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Application of Derivatives question

2024 · 9 Apr · Shift 2 · Q57
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  5. /2024 · 9 Apr · Shift 2 · Q57

Application of Derivatives question

2024 · 9 Apr · Shift 2 · Q57

JEE MainMathematicsApplication of DerivativesNumerical+4 / −1
Let the set of all values of ppp, for which f(x)=(p2−6p+8)(sin⁡22x−cos⁡22x)+2(2−p)x+7f(x)=\left(p^2-6 p+8\right)\left(\sin ^2 2 x-\cos ^2 2 x\right)+2(2-p) x+7f(x)=(p2−6p+8)(sin22x−cos22x)+2(2−p)x+7 does not have any critical point, be the interval (a,b)(a, b)(a,b). Then 16ab16 a b16ab is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 252

  1. Given function

We have

f(x)=(p2−6p+8)(sin⁡22x−cos⁡22x)+2(2−p)x+7.f(x)=\left(p^2-6p+8\right)(\sin^2 2x-\cos^2 2x)+2(2-p)x+7.f(x)=(p2−6p+8)(sin22x−cos22x)+2(2−p)x+7.

We need all values of ppp for which f(x)f(x)f(x) has no critical point.

A critical point occurs where f′(x)=0f'(x)=0f′(x)=0 (since this function is differentiable for all xxx).

  1. Simplify the trigonometric part

Use

sin⁡2θ−cos⁡2θ=−cos⁡2θ.\sin^2\theta-\cos^2\theta=-\cos 2\theta.sin2θ−cos2θ=−cos2θ.

With θ=2x\theta=2xθ=2x,

sin⁡22x−cos⁡22x=−cos⁡4x.\sin^2 2x-\cos^2 2x=-\cos 4x.sin22x−cos22x=−cos4x.

Also,

p2−6p+8=(p−2)(p−4).p^2-6p+8=(p-2)(p-4).p2−6p+8=(p−2)(p−4).

So

f(x)=−(p−2)(p−4)cos⁡4x+2(2−p)x+7.f(x)=-(p-2)(p-4)\cos 4x+2(2-p)x+7.f(x)=−(p−2)(p−4)cos4x+2(2−p)x+7.
  1. Differentiate

Differentiate term by term:

ddx[−(p−2)(p−4)cos⁡4x]=−(p−2)(p−4)(−4sin⁡4x)=4(p−2)(p−4)sin⁡4x.\frac{d}{dx}[-(p-2)(p-4)\cos 4x]=-(p-2)(p-4)(-4\sin 4x)=4(p-2)(p-4)\sin 4x.dxd​[−(p−2)(p−4)cos4x]=−(p−2)(p−4)(−4sin4x)=4(p−2)(p−4)sin4x.

And

ddx[2(2−p)x]=2(2−p).\frac{d}{dx}[2(2-p)x]=2(2-p).dxd​[2(2−p)x]=2(2−p).

Hence,

f′(x)=4(p−2)(p−4)sin⁡4x+2(2−p).f'(x)=4(p-2)(p-4)\sin 4x+2(2-p).f′(x)=4(p−2)(p−4)sin4x+2(2−p).

Factor out (p−2)(p-2)(p−2):

f′(x)=2(p−2)(2(p−4)sin⁡4x−1).f'(x)=2(p-2)\big(2(p-4)\sin 4x-1\big).f′(x)=2(p−2)(2(p−4)sin4x−1).
  1. Condition for no critical point

We want f′(x)≠0f'(x)\neq 0f′(x)=0 for every real xxx.

So we analyze cases.

Case 1: p=2p=2p=2

Then

f′(x)=0∀x,f'(x)=0\quad \forall x,f′(x)=0∀x,

so every point is critical. Not allowed.

Case 2: p≠2p\neq 2p=2

Then we need

2(p−4)sin⁡4x−1≠02(p-4)\sin 4x-1\neq 02(p−4)sin4x−1=0

for every xxx.

That means the equation

2(p−4)sin⁡4x=12(p-4)\sin 4x=12(p−4)sin4x=1

should have no real solution.

Equivalently,

sin⁡4x=12(p−4)\sin 4x=\frac{1}{2(p-4)}sin4x=2(p−4)1​

should have no real solution.

But sin⁡4x\sin 4xsin4x can take all values in [−1,1][-1,1][−1,1]. Therefore, a real solution exists iff

∣12(p−4)∣≤1.\left|\frac{1}{2(p-4)}\right|\le 1.​2(p−4)1​​≤1.

So for no solution, we need

∣12(p−4)∣>1.\left|\frac{1}{2(p-4)}\right|>1.​2(p−4)1​​>1.

That is,

1>2∣p−4∣⟹∣p−4∣<12.1>2|p-4| \quad\Longrightarrow\quad |p-4|<\frac12.1>2∣p−4∣⟹∣p−4∣<21​.

Thus,

72<p<92.\frac72<p<\frac92.27​<p<29​.

Also p=2p=2p=2 is outside this interval, so no extra exclusion is needed.

Hence the required interval is

(a,b)=(72,92).(a,b)=\left(\frac72,\frac92\right).(a,b)=(27​,29​).
  1. Compute 16ab16ab16ab
a=72,b=92.a=\frac72,\qquad b=\frac92.a=27​,b=29​.

So

ab=72⋅92=634.ab=\frac72\cdot\frac92=\frac{63}{4}.ab=27​⋅29​=463​.

Therefore,

16ab=16⋅634=4⋅63=252.16ab=16\cdot\frac{63}{4}=4\cdot 63=252.16ab=16⋅463​=4⋅63=252.
  1. Comparison with stored answer

Our derived answer is 252252252, which matches the stored correct answer.

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