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Limits Continuity and Differentiability question

2002 · Shift 0 · Q62
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  5. /2002 · Shift 0 · Q62

Limits Continuity and Differentiability question

2002 · Shift 0 · Q62

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
f(x) and g(x) are two differentiable functions on [0, 2] such that f''(x) - g''(x) = 0, f'(1) = 2, g'(1) = 4, f(2) = 3, g(2) = 9 then f(x) - g(x) at x = 32{3 \over 2}23​ is
  1. A
    0
  2. B
    2
  3. C
    10
  4. D
    -5
View written solutionFree

Correct answer: D

Let h(x)=f(x)−g(x).h(x)=f(x)-g(x).h(x)=f(x)−g(x). Then h′′(x)=f′′(x)−g′′(x)=0.h''(x)=f''(x)-g''(x)=0.h′′(x)=f′′(x)−g′′(x)=0.

So h(x)h(x)h(x) is a linear function: h(x)=ax+b.h(x)=ax+b.h(x)=ax+b.

1. Use the derivative condition

Given f′(1)=2, g′(1)=4,f'(1)=2,\, g'(1)=4,f′(1)=2,g′(1)=4, so h′(1)=f′(1)−g′(1)=2−4=−2.h'(1)=f'(1)-g'(1)=2-4=-2.h′(1)=f′(1)−g′(1)=2−4=−2.

But for h(x)=ax+bh(x)=ax+bh(x)=ax+b, we have h′(x)=a.h'(x)=a.h′(x)=a. Hence a=−2.a=-2.a=−2.

Therefore, h(x)=−2x+b.h(x)=-2x+b.h(x)=−2x+b.

2. Use the value at x=2x=2x=2

Given f(2)=3, g(2)=9,f(2)=3,\, g(2)=9,f(2)=3,g(2)=9, so h(2)=f(2)−g(2)=3−9=−6.h(2)=f(2)-g(2)=3-9=-6.h(2)=f(2)−g(2)=3−9=−6.

Now substitute in h(x)=−2x+bh(x)=-2x+bh(x)=−2x+b: −2(2)+b=−6-2(2)+b=-6−2(2)+b=−6 −4+b=−6-4+b=-6−4+b=−6 b=−2.b=-2.b=−2.

Thus, h(x)=−2x−2.h(x)=-2x-2.h(x)=−2x−2.

3. Find the value at x=32x=\tfrac32x=23​

h(32)=−2(32)−2=−3−2=−5.h\left(\frac32\right)=-2\left(\frac32\right)-2=-3-2=-5.h(23​)=−2(23​)−2=−3−2=−5.

So, f(32)−g(32)=−5.f\left(\frac32\right)-g\left(\frac32\right)=-5.f(23​)−g(23​)=−5.

4. Check options

  • A: 000 ❌
  • B: 222 ❌
  • C: 101010 ❌
  • D: −5-5−5 ✅

Hence the correct option is D.

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