JEE MATHEMATICS MCQ: MEAN VALUES THEOREM

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Application +4 | -1

The function

f(x)=(x-3)^2

satisfies all the conditions of mean value theorem in

\{3,4\}

. A point on

y=(x-3)^2

, where the tangent is parallel to the chord joining

(3,0)

and

(4,1)

View answer and explanation

Correct option: B

\left(\frac{7}{2}, \frac{1}{4}\right)

Option B

\left(\frac{7}{2}, \frac{1}{4}\right)

is correct because it satisfies the Mean Value Theorem, where the tangent's slope equals the chord's slope.

More Questions from MEAN VALUES THEOREM

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0 < \mathrm{c} < 1

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f(x)

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f'\left( x \right)

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f(x)

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\int\limits_3^5 {f\left( x \right)} \,dx

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f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right)