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JEE MATHEMATICS MCQ: MAXIMA & MINIMA

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Application +4 | -1

The marks obtained by a stu first and seound terminal examinamuns are 62 antu-n 48 , respectively. The minimum marks he should get in the annual examination to have an average of at least 60 marks, are

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Correct option: A

70

The correct answer is A (70) because it represents the minimum marks needed to achieve an average of at least 60, calculated using the average formula.

More Questions from MAXIMA & MINIMA

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f(x)=\left(1+b^2\right) x^2+2 b x+1

m(b)

be the minimum value of

f(x)

(that is least value). As

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varies, the range of

m(b)

f: R \rightarrow R

be a function defined by

f(x)=(x-3)^{n_1}(x-5)^{n_2}, n_1, n_2 \in N

. Then, which of the following is NOT true? The equation

|2 a x-3|+|a x+1|+|5-a x|=\frac{1}{2}

possesses Let the function

f(x)=2 x^{3}+(2 p-7) x^{2}+3(2 p-9) x

-6 have a maxima for some value of

x<0

and a minima for some value of

x>0

. Then, the set of all values of

p

is The least value of the

f\left( x \right)

f\left( x \right) = {\tan ^{ - 1}}x - \frac{1}{2}{\log _e}x

in the interval

\left[ {1/\sqrt 3 ,\;\sqrt 3 } \right]

, is If the total maximum value of the function

f(x)=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^{2} x}, x \in\left(0, \frac{\pi}{2}\right),

\frac{k}{e}

\left(\frac{k}{e}\right)^{8}+\frac{k^{8}}{e^{5}}+k^{8}