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Functions question

2024 · 30 Jan · Shift 2 · Q48
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  5. /2024 · 30 Jan · Shift 2 · Q48

Functions question

2024 · 30 Jan · Shift 2 · Q48

JEE MainMathematicsFunctionsMCQ+4 / −1
If the domain of the function f(x)=log⁡e(2x+34x2+x−3)+cos⁡−1(2x−1x+2)f(x)=\log _e\left(\frac{2 x+3}{4 x^2+x-3}\right)+\cos ^{-1}\left(\frac{2 x-1}{x+2}\right)f(x)=loge​(4x2+x−32x+3​)+cos−1(x+22x−1​) is (α,β](\alpha, \beta](α,β], then the value of 5β−4α5 \beta-4 \alpha5β−4α is equal to
  1. A
    9
  2. B
    12
  3. C
    11
  4. D
    10
View written solutionFree

Correct answer: B

  1. Given function
f(x)=ln⁡(2x+34x2+x−3)+cos⁡−1(2x−1x+2)f(x)=\ln\left(\frac{2x+3}{4x^2+x-3}\right)+\cos^{-1}\left(\frac{2x-1}{x+2}\right)f(x)=ln(4x2+x−32x+3​)+cos−1(x+22x−1​)

We need the domain of f(x)f(x)f(x).

For the domain:

  • the logarithm argument must be positive,
  • the argument of cos⁡−1\cos^{-1}cos−1 must lie in [−1,1][-1,1][−1,1],
  • and denominators must be nonzero.

  1. Condition from the logarithm

We need

2x+34x2+x−3>0\frac{2x+3}{4x^2+x-3}>04x2+x−32x+3​>0

Factor the denominator:

4x2+x−3=(4x−3)(x+1)4x^2+x-3=(4x-3)(x+1)4x2+x−3=(4x−3)(x+1)

So the inequality is

2x+3(4x−3)(x+1)>0\frac{2x+3}{(4x-3)(x+1)}>0(4x−3)(x+1)2x+3​>0

Critical points are:

x=−32,x=−1,x=34x=-\frac32, \quad x=-1, \quad x=\frac34x=−23​,x=−1,x=43​

Now check signs on intervals:

  • For x<−32x< -\frac32x<−23​, expression is negative.
  • For −32<x<−1-\frac32 < x < -1−23​<x<−1, expression is positive.
  • For −1<x<34-1 < x < \frac34−1<x<43​, expression is negative.
  • For x>34x>\frac34x>43​, expression is positive.

Hence logarithm condition gives

x∈(−32,−1)∪(34,∞)x\in\left(-\frac32,-1\right)\cup\left(\frac34,\infty\right)x∈(−23​,−1)∪(43​,∞)
  1. Condition from the inverse cosine

We need

−1≤2x−1x+2≤1,x≠−2-1\le \frac{2x-1}{x+2}\le 1, \qquad x\ne -2−1≤x+22x−1​≤1,x=−2

We solve both inequalities.

(i) Solve

2x−1x+2≥−1\frac{2x-1}{x+2}\ge -1x+22x−1​≥−1

Bring to one side:

2x−1x+2+1≥0\frac{2x-1}{x+2}+1\ge 0x+22x−1​+1≥0 2x−1+x+2x+2≥0\frac{2x-1+x+2}{x+2}\ge 0x+22x−1+x+2​≥0 3x+1x+2≥0\frac{3x+1}{x+2}\ge 0x+23x+1​≥0

Critical points: x=−2,−13x=-2,-\frac13x=−2,−31​.

Sign analysis gives

3x+1x+2≥0  ⟺  x∈(−∞,−2)∪[−13,∞)\frac{3x+1}{x+2}\ge 0 \iff x\in (-\infty,-2)\cup\left[-\frac13,\infty\right)x+23x+1​≥0⟺x∈(−∞,−2)∪[−31​,∞)

(ii) Solve

2x−1x+2≤1\frac{2x-1}{x+2}\le 1x+22x−1​≤1 2x−1−(x+2)x+2≤0\frac{2x-1-(x+2)}{x+2}\le 0x+22x−1−(x+2)​≤0 x−3x+2≤0\frac{x-3}{x+2}\le 0x+2x−3​≤0

Critical points: x=−2,3x=-2,3x=−2,3.

Sign analysis gives

x−3x+2≤0  ⟺  x∈(−2,3]\frac{x-3}{x+2}\le 0 \iff x\in (-2,3]x+2x−3​≤0⟺x∈(−2,3]

Now intersect the two results:

[(−∞,−2)∪(−13,∞)]∩(−2,3]=[−13,3]\left[(-\infty,-2)\cup\left(-\frac13,\infty\right)\right]\cap (-2,3] =\left[-\frac13,3\right][(−∞,−2)∪(−31​,∞)]∩(−2,3]=[−31​,3]

So the cos⁡−1\cos^{-1}cos−1 condition gives

x∈[−13,3]x\in \left[-\frac13,3\right]x∈[−31​,3]
  1. Intersect both domain conditions

From logarithm:

(−32,−1)∪(34,∞)\left(-\frac32,-1\right)\cup\left(\frac34,\infty\right)(−23​,−1)∪(43​,∞)

From inverse cosine:

[−13,3]\left[-\frac13,3\right][−31​,3]

Their intersection is

(34,3]\left(\frac34,3\right](43​,3]

Thus,

(α,β]=(34,3](\alpha,\beta]=\left(\frac34,3\right](α,β]=(43​,3]

So,

α=34,β=3\alpha=\frac34, \qquad \beta=3α=43​,β=3
  1. Compute 5β−4α5\beta-4\alpha5β−4α
5β−4α=5(3)−4(34)=15−3=125\beta-4\alpha=5(3)-4\left(\frac34\right)=15-3=125β−4α=5(3)−4(43​)=15−3=12
  1. Final answer
12\boxed{12}12​

So the correct option is B.

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