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  • Provide solution for RD Sharma maths Class 12 Chapter Functions Exercise 2.3 Question 13 maths text book solution.

Provide solution for RD Sharma maths Class 12 Chapter Functions Exercise 2.3 Question 13 maths text book solution.

Answers (1)

Answer : f \circ g(x)=\left\{\begin{array}{ll} 0, & x \geq 0 \\ -4 x, & x<0 \end{array}\right.

\begin{aligned} &g \circ f(x)=0 \text { for all } \mathrm{x} \\ &f \circ g(-3)=12, f \circ g(5)=0 \text { and } g \circ f(-2)=0 \end{aligned}

Hint : We know about modulus function,

\begin{aligned} &\mathrm{f}(\mathrm{x})=|x|+\mathrm{x} \\ &\Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l} x+x, \text { if } x \geq 0 \\ -x+x, \text { if } x<0 \end{array}\right. \end{aligned}

Given : Let f, g:R\rightarrow R be two function defined as :

                     \begin{aligned} &\mathrm{f}(\mathrm{x})=|x|+\mathrm{x} \\ &g(\mathrm{x})=|x|-\mathrm{x} \end{aligned}

Here, we have to find out f \circ g(x), g \circ f(x), f \circ g(-3), f \circ g(5) \text { and } g \circ f(-2)

Solution :

Here,

         \begin{aligned} &\mathrm{f}(\mathrm{x})=|x|+\mathrm{x} \text { and } \mathrm{g}(\mathrm{x})=|x|-\mathrm{x} \\ &\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l} x+x, \text { if } x \geq 0 \\ -x+x, \text { if } x<0 \end{array}\right. \\ &\Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{c} 2 x, \text { if } x \geq 0 \\ 0, \text { if } x<0 \end{array}\right. \\ &\mathrm{g}(\mathrm{x})=\left\{\begin{array}{l} x-x, \text { if } x \geq 0 \\ -x-x, \text { if } x<0 \end{array}\right. \\ &\Rightarrow g(\mathrm{x})=\left\{\begin{array}{c} 0, \text { if } x \geq 0 \\ -2 x, \text { if } x<0 \end{array}\right. \end{aligned}

Thus for x \geq 0, g \circ f(x)=g(f(x))

                                               \begin{aligned} &=g(2 x) \\ &=0 \end{aligned}

For x<0, g \circ f(x)=g(f(x))

                      \begin{gathered} =\mathrm{g}(0) \\ =0 \\ \Rightarrow g \circ f(x)=0 \forall \mathrm{x} \in \mathrm{R} \end{gathered}

Thus for  x \geq 0, f \circ g(x)=\mathrm{f}(\mathrm{g}(\mathrm{x}))

                                                \begin{aligned} &=\mathrm{f}(0) \\ &=2(0) \\ &=0 \end{aligned}

For x<0, f \circ g(x)=\mathrm{f}(\mathrm{g}(\mathrm{x}))

                                       \begin{aligned} &=f(-2 x) \\ &=-4 x \end{aligned}

\Rightarrow f \circ g(x)=\left\{\begin{array}{ll} 0, & x \geq 0 \\ -4 x, & x<0 \end{array}\right.

Again,

\begin{aligned} &\mathrm{f}(\mathrm{x})=|x|+\mathrm{x} \text { and } \mathrm{g}(\mathrm{x})=|x|-\mathrm{x} \\ &\therefore \begin{aligned} \therefore \circ g(x) &=\mathrm{f}(\mathrm{g}(\mathrm{x})) \\ &=\lg (\mathrm{x}) \mid+\mathrm{g}(\mathrm{x}) \\ &=|| x|-x|+|x|-\mathrm{x} \end{aligned} \\ &\begin{aligned} \therefore g \circ f(x) &=\mathrm{g}(\mathrm{f}(\mathrm{x})) \\ &=|\mathrm{f}(\mathrm{x})|-\mathrm{f}(\mathrm{x}) \\ &=|| x|+x|-[|x|+\mathrm{x}] \end{aligned} \end{aligned}

Then,

\begin{aligned} &f \circ g(-3)=||-3|-(-3)|+|-3|-(-3) \\ &=|3+3|+3+3 \\ &=6+6 \\ &\quad=12 \\ &\therefore f \circ g(-3)=12 \end{aligned}

\begin{aligned} &\begin{aligned} f \circ g(5) &=|| 5|-(5)|+|5|-(5) \\ &=5-5+5-5 \\ &=0 \\ \therefore f \circ g &(5)=0 \end{aligned} \end{aligned}

\begin{aligned} &\begin{aligned} \therefore g \circ f(-2) &=||-2|+(-2)|-[|-2|+(-2)] \\ &=|2+2|-[2+2] \\ &=4-4 \\ &=0 \\ \therefore g \circ f(-2) &=0 \end{aligned} \end{aligned}

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