JEE MATHEMATICS MCQ: MAXIMA & MINIMA

1 Medium

1 of 1

Analyzing +4 | -1

Let

p\left( x \right) = {a_0} + {a_1}{x^2} + {a_2}{x^4} + \ldots + {a_n}{x^2}^n

be a polynomial in a real variable

x

with

0 < {a_0} < {a_1} < {a_2} \ldots < {a_n}.\;

The function

p\left( x \right)

has

View answer and explanation

Correct option: C

Only one minimum

Option C (Only one minimum) is correct because the derivative changes sign only once, from negative to positive, at x=0.

More Questions from MAXIMA & MINIMA

Practice All Chapter MCQs →

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

m(b)

f(x)

b

m(b)

f: R \rightarrow R

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

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

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

x<0

x>0

p

f\left( x \right)

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

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

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}