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JEE Mains · Maths · STD 12 - 1. relation and function

જો \(f(x)=\left\{\begin{array}{ll}x+a, & x \leq 0 \\ |x-4|, & x>0\end{array}\right.\) અને \(g(x)=\left\{\begin{array}{ll}x+1 & x<0 \\ (x-4)^{2}+b, & x \geq 0\end{array}\right.\) એ  \(R\) પર સતત હોય તો \((gof) (2)+( fog) (-2)\) ની કિમંત મેળવો.

  1. A \(-10\)
  2. B \(10\)
  3. C \(8\)
  4. D \(-8\)
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Correct Answer

(D) \(-8\)

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\((x)=\left\{\begin{array}{l} x+a ; x \leq 0 \\ |x-4| ; x>0 \end{array} ; g(x)=\left\{\begin{array}{ll} x+1 & ; x<0 \\ (x-4)^{2}+b ; & x \geq 0 \end{array}\right.\right.\) For continuity \(a =4\) and \(b =-15\) \(g(f(2))+f(g(-2))\) \(=g(2)+f(-1)=-8\)
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