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Q.

The equation e4x+8e3x+13e2x8ex+1=0,x has

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a

four solutions two of which are negative

b

no solution

c

two solutions and only one of them is negative

d

two solutions and both are negative

answer is B.

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Detailed Solution

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ex=t,(t>0)t4+8t3+13t28t+1=0 divide with t2t2+8t+138t+1t2=0t2+1t2+8t1t+13=0t1t=z,z2+2+8z+13=0z2+8z+15=0(z+5)(z+3)=0t1t+5t1t+3=0t2+5t1t2+3t1=0t2+5t1=0 (or) t2+3t1=0t=5±262,t=3±122 let t=ex>0ex=2652( or )1232x=loge2652<0  (or =loge1232<0
both roots are negative
(or) usef(t),f′′(t)

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The equation e4x+8e3x+13e2x−8ex+1=0,x∈ℝ has