Question

(1) The minimum value of e(2:22+1)sin’r
is
Оe
1/e
0
1​

Answer 1

Answer:

y =0 because it is an exponential function and its power must reach infinity to make it zero which is only possible if x tends to infinity. Hence.

Answer 2

Answer:

answer =1

12th

Maths

Application of Derivatives

Maxima and Minima

The minimum value of e^(2x^...

MATHS

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Asked on October 15, 2019 by

Daizzy Motwani

The minimum value of e

(2x

2

−2x+1)sin

2

x

is

HARD

ANSWER

The given equation is:

y=e

(2x

2

−2x+1)sin

2

x

Taking logarithm on both sides we get,

⇒lny=(2x

2

−2x+1)sin

2

x

Differentiating once and equating it to zero we get,

⇒

y

1

.y

′

=sin2x(2x

2

−2x+1)+(4x−2)sin

2

x

⇒y

′

=y×sin2x(2x

2

−2x+1)+(4x−2)sin

2

x

y

′

=0

y



=0 because it is an exponential function and its power must reach infinity to make it zero which is only possible if x tends to infinity. Hence

⇒sin2x(2x

2

−2x+1)+(4x−2)sin

2

x=0

∵2x

2

−2x+1>0 as Its D=

b

2

−4ac

is negative hence the function is always positive.

⇒sin

2

x=0

⇒x=nπ

⇒sin2x=0

⇒x=

2

nπ

The intersection of both the domains of the point where we have a critical point gives that the critical points occurs at x=nπ

⇒y(nπ)=e

(2(nπ)

2

−2nπ+1)sin

2

nπ

⇒y(nπ)=e

0

⇒y(nπ)=1 .....Answer