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Question : 65 of 160
Marks:
+1,
-0
Solution:
Given equation
(x2−3x+2)e‌=x+2Differentiate both sides w.r.t., we get
‌‌((x2−3x+2)⋅e‌)=‌(x+2)‌⇒(2x−3)e‌+(x2−3x+2)⋅e‌‌⇒(2x−3)e‌+(x2−3x+2)⋅e‌(‌)=1‌‌‌[‌]=1Now, substitute
x=0 into the given equation
‌(02−3(0)+2)e‌=0+2⇒2e−y=2⇒e−y=1⇒−y=ln(1)⇒−y=0⇒y=0Substitute
x=0 and
y=0 in Eq. (i), we get
‌(2(0)−3)e‌+(02−3(0)+2)e‌‌⇒−‌]=1‌⇒−3e0+2e0(‌)=1‌⇒−3+2(‌)=1‌⇒‌‌−2‌=1+3=4⇒‌=‌=−2‌∴‌‌(‌)x=0=−2
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