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JEE MATHEMATICS MCQ: MAXIMA & MINIMA

1 Easy

1 of 1

Analyzing +4 | -1

\sin x+\cos x=y^2-y+a

has no value of

x

for any

y

, if

a

belongs

View answer and explanation

Correct option: D

(\sqrt{3}, \infty)

Option D (

(\sqrt{3}, \infty)

) is correct because the range of

\sin x + \cos x

is

[-\sqrt{2}, \sqrt{2}]

, and

y^2 - y + a

must be outside this range for all

y

.

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

m(b)

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f(x)

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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

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

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and a minima for some value of

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. 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

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

, is If the total maximum value of the function

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\frac{k}{e}

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