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MCQMediumJEE 2023Applications of Derivatives (Monotonicity, Extrema)

JEE Mathematics 2023 Question with Solution

If f(x)=[(tan⁡1∘)x+log⁡e(123)][xlog⁡e(1234)−(tan⁡1∘)]f(x) = \frac{[(\tan 1^\circ)x + \log_e(123)]}{[x \log_e(1234) - (\tan 1^\circ)]}, x>0x > 0, then the least value of f(f(x))+f(f(4/x))f(f(x)) + f(f(4/x)) is:

  • A

    22

  • B

    44

  • C

    88

  • D

    00

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

f(x)=(tan⁡1)x+log⁡123xlog⁡1234−tan⁡1f(x) = \frac{(\tan 1)x + \log_{12}3}{x \log_{12}34 - \tan 1}

Let

A=tan⁡1,B=log⁡123,C=log⁡1234A = \tan 1, \quad B = \log_{12}3, \quad C = \log_{12}34

Find: The least value of f(f(x))+f(f(4/x))f(f(x)) + f(f(4/x)).

Rewrite the function as

f(x)=Ax+BxC−Af(x) = \frac{Ax + B}{xC - A}

Now evaluate

f(f(x))=f(Ax+BxC−A)f(f(x)) = f\left(\frac{Ax+B}{xC-A}\right)

Substituting into ff,

f(f(x))=A(Ax+BxC−A)+BC(Ax+BxC−A)−Af(f(x)) = \frac{A\left(\frac{Ax+B}{xC-A}\right) + B}{C\left(\frac{Ax+B}{xC-A}\right) - A}

On simplifying,

f(f(x))=A(Ax+B)+B(xC−A)C(Ax+B)−A(xC−A)f(f(x)) = \frac{A(Ax+B) + B(xC-A)}{C(Ax+B) - A(xC-A)}

Expand numerator and denominator:

Numerator=A2x+AB+xBC−AB=x(A2+BC)\text{Numerator} = A^2x + AB + xBC - AB = x(A^2 + BC) Denominator=ACx+BC−ACx+A2=A2+BC\text{Denominator} = ACx + BC - ACx + A^2 = A^2 + BC

Hence,

f(f(x))=x(A2+BC)A2+BC=xf(f(x)) = \frac{x(A^2 + BC)}{A^2 + BC} = x

Similarly,

f(f(4/x))=4xf(f(4/x)) = \frac{4}{x}

Therefore,

f(f(x))+f(f(4/x))=x+4xf(f(x)) + f(f(4/x)) = x + \frac{4}{x}

Apply AM-GM inequality:

x+4x2≥x⋅4x\frac{x + \frac{4}{x}}{2} \geq \sqrt{x \cdot \frac{4}{x}}

So,

x+4x≥4x + \frac{4}{x} \geq 4

Therefore, the least value is 44 and the correct option is C.

Key Observation

The solution shows that the transformation

f(x)=Ax+BxC−Af(x) = \frac{Ax+B}{xC-A}

is an involution, because applying it twice gives back the original input:

f(f(x))=xf(f(x)) = x

Therefore the expression becomes

x+4xx + \frac{4}{x}

for x>0x > 0. Its minimum is obtained from AM-GM, giving

x+4x≥4x + \frac{4}{x} \geq 4

So the least value is 44.

Common mistakes

  • Assuming the answer key is correct without checking the worked solution. The solution explicitly concludes the least value is 44, but labels the option as C, which conflicts with the given options. Use the worked value and then map it to the matching option, which is B.

  • Making an algebraic error while simplifying f(f(x))f(f(x)). The cancellation occurs only after correctly expanding both numerator and denominator. Keep ABAB terms and ACxACx terms carefully so that f(f(x))=xf(f(x)) = x is obtained.

  • Using AM-GM incorrectly on x+4xx + \frac{4}{x} for x>0x > 0. The inequality must be applied to two positive terms, giving x+4x≥24=4x + \frac{4}{x} \geq 2\sqrt{4} = 4. Do not conclude a smaller bound.

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