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Question

For equation $$(x^2-5x+5)^{x^2+4x-60}=1$$


A
Number of real value of x=4
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B
Number of real value of x=5
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C
Sum of real value of x=1
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D
Sum of real value of x=3
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Solution

The correct option is A Number of real value of $$x=4$$
$${ { \left( { x }^{ 2 }-5x+5 \right)  } }^{ { x }^{ 2 }+4x-60 }=1\longrightarrow (1)\\ \Rightarrow log{ { \left( { x }^{ 2 }-5x+5 \right)  } }^{ { x }^{ 2 }+4x-60 }=log1\quad (taking\quad log\quad on\quad both\quad side)\\ \Rightarrow \left( { x }^{ 2 }+4x-60 \right) log{ { \left( { x }^{ 2 }-5x+5 \right)  } }=0\quad (since\quad log1=0)\\ so,\quad \left( { x }^{ 2 }+4x-60 \right) =0\quad (0r)\quad log{ { \left( { x }^{ 2 }-5x+5 \right)  } }=0\\ for,\quad \left( { x }^{ 2 }+4x-60 \right) =0\\ \Rightarrow { b }^{ 2 }-4ac={ 4 }^{ 2 }-4.1.(-60)=256>0\\ \therefore \quad x\quad has\quad 2\quad distinct\quad real\quad roots\\ also\quad for,\quad log{ { \left( { x }^{ 2 }-5x+5 \right)  } }=0\\ \Rightarrow log{ { \left( { x }^{ 2 }-5x+5 \right)  } }=log1\quad (since\quad log1=0)\\ \Rightarrow \left( { x }^{ 2 }-5x+5 \right) =0\\ \Rightarrow { b }^{ 2 }-4ac={ (-5) }^{ 2 }-4.1.4=9>0\\ \therefore \quad x\quad has\quad 2\quad distinct\quad real\quad roots\\ \therefore \quad for\quad equation\quad (1),\quad number\quad of\quad real\quad values\quad of\quad x=4\\ \\ $$

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