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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Differentiation

JEE Advanced · Mathematics · Differentiation

Let f: R arrow R be a function defined by f(x)=x^3+2 x^2+3 x+2 and...

Let f:R→Rbeafunctiondefinedbyf: R \rightarrow R be a function defined byf(x)=x^3+2 x^2+3 x+2andandgbetheinversefunctionofbe the inverse function off.Ifddx(g(g(g(x))))∣x=24=pq. If \left.\frac{d}{d x}(g(g(g(x))))\right|_{x=24}=\frac{p}{q}, where p,qp, q are co-prime, then find the value of (p+q)(p+q).

Correct answer

139

Step-by-step solution

 9) f(−1)=0,f(0)=2 and f(2)=24,f′(x)=3x2+4x+3ddx(g(g(g(x))))=g′(g(g(x)))⋅g′(g(x))⋅g′(x)ddx(g(g(g(x))))∣x=24=g′(g(g(24)))⋅g′(g(24))⋅g′(24)=g′(g(2))⋅g′(2)⋅1f′(2)=g′(0)⋅1f′(0)⋅1f′(2)=1f′(−1)⋅1f′(0)⋅1f′(2)=12×13×123=1138≡pq∴pp=139\begin{aligned} & \text { 9) } f(-1)=0, f(0)=2 \text { and } f(2)=24, f^{\prime}(x)=3 x^2+4 x+3 \\ & \frac{d}{d x}(g(g(g(x))))=g^{\prime}(g(g(x))) \cdot g^{\prime}(g(x)) \cdot g^{\prime}(x) \\ & \left.\frac{d}{d x}(g(g(g(x))))\right|_{x=24}=g^{\prime}(g(g(24))) \cdot g^{\prime}(g(24)) \cdot g^{\prime}(24)=g^{\prime}(g(2)) \cdot g^{\prime}(2) \cdot \frac{1}{f^{\prime}(2)} \\ & =g^{\prime}(0) \cdot \frac{1}{f^{\prime}(0)} \cdot \frac{1}{f^{\prime}(2)}=\frac{1}{f^{\prime}(-1)} \cdot \frac{1}{f^{\prime}(0)} \cdot \frac{1}{f^{\prime}(2)}=\frac{1}{2} \times \frac{1}{3} \times \frac{1}{23}=\frac{1}{138} \equiv \frac{p}{q} \\ & \therefore \quad p \quad p=139\end{aligned}
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