Question

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5. Verify Lagranges' Mean Value Theorem for the function f(x) = 2 sin x + sin 2x on (0,7).डिफरेंट ch4 कॉस क्यूब माइनस टू गो टू सी इस इक्वल टू ​

Answer 1

Answer:

Given, f(x)=2sinx+sin2x,x∈[0,π]

f(x) is continuous is [0,π]

f(x) is differentiable in (0,π)

Thus, both the conditions of Lagrange's man value theorem are satisfied by the function f(x) in [0,π], therefore, there exists at least one real number c in [0,π] such that

f

′

(c)=

π−0

f(π)−f(0)

fπ=2sinπ+sin2π=0

f(0)=2sin0+sin0=0

Differentiating f(x) w.r.t. x, we get

f

′

(x)=2cosx+2cos2x

Now, 2cosx+2cos2x=0

⇒2cos

2

x+cosx−1=0 (∵cos2x=2cos

2

x−1)

⇒2cos

2

x+2cosx−cosx−1=0

⇒2cosx(cosx+1)−1(cosx+1)=0

⇒(2cosx−1)(cosx+1)=0

2cos−1=0

or cosx+1=0

2cosx=1 or cosx=−1

cosx=

2

1

or cosx=−1

⇒x=

3

π

,π

∴x=

3

π

∵

3

π

ϵ(0,π)

Thus Lagrange's mean value theorem is verified