JEE Main 2024 5 April Morning Shift
Exam JEE Main
Year 2024
Date 5 April
Shift Morning Shift
Subject Mathematics
Chapter Functions JEE Main 2024 · 5 April · Morning Shift · Mathematics · Functions
Let f(x)=x^5+2 x^3+3 x+1, x ∈ R , and g(x) be a function such that...
Let
f ( x ) = x 5 + 2 x 3 + 3 x + 1 , x ∈ R f(x)=x^5+2 x^3+3 x+1, x \in \mathbf{R} f ( x ) = x 5 + 2 x 3 + 3 x + 1 , x ∈ R , and
g ( x ) g(x) g ( x ) be a function such that
g ( f ( x ) ) = x g(f(x))=x g ( f ( x )) = x for all
x ∈ R x \in \mathbf{R} x ∈ R . Then
g ( 7 ) g ′ ( 7 ) \frac{g(7)}{g^{\prime}(7)} g ′ ( 7 ) g ( 7 ) is equal to :
Step-by-step solution
f ( x ) = x 5 + 2 x 3 + 3 x + 1 f ′ ( x ) = 5 x 4 + 6 x 2 + 3 f ′ ( 1 ) = 5 + 6 + 3 = 14 g ( f ( x ) ) = x g ′ ( f ( x ) ) f ′ ( x ) = 1 for f ( x ) = 7 ⇒ x 5 + 2 x 3 + 3 x + 1 = 7 ⇒ x = 1 g ′ ( 7 ) f ′ ( 1 ) = 1 ⇒ g ′ ( 7 ) = 1 f ′ ( 1 ) = 1 14 \begin{aligned} & f(x)=x^5+2 x^3+3 x+1 \\ & f^{\prime}(x)=5 x^4+6 x^2+3 \\ & f^{\prime}(1)=5+6+3=14 \\ & g(f(x))=x \\ & g^{\prime}(f(x)) f^{\prime}(x)=1 \\ & \text{ for } f(x)=7 \\ & \Rightarrow x^5+2 x^3+3 x+1=7 \\ & \Rightarrow x=1 \\ & g^{\prime}(7) f^{\prime}(1)=1 \Rightarrow g^{\prime}(7)=\frac{1}{f^{\prime}(1)}=\frac{1}{14}\end{aligned} f ( x ) = x 5 + 2 x 3 + 3 x + 1 f ′ ( x ) = 5 x 4 + 6 x 2 + 3 f ′ ( 1 ) = 5 + 6 + 3 = 14 g ( f ( x )) = x g ′ ( f ( x )) f ′ ( x ) = 1 for f ( x ) = 7 ⇒ x 5 + 2 x 3 + 3 x + 1 = 7 ⇒ x = 1 g ′ ( 7 ) f ′ ( 1 ) = 1 ⇒ g ′ ( 7 ) = f ′ ( 1 ) 1 = 14 1 x = 1 , f ( x ) = 7 ⇒ g ( 7 ) = 1 g ( 7 ) g ′ ( 7 ) = 1 1 / 14 = 14 \begin{aligned} & x=1, f(x)=7 \Rightarrow g(7)=1 \\ & \frac{g(7)}{g^{\prime}(7)}=\frac{1}{1 / 14}=14\end{aligned} x = 1 , f ( x ) = 7 ⇒ g ( 7 ) = 1 g ′ ( 7 ) g ( 7 ) = 1/14 1 = 14 $
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