Q.

The number of real solutions of the equation e4x+4e3x58e2x+4ex+1=0 is _______.

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answer is 2.

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Detailed Solution

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e4x+4e3x58e2x+4ex+1=0 Let f(x)=e2xe2x+1e2x+4ex+1ex58ex+1ex=t Let h(t)=t2+4t58=0t=4±16+4.5824±2622t1=2+262t2=2262( not possible ) since t2 ex+1ex=2+262e2x(2+262)ex+1=0b2-4ac=(2+262)24                 =4+4.628624                  =248862>0and b2a>0 both roots are positiveNumber of real roots =2

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