}{2+\sin^4(\pi x)}f′(x)=2+sin4(πx)192x3​

for all real xxx, and

f(12)=0.f\left(\frac12\right)=0.f(21​)=0.

We need bounds for

I=∫1/21f(x) dx.I=\int_{1/2}^1 f(x)\,dx.I=∫1/21​f(x)dx.
  1. Bound f′(x)f'(x)f′(x)

Since 0≤sin⁡4(πx)≤1,0\le \sin^4(\pi x)\le 1,0≤sin4(πx)≤1, we get

Therefore,

192x33≤f′(x)≤192x32\frac{192x^3}{3}\le f'(x)\le \frac{192x^3}{2}3192x3​≤f′(x)≤2192x3​

that is,

64x3≤f′(x)≤96x3.64x^3\le f'(x)\le 96x^3.64x3≤f′(x)≤96x3.
  1. Integrate to bound f(x)f(x)f(x)

Using f(1/2)=0f(1/2)=0f(1/2)=0, for x∈[1/2,1]x\in[1/2,1]x∈[1/2,1],

f(x)=∫1/2xf′(t) dt.f(x)=\int_{1/2}^x f'(t)\,dt.f(x)=∫1/2x​f′(t)dt.

Hence,

∫1/2x64t3 dt≤f(x)≤∫1/2x96t3 dt.\int_{1/2}^x 64t^3\,dt \le f(x) \le \int_{1/2}^x 96t^3\,dt.∫1/2x​64t3dt≤f(x)≤∫1/2x​96t3dt.

Now,

∫64t3dt=16t4,\int 64t^3dt=16t^4,∫64t3dt=16t4,

so

∫1/2x64t3dt=16(x4−(12)4)=16x4−1.\int_{1/2}^x 64t^3dt=16\left(x^4-\left(\frac12\right)^4\right)=16x^4-1.∫1/2x​64t3dt=16(x4−(21​)4)=16x4−1.

Similarly,

∫1/2x96t3dt=24(x4−(12)4)=24x4−32.\int_{1/2}^x 96t^3dt=24\left(x^4-\left(\frac12\right)^4\right)=24x^4-\frac32.∫1/2x​96t3dt=24(x4−(21​)4)=24x4−23​.

Thus,

16x4−1≤f(x)≤24x4−32.16x^4-1\le f(x)\le 24x^4-\frac32.16x4−1≤f(x)≤24x4−23​.
  1. Integrate from 1/21/21/2 to 111

So,

∫1/21(16x4−1) dx≤∫1/21f(x) dx≤∫1/21(24x4−32)dx.\int_{1/2}^1 (16x^4-1)\,dx \le \int_{1/2}^1 f(x)\,dx \le \int_{1/2}^1 \left(24x^4-\frac32\right)dx.∫1/21​(16x4−1)dx≤∫1/21​f(x)dx≤∫1/21​(24x4−23​)dx.

Lower bound

∫1/21(16x4−1)dx=16∫1/21x4dx−∫1/211 dx.\int_{1/2}^1 (16x^4-1)dx =16\int_{1/2}^1 x^4dx-\int_{1/2}^1 1\,dx.∫1/21​(16x4−1)dx=16∫1/21​x4dx−∫1/21​1dx.

Now,

∫1/21x4dx=[x55]1/21=15(1−132)=15⋅3132=31160.\int_{1/2}^1 x^4dx=\left[\frac{x^5}{5}\right]_{1/2}^1 =\frac15\left(1-\frac1{32}\right) =\frac15\cdot\frac{31}{32} =\frac{31}{160}.∫1/21​x4dx=[5x5​]1/21​=51​(1−321​)=51​⋅3231​=16031​.

Therefore,

16⋅31160−12=3110−12=31−510=2610=135.16\cdot \frac{31}{160}-\frac12 =\frac{31}{10}-\frac12 =\frac{31-5}{10} =\frac{26}{10} =\frac{13}{5}.16⋅16031​−21​=1031​−21​=1031−5​=1026​=513​.

Upper bound

∫1/21(24x4−32)dx=24∫1/21x4dx−32⋅12.\int_{1/2}^1 \left(24x^4-\frac32\right)dx =24\int_{1/2}^1 x^4dx-\frac32\cdot \frac12.∫1/21​(24x4−23​)dx=24∫1/21​x4dx−23​⋅21​.

So,

24⋅31160−34=9320−34=279−4560=23460=3910.24\cdot \frac{31}{160}-\frac34 =\frac{93}{20}-\frac34 =\frac{279-45}{60} =\frac{234}{60} =\frac{39}{10}.24⋅16031​−43​=2093​−43​=60279−45​=60234​=1039​.

Hence,

135≤∫1/21f(x) dx≤3910.\frac{13}{5}\le \int_{1/2}^1 f(x)\,dx\le \frac{39}{10}.513​≤∫1/21​f(x)dx≤1039​.

So any valid pair (m,M)(m,M)(m,M) must satisfy

m≤135,M≥3910.m\le \frac{13}{5},\qquad M\ge \frac{39}{10}.m≤513​,M≥1039​.
  1. Check options

We need an option whose stated interval contains all possible values of the integral.

  • A: m=13,M=24m=13, M=24m=13,M=24 means 13≤I≤24,13\le I\le 24,13≤I≤24, which is impossible since actually I≤3.9I\le 3.9I≤3.9. So A is false.

  • B: m=14,M=12m=\frac14, M=\frac12m=41​,M=21​ means 14≤I≤12,\frac14\le I\le \frac12,41​≤I≤21​, but actually I≥2.6I\ge 2.6I≥2.6. So B is false.

  • C: m=−11,M=0m=-11, M=0m=−11,M=0 means −11≤I≤0,-11\le I\le 0,−11≤I≤0, but I>0I>0I>0. So C is false.

  • D: m=1,M=12m=1, M=12m=1,M=12 means 1≤I≤12,1\le I\le 12,1≤I≤12, and this is certainly true because 135=2.6≥1,3910=3.9≤12.\frac{13}{5}=2.6 \ge 1, \qquad \frac{39}{10}=3.9 \le 12.513​=2.6≥1,1039​=3.9≤12. So D is true.

Thus the correct option is D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.

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