QBManager

Application of Derivatives

85698 Let $f (x) = {a x}^{3} + {b x}^{2} + c x + 1$ have extrema at $x$ $= α, β$ such that $α β < 0$ and $f (α) f (β) < 0$ . Then the equation $f (x) = 0$ has

1 three equal roots

2 one negative root if

f (α) < 0

and

f (β) > 0

3 one positive root if

f (α) > 0

and

f (β) < 0

4 None of these

Explanation:

(D) : Given,
$f (x) = a x^{3} + b x^{2} + c x + 1$
Then, $f (x) = 3 a x^{2} + 2 b x + c$
It have extreme at $x = α, β$ such that $α β < 0$
Then,
$f^{'} (α) = 3 a α^{2} + 2 b α + c$
and, $f^{'} (β) = 3 α β^{2} + 2 b β + c$
And also $f (α) f (β) < 0$
$(3 α^{3} + b α^{2} + c α + 1) (a β^{3} + b β^{2} + c β + 1) < 0$
As $α β < 0$ , there will be a root $γ$ between $α$ and $β$ .
So, $γ$ is the third root and may be positive or negative.
Hence, none of these is the correct answer.

AMU-2019

Application of Derivatives

85699 The minimum value of $\sec θ + cosec θ$ is

1 2

2

2 \sqrt{2}

3 4

4

4 \sqrt{2}

Explanation:

(B) : Find, minimum value of $\sec θ + cosec θ$
Let, $y = \sec θ + cosec θ$
$\frac{d y}{d x} = \sec θ \cdot \tan θ - cosec θ \cdot \cot θ$
For minimum $\frac{dy}{dx} = 0$
$\sec θ \tan θ - cosec θ \cot θ = 0$
$\sec θ \tan θ = cosec θ \cot θ$
$\sin^{3} θ = \cos^{3} θ$
$\Rightarrow \tan^{3} θ = 1$
$\Rightarrow \tan θ = 1$
$θ = \tan^{- 1} (π / 4)$
$θ = (π / 4)$
Minimum value of $\sec θ + cosec θ$
$= \sec π / 4 + cosec π / 4$
$= \sqrt{2} + \sqrt{2} = 2 \sqrt{2}$

AMU-2006

Application of Derivatives

85700 Maximum value of $f (x) = \sin x + \cos x$ is

1

\sqrt{2}

2 2

3

1 / \sqrt{2}

4 1

Explanation:

(A) : Find maximum value of
$f (x) = \sin x + \cos x$
We know that,
$- \sqrt{a^{2} + b^{2}} \leq a \sin x + b \cos x \leq \sqrt{a^{2} + b^{2}}$
$\sin x + \cos x \leq \sqrt{1^{2} + 1^{2}}$
$\sin x + \cos x \leq \sqrt{2}$
Maximum value of $\sin x + \cos x$ is $\sqrt{2}$

AMU-2002

Application of Derivatives

85701 If the minimum value of $f (x) = \frac{5 x^{2}}{2} + \frac{α}{x^{5}}, x > 0$ , is 14 then the value of $α$ is equal to :

1 32

2 64

3 128

4 256

Explanation:

(C) : Given,
$f (x) = \frac{5 x^{2}}{2} + \frac{α}{x^{5}}, x > 0$
$f^{'} (x) = \frac{2 \times 5 \times x}{2} - \frac{5 α}{x^{6}}$
$f^{'} (x) = 5 x - \frac{5 α}{x^{6}}$
For maxima $/$ minima, $f^{'} (x) = 0$
$5 x - \frac{5 α}{x^{6}} = 0 \Rightarrow 5 x = \frac{5 α}{x^{6}}$
$5 x^{7} = 5 α$
$x^{7} = α$
$x = α^{1 / 7}$
And, also given, $f (α^{1 / 7}) = 14$
$\frac{5 α^{2 / 7}}{2} + \frac{α}{α^{5 / 7}} = 14$
* $\frac{7 α^{2 / 7}}{α^{2 / 7} = 4} = 14$
$α = 4^{7 / 2} = 2^{7} = 128$

JEE Main-2022-28.07.2022

Application of Derivatives

85698 Let $f (x) = {a x}^{3} + {b x}^{2} + c x + 1$ have extrema at $x$ $= α, β$ such that $α β < 0$ and $f (α) f (β) < 0$ . Then the equation $f (x) = 0$ has

1 three equal roots

2 one negative root if

f (α) < 0

and

f (β) > 0

3 one positive root if

f (α) > 0

and

f (β) < 0

4 None of these

Explanation:

(D) : Given,
$f (x) = a x^{3} + b x^{2} + c x + 1$
Then, $f (x) = 3 a x^{2} + 2 b x + c$
It have extreme at $x = α, β$ such that $α β < 0$
Then,
$f^{'} (α) = 3 a α^{2} + 2 b α + c$
and, $f^{'} (β) = 3 α β^{2} + 2 b β + c$
And also $f (α) f (β) < 0$
$(3 α^{3} + b α^{2} + c α + 1) (a β^{3} + b β^{2} + c β + 1) < 0$
As $α β < 0$ , there will be a root $γ$ between $α$ and $β$ .
So, $γ$ is the third root and may be positive or negative.
Hence, none of these is the correct answer.

