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Limits Continuity and Differentiability question

2022 · 28 Jun · Shift 2 · Q44
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  5. /2022 · 28 Jun · Shift 2 · Q44

Limits Continuity and Differentiability question

2022 · 28 Jun · Shift 2 · Q44

JEE MainMathematicsLimits Continuity and DifferentiabilityNumerical+4 / −1
If lim⁡x→1sin⁡(3x2−4x+1)−x2+12x3−7x2+ax+b=−2\mathop {\lim }\limits_{x \to 1} {{\sin (3{x^2} - 4x + 1) - {x^2} + 1} \over {2{x^3} - 7{x^2} + ax + b}} = - 2x→1lim​2x3−7x2+ax+bsin(3x2−4x+1)−x2+1​=−2, then the value of (a −-− b) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 11

  1. We need
lim⁡x→1sin⁡(3x2−4x+1)−x2+12x3−7x2+ax+b=−2.\lim_{x\to 1} \frac{\sin(3x^2-4x+1)-x^2+1}{2x^3-7x^2+ax+b}=-2.x→1lim​2x3−7x2+ax+bsin(3x2−4x+1)−x2+1​=−2.
  1. First check the numerator at x=1x=1x=1:
3(1)2−4(1)+1=0,3(1)^2-4(1)+1=0,3(1)2−4(1)+1=0,

so

sin⁡(3x2−4x+1)−x2+1∣x=1=sin⁡0−1+1=0.\sin(3x^2-4x+1)-x^2+1\Big|_{x=1}=\sin 0-1+1=0.sin(3x2−4x+1)−x2+1​x=1​=sin0−1+1=0.

For the limit to be finite, denominator must also vanish at x=1x=1x=1:

2−7+a+b=0  ⟹  a+b=5.(1)2-7+a+b=0 \implies a+b=5. \qquad (1)2−7+a+b=0⟹a+b=5.(1)
  1. Since both numerator and denominator go to 000, apply L'Hospital's Rule.

Let

N(x)=sin⁡(3x2−4x+1)−x2+1,N(x)=\sin(3x^2-4x+1)-x^2+1,N(x)=sin(3x2−4x+1)−x2+1, D(x)=2x3−7x2+ax+b.D(x)=2x^3-7x^2+ax+b.D(x)=2x3−7x2+ax+b.

Then

N′(x)=cos⁡(3x2−4x+1)(6x−4)−2x,N'(x)=\cos(3x^2-4x+1)(6x-4)-2x,N′(x)=cos(3x2−4x+1)(6x−4)−2x, D′(x)=6x2−14x+a.D'(x)=6x^2-14x+a.D′(x)=6x2−14x+a.

At x=1x=1x=1,

N′(1)=cos⁡0⋅(6−4)−2=1⋅2−2=0.N'(1)=\cos 0\cdot (6-4)-2=1\cdot 2-2=0.N′(1)=cos0⋅(6−4)−2=1⋅2−2=0.

So again we get 0/00/00/0.

Also,

D′(1)=6−14+a=a−8.D'(1)=6-14+a=a-8.D′(1)=6−14+a=a−8.

For the limit to exist as a finite nonzero number, we must have

D′(1)=0  ⟹  a=8.(2)D'(1)=0 \implies a=8. \qquad (2)D′(1)=0⟹a=8.(2)

Using (1):

a+b=5  ⟹  8+b=5  ⟹  b=−3.a+b=5 \implies 8+b=5 \implies b=-3.a+b=5⟹8+b=5⟹b=−3.
  1. Now use L'Hospital's Rule again.

Differentiate once more:

N′′(x)=−sin⁡(3x2−4x+1)(6x−4)2+cos⁡(3x2−4x+1)⋅6−2.N''(x)= -\sin(3x^2-4x+1)(6x-4)^2 + \cos(3x^2-4x+1)\cdot 6 -2.N′′(x)=−sin(3x2−4x+1)(6x−4)2+cos(3x2−4x+1)⋅6−2.

At x=1x=1x=1,

N′′(1)=−sin⁡0⋅22+cos⁡0⋅6−2=0+6−2=4.N''(1)= -\sin 0\cdot 2^2 + \cos 0\cdot 6 -2 = 0+6-2=4.N′′(1)=−sin0⋅22+cos0⋅6−2=0+6−2=4.

And

D′′(x)=12x−14,D''(x)=12x-14,D′′(x)=12x−14,

so

D′′(1)=12−14=−2.D''(1)=12-14=-2.D′′(1)=12−14=−2.

Therefore,

lim⁡x→1N(x)D(x)=N′′(1)D′′(1)=4−2=−2,\lim_{x\to 1} \frac{N(x)}{D(x)}=\frac{N''(1)}{D''(1)}=\frac{4}{-2}=-2,x→1lim​D(x)N(x)​=D′′(1)N′′(1)​=−24​=−2,

which matches the given limit.

  1. Hence,
a−b=8−(−3)=11.a-b=8-(-3)=11.a−b=8−(−3)=11.

So the required integer is

11.\boxed{11}.11​.

Comparison with stored correct answer: stored answer is 111111, which matches our result.

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