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The equation e^x-x-1=0 has...

The equation `e^x-x-1=0` has

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The equation e^(x-1)+x-2=0 as

Show that the equation e^(x-1)+x-2=0 has no real root greater than 1.

Show that the equation e^(x-1)+x-2=0 has no real root which is less than 1.

Statement-1: The equation e^(x-1) +x-2=0 has only one real root. Statement-2 : Between any two root of an equation f(x)=0 there is a root of its derivative f'(x)=0

Statement-1: The equation e^(x-1) +x-2=0 has only one real root. Statement-2 : Between any two root of an equation f(x)=0 there is a root of its derivative f'(x)=0

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Between any two real roots of the equation e^(x)sinx-1=0 the equation e^(x)cosx+1=0 has

Between any two real roots of the equation e^(x)sinx-1=0 the equation e^(x)cosx+1=0 has

Between any two real roots of the equation e^(x)sinx-1=0 the equation e^(x)cosx+1=0 has

The equation e^(x-8)+2x-17=0 has :