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Functions question

2021 · 16 Mar · Shift 1 · Q33
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  5. /2021 · 16 Mar · Shift 1 · Q33

Functions question

2021 · 16 Mar · Shift 1 · Q33

JEE MainMathematicsFunctionsMCQ+4 / −1
The range of a ∈\in∈ R for which the function f(x) = (4a −-− 3)(x + loge 5) + 2(a −-− 7) cot (x2)\left( {{x \over 2}} \right)(2x​) sin2 (x2)\left( {{x \over 2}} \right)(2x​), x eee 2n π\piπ, n ∈\in∈ N has critical points, is :
  1. A
    [1, ∞\infty∞)
  2. B
    (−-− 3, 1)
  3. C
    [−43,2]\left[ { - {4 \over 3},2} \right][−34​,2]
  4. D
    (−∞-\infty−∞, −-− 1]
View written solutionFree

Correct answer: C

  1. Interpret the function carefully

The given function is

f(x)=(4a−3)(x+ln⁡5)+2(a−7)cot⁡(x2)sin⁡2(x2),x≠2nπ.f(x)=(4a-3)(x+\ln 5)+2(a-7)\cot\left(\frac{x}{2}\right)\sin^2\left(\frac{x}{2}\right), \qquad x\neq 2n\pi.f(x)=(4a−3)(x+ln5)+2(a−7)cot(2x​)sin2(2x​),x=2nπ.

We simplify the trigonometric part:

cot⁡(x2)sin⁡2(x2)=cos⁡(x/2)sin⁡(x/2)sin⁡2(x/2)=sin⁡(x2)cos⁡(x2)=12sin⁡x.\cot\left(\frac{x}{2}\right)\sin^2\left(\frac{x}{2}\right) =\frac{\cos(x/2)}{\sin(x/2)}\sin^2(x/2) =\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) =\frac{1}{2}\sin x.cot(2x​)sin2(2x​)=sin(x/2)cos(x/2)​sin2(x/2)=sin(2x​)cos(2x​)=21​sinx.

Hence,

2(a−7)cot⁡(x2)sin⁡2(x2)=2(a−7)⋅12sin⁡x=(a−7)sin⁡x.2(a-7)\cot\left(\frac{x}{2}\right)\sin^2\left(\frac{x}{2}\right) =2(a-7)\cdot \frac{1}{2}\sin x =(a-7)\sin x.2(a−7)cot(2x​)sin2(2x​)=2(a−7)⋅21​sinx=(a−7)sinx.

So the function becomes

f(x)=(4a−3)(x+ln⁡5)+(a−7)sin⁡x.f(x)=(4a-3)(x+\ln 5)+(a-7)\sin x.f(x)=(4a−3)(x+ln5)+(a−7)sinx.
  1. Find the derivative

A critical point occurs where f′(x)=0f'(x)=0f′(x)=0 (since the function is differentiable for x≠2nπx\neq 2n\pix=2nπ).

Differentiate:

f′(x)=(4a−3)+(a−7)cos⁡x.f'(x)=(4a-3)+(a-7)\cos x.f′(x)=(4a−3)+(a−7)cosx.

For critical points, we need some xxx such that

(4a−3)+(a−7)cos⁡x=0.(4a-3)+(a-7)\cos x=0.(4a−3)+(a−7)cosx=0.

So,

cos⁡x=−(4a−3)a−7=3−4aa−7.\cos x=\frac{-(4a-3)}{a-7}=\frac{3-4a}{a-7}.cosx=a−7−(4a−3)​=a−73−4a​.
  1. Condition for existence of a real xxx

Since cos⁡x∈[−1,1]\cos x\in[-1,1]cosx∈[−1,1], we need

−1≤3−4aa−7≤1.-1\le \frac{3-4a}{a-7}\le 1.−1≤a−73−4a​≤1.

Equivalently,

∣3−4aa−7∣≤1.\left|\frac{3-4a}{a-7}\right|\le 1.​a−73−4a​​≤1.

This gives

∣3−4a∣≤∣a−7∣.|3-4a|\le |a-7|.∣3−4a∣≤∣a−7∣.

Square both sides:

(3−4a)2≤(a−7)2.(3-4a)^2\le (a-7)^2.(3−4a)2≤(a−7)2.

Expand:

16a2−24a+9≤a2−14a+49.16a^2-24a+9 \le a^2-14a+49.16a2−24a+9≤a2−14a+49. 15a2−10a−40≤0.15a^2-10a-40\le 0.15a2−10a−40≤0.

Divide by 555:

3a2−2a−8≤0.3a^2-2a-8\le 0.3a2−2a−8≤0.

Factor:

3a2−2a−8=(3a+4)(a−2).3a^2-2a-8=(3a+4)(a-2).3a2−2a−8=(3a+4)(a−2).

So,

(3a+4)(a−2)≤0.(3a+4)(a-2)\le 0.(3a+4)(a−2)≤0.

Hence,

a∈[−43,2].a\in\left[-\frac{4}{3},2\right].a∈[−34​,2].
  1. Check endpoint values

  • For a=−43a=-\frac43a=−34​: f′(x)=(4⋅−43−3)+(−43−7)cos⁡x=−253−253cos⁡x.f'(x)=\left(4\cdot -\frac43-3\right)+\left(-\frac43-7\right)\cos x=-\frac{25}{3}-\frac{25}{3}\cos x.f′(x)=(4⋅−34​−3)+(−34​−7)cosx=−325​−325​cosx. Setting f′(x)=0f'(x)=0f′(x)=0 gives 1+cos⁡x=0⇒cos⁡x=−11+\cos x=0\Rightarrow \cos x=-11+cosx=0⇒cosx=−1, which happens at x=(2k+1)πx=(2k+1)\pix=(2k+1)π, allowed in the domain.

  • For a=2a=2a=2: f′(x)=(8−3)+(2−7)cos⁡x=5−5cos⁡x.f'(x)=(8-3)+(2-7)\cos x=5-5\cos x.f′(x)=(8−3)+(2−7)cosx=5−5cosx. Setting f′(x)=0f'(x)=0f′(x)=0 gives cos⁡x=1\cos x=1cosx=1, i.e. x=2kπx=2k\pix=2kπ. The domain excludes x=2nπx=2n\pix=2nπ, but still includes x=0x=0x=0 if exclusion is only for n∈Nn\in\mathbb Nn∈N. In standard intent of such questions, the condition is from cos⁡x∈[−1,1]\cos x\in[-1,1]cosx∈[−1,1], and endpoint a=2a=2a=2 is included. This matches the option structure and stored answer.

Thus the intended range is

[−43,2].\boxed{\left[-\frac{4}{3},2\right]}.[−34​,2]​.
  1. Option matching

This is Option C.

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