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Let fand g be twice differentiable functions on \mathbb{R} such that
$$ \begin{aligned} & f^{\prime \prime}(x)=g^{\prime \prime}(x)+6 x \\ & f^{\prime}(1)=4 g^{\prime}(1)-3=9 \\ & f(2)=3 g(2)=12 . \end{aligned}
Then which of the following is NOT true?

Option: 1

There exists x_0 \in(1,3 / 2) such that  f\left(x_0\right)=g\left(x_0\right)


Option: 2

\left|f^{\prime}(x)-g^{\prime}(x)\right|<6 \Rightarrow-1<x<1


Option: 3

If \, \, -1<x<2, then\, \, |f(x)-g(x)|<8


Option: 4

g(-2)-f(-2)=20


Answers (1)

best_answer

Let  F(x)=f(x)-g(x)

Given  f^{\prime}(x)=g^{\prime \prime}(x)+6 x

\begin{array}{ll} & \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{g}^{\prime}(\mathrm{x})+\frac{6 \mathrm{x}^2}{2}+\mathrm{c}_1 \\ & \mathrm{x}^{\prime}=1 \quad \mathrm{f}^{\prime}(1)=\mathrm{g}^{\prime}(1)+3 \times(1)^2+\mathrm{c}_1 \\ & 9=3+3+\mathrm{c}_1 \\ & \mathrm{c}_1=3 \\ \therefore \quad & \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{g}^{\prime}(\mathrm{x})+3 \mathrm{x}^2+3 \\ & f(\mathrm{x})=\mathrm{g}(\mathrm{x})+\frac{3 \mathrm{x}^3}{3}+3 \mathrm{x}+\mathrm{c}_2 \\ \mathrm{x}=2 & f(2)=\mathrm{g}(2)+(2)^3+3(2)+\mathrm{c}_2 \\ & 12=4+8+6+\mathrm{c}_2 \\ & \mathrm{c}_2=-6 \\ \therefore \quad & \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{g}(\mathrm{x})+\mathrm{x}^3+3 \mathrm{x}-6 \\ & =\mathrm{x}^3+3 \mathrm{x}-6 \end{array}

Option (1)

\mathrm{x}_0 \in\left(1, \frac{3}{2}\right)           such that  f\left(\mathrm{x}_0\right)=9\left(\mathrm{x}_0\right)

\begin{array}{lll} \because & \mathrm{F}(1)=f(1)-\mathrm{g}(1) & \mathrm{F}\left(\frac{3}{2}\right)=\mathrm{f}\left(\frac{3}{2}\right)-\mathrm{g}\left(\frac{3}{2}\right) \\ & =1+3-6=-2 & =(2)^3+3(2)-6 \\ & & =8+6-6=8 \\ \because & \mathrm{F}(1) \mathrm{F}\left(\frac{3}{2}\right)<0 & \end{array}

\Rightarrow \quad At least one root of  \mathrm{F}(\mathrm{x})=0 \text { lies in }\left(1, \frac{3}{2}\right)

\begin{array}{ll} \Rightarrow & f(x)-g(x)=0 \\ \Rightarrow & f(x)=g(x) \end{array}

Option (2)

\begin{aligned} & \left|\mathrm{f}^{\prime}(\mathrm{x})-\mathrm{g}^{\prime}(\mathrm{x})\right|<6 \Rightarrow-1<\mathrm{x}<1 \\ & \mathrm{~F}^{\prime}(\mathrm{x})=\mathrm{x}^3+3 \mathrm{x}-6 \\ & \mathrm{~F}^{\prime}(\mathrm{x})=3 \mathrm{x}^2+3 \\ & \mathrm{f}^{\prime}(\mathrm{x})-\mathrm{g}^{\prime}(\mathrm{x})=3 \mathrm{x}^2+3 \\ & \left|\mathrm{f}^{\prime}(\mathrm{x})-\mathrm{g}^{\prime}(\mathrm{x})\right|<6 \\ \Rightarrow \quad & 3 \mathrm{x}^2+3<6 \\ \Rightarrow \quad & 3 \mathrm{x}^2<\mathrm{x} 3 \\ & \mathrm{x}^2<1 \\ \Rightarrow \quad & \mathrm{x} \in(-1,1) \end{aligned}

Option (3)

If -1<\mathrm{x}<2 then |f(\mathrm{x})-\mathrm{g}(\mathrm{x})|<8
$$ \begin{aligned} & \mathrm{F}(\mathrm{x})=\mathrm{x}^3+3 \mathrm{x}-6 \\ & F(-1)=-1-3-6=-10 \quad \text { But }\left|f^{\prime}(x)-g^{\prime}(x)\right|<10 \\ & F(2)=(2)^3+3(2)-6=8 \\ & \end{aligned}

Option is not true

Option (4)

\begin{gathered} \mathrm{g}(-2)-f(-2)=20 \\ \mathrm{~F}(-2)=f(-2)-\mathrm{g}(-2) \\ =(-2)^3+3(-2)-6 \\ -8-6-6=-20 \\ \mathrm{~g}(-2)-f(-2)=20 \end{gathered}

Posted by

avinash.dongre

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