AMU-2019

Application of Derivatives

85699 The minimum value of $\sec θ + cosec θ$ is

1 2

2

2 \sqrt{2}

3 4

4

4 \sqrt{2}

Explanation:

(B) : Find, minimum value of $\sec θ + cosec θ$
Let, $y = \sec θ + cosec θ$
$\frac{d y}{d x} = \sec θ \cdot \tan θ - cosec θ \cdot \cot θ$
For minimum $\frac{dy}{dx} = 0$
$\sec θ \tan θ - cosec θ \cot θ = 0$
$\sec θ \tan θ = cosec θ \cot θ$
$\sin^{3} θ = \cos^{3} θ$
$\Rightarrow \tan^{3} θ = 1$
$\Rightarrow \tan θ = 1$
$θ = \tan^{- 1} (π / 4)$
$θ = (π / 4)$
Minimum value of $\sec θ + cosec θ$
$= \sec π / 4 + cosec π / 4$
$= \sqrt{2} + \sqrt{2} = 2 \sqrt{2}$

AMU-2006

Application of Derivatives

85700 Maximum value of $f (x) = \sin x + \cos x$ is

1

\sqrt{2}

2 2

3

1 / \sqrt{2}

4 1

Explanation:

(A) : Find maximum value of
$f (x) = \sin x + \cos x$
We know that,
$- \sqrt{a^{2} + b^{2}} \leq a \sin x + b \cos x \leq \sqrt{a^{2} + b^{2}}$
$\sin x + \cos x \leq \sqrt{1^{2} + 1^{2}}$
$\sin x + \cos x \leq \sqrt{2}$
Maximum value of $\sin x + \cos x$ is $\sqrt{2}$

AMU-2002

Application of Derivatives

85701 If the minimum value of $f (x) = \frac{5 x^{2}}{2} + \frac{α}{x^{5}}, x > 0$ , is 14 then the value of $α$ is equal to :

1 32

2 64

3 128

4 256

Explanation:

(C) : Given,
$f (x) = \frac{5 x^{2}}{2} + \frac{α}{x^{5}}, x > 0$
$f^{'} (x) = \frac{2 \times 5 \times x}{2} - \frac{5 α}{x^{6}}$
$f^{'} (x) = 5 x - \frac{5 α}{x^{6}}$
For maxima $/$ minima, $f^{'} (x) = 0$
$5 x - \frac{5 α}{x^{6}} = 0 \Rightarrow 5 x = \frac{5 α}{x^{6}}$
$5 x^{7} = 5 α$
$x^{7} = α$
$x = α^{1 / 7}$
And, also given, $f (α^{1 / 7}) = 14$
$\frac{5 α^{2 / 7}}{2} + \frac{α}{α^{5 / 7}} = 14$
* $\frac{7 α^{2 / 7}}{α^{2 / 7} = 4} = 14$
$α = 4^{7 / 2} = 2^{7} = 128$

JEE Main-2022-28.07.2022

Application of Derivatives

85698 Let $f (x) = {a x}^{3} + {b x}^{2} + c x + 1$ have extrema at $x$ $= α, β$ such that $α β < 0$ and $f (α) f (β) < 0$ . Then the equation $f (x) = 0$ has

1 three equal roots

2 one negative root if

f (α) < 0

and

f (β) > 0

3 one positive root if

f (α) > 0

and

f (β) < 0

4 None of these

Explanation:

(D) : Given,
$f (x) = a x^{3} + b x^{2} + c x + 1$
Then, $f (x) = 3 a x^{2} + 2 b x + c$
It have extreme at $x = α, β$ such that $α β < 0$
Then,
$f^{'} (α) = 3 a α^{2} + 2 b α + c$
and, $f^{'} (β) = 3 α β^{2} + 2 b β + c$
And also $f (α) f (β) < 0$
$(3 α^{3} + b α^{2} + c α + 1) (a β^{3} + b β^{2} + c β + 1) < 0$
As $α β < 0$ , there will be a root $γ$ between $α$ and $β$ .
So, $γ$ is the third root and may be positive or negative.
Hence, none of these is the correct answer.

AMU-2019

Application of Derivatives

85699 The minimum value of $\sec θ + cosec θ$ is

1 2

2

2 \sqrt{2}

3 4

4

4 \sqrt{2}

Explanation:

(B) : Find, minimum value of $\sec θ + cosec θ$
Let, $y = \sec θ + cosec θ$
$\frac{d y}{d x} = \sec θ \cdot \tan θ - cosec θ \cdot \cot θ$
For minimum