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KVPY Sample Paper KVPY Stream-SX Model Paper-9

  • question_answer
    \[let\,f:\mathbb{R}\to \mathbb{R},g:\mathbb{R}\to \mathbb{R}\] and h:\[\mathbb{R}\to \mathbb{R}\,\]be differentiable functions such that \[f(x)\,={{x}^{3}}+3x+2,\] g(f(x))=x  and h (g(g(x)))=x  for all x\[\in \mathbb{R}.\] then, which of the following is correct.

    A) \[g'(2)=\frac{1}{15}\]

    B) \[h'(1)=666\]

    C) \[h(g(3))=36\]                

    D) none of these

    Correct Answer: B

    Solution :

    \[f\left( x \right)\text{ }={{x}^{3}}+3x+2\Rightarrow f'(x)=3{{x}^{2}}+3\]
    \[\begin{align}   & ~Also\text{ }f\left( 0 \right)=2,\text{ }f\left( 1 \right)=6,\text{ }f\left( 2 \right)=16,\text{ }f\left( 3 \right)=38, \\  & \text{ }f\left( 6 \right)=0236 \\ \end{align}\]
    \[\begin{align}   & And\text{ }g\left( f\left( x \right) \right)=x\Rightarrow g(2)=0,\,(6)=1, \\  & g(16)=2,\,g(38)=3,g(236)=6 \\ \end{align}\]
    [a] \[g(f(x))=x\Rightarrow g'(f(x)).f'(x)=1\]
    For h?(2), f(x)=2\[\Rightarrow \]x=0
    \[\Rightarrow g(x)\,=6\Rightarrow x=236\]
    \[\therefore \]Putting\[x=236\], we get 
    \[h'(g(g(236))=\frac{1}{g'(g(236)).g'(236)}\]
    \[\Rightarrow h'(g(6))=\frac{1}{g'(6)g'(236)}\]
    \[\Rightarrow h'(g(1)=\frac{1}{g'(f(1))g'(f(6))}=f'(1)\,f'(6)\]
    \[\Rightarrow 6\times 111=666\]
    [c] \[h\left[ g(g(x)) \right]=x\]
    For \[h(g(g(3)),\]we need g\[\left( x \right)\text{ }=3\Rightarrow x=38\]
    \[\therefore \] Putting x=38, we get
    \[h\left[ g(g(38)) \right]=38\Rightarrow h(g(3))=38\]


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