Let f, g : R $\to$ R be two real valued functions defined as $$f(x) = \left\{ {\matrix{ { - |x + 3|} & , & {x < 0} \cr {{e^x}} & , & {x \ge 0} \cr } } \right.$$ and $$g(x) = \left\{ {\matrix{ {{x^2} + {k_1}x} & , & {x < 0} \cr {4x + {k_2}} & , & {x \ge 0} \cr } } \right.$$, where k1 and k2 are real constants. If (gof) is differentiable at x = 0, then (gof) ($-$ 4) + (gof) (4) is equal to :
Solution
<p>$\because$ gof is differentiable at x = 0</p>
<p>So R.H.D = L.H.D</p>
<p>$${d \over {dx}}(4{e^x} + {k_2}) = {d \over {dx}}\left( {{{( - |x + 3|)}^2} - {k_1}|x + 3|} \right)$$</p>
<p>$\Rightarrow 4 = 6 - {k_1} \Rightarrow {k_1} = 2$</p>
<p>Also $f(f({0^ + })) = g(f({0^ - }))$</p>
<p>$\Rightarrow 4 + {k_2} = 9 - 3{k_1} \Rightarrow {k_2} = - 1$</p>
<p>Now $g(f( - 4)) + g(f(4))$</p>
<p>$= g( - 1) + g({e^4}) = (1 - {k_1}) + (4{e^4} + {k_2})$</p>
<p>$= 4{e^4} - 2$</p>
<p>$= 2(2{e^4} - 1)$</p>
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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