The graph of a function f(x) is shown in the figure. For f(x) to be a valid probability density function, the value of h is

GATE CE 2018 Shift 2 Assignment Solution 14 D1

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  1. 1/3
  2. 2/3
  3. 1
  4. 3

Answer (Detailed Solution Below)

Option 1 : 1/3
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Form graph

\(\frac{{{{\rm{y}}_1} - 0}}{{{\rm{x}} - 0}} = \frac{{{\rm{h}} - 0}}{{1 - 0}}\)

⇒ y1 = hx = f(x1)

Similarly

f(x2) = 2h (x-1)

f(x3) = 3h (x-2)

Since for probability density function

\(\mathop \smallint \limits_0^3 {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = 1\)

\(\Rightarrow \mathop \smallint \limits_0^1 {\rm{f}}\left( {{{\rm{x}}_1}} \right){\rm{dx}}\mathop \smallint \limits_1^2 2{\rm{h}}\left( {{\rm{x}} - 1} \right){\rm{dx}} \Rightarrow \mathop \smallint \limits_3^1 {\rm{f}}\left( {{{\rm{x}}_1}} \right){\rm{dx}} + \mathop \smallint \limits_1^2 {\rm{f}}\left( {{{\rm{x}}_2}} \right){\rm{dx}} + \mathop \smallint \limits_2^3 {\rm{f}}\left( {{{\rm{x}}_3}} \right){\rm{dx}}\)

\(\Rightarrow \mathop \smallint \limits_0^1 \left( {{\rm{hx}}} \right){\rm{dx}} + \mathop \smallint \limits_1^2 \left[ {2{\rm{h}}\left( {{\rm{x}} - 1} \right){\rm{dx}}} \right]{\rm{dx}} + \mathop \smallint \limits_3^1 \left[ {3{\rm{h}}\left( {{\rm{x}} - 2} \right)} \right]{\rm{dx}} = 1\)

\(\Rightarrow \left. {\frac{{{\rm{h}}{{\rm{x}}^2}}}{2}} \right|_0^1 + 2{\rm{h}}\left. {\frac{{{{\left( {{\rm{x}} - 1} \right)}^2}}}{2}} \right|_1^2 + 3{\rm{h}}\left. {\frac{{{{\left( {{\rm{x}} - 2} \right)}^2}}}{2}} \right|_1^3 = 1\)

\(\Rightarrow \frac{{\rm{h}}}{2} + \frac{{2{\rm{h}}}}{2} + \frac{{3{\rm{h}}}}{2} = 1\)

\(\Rightarrow {\rm{h}} = \frac{1}{3}\)

∴ The value of h is 1/3
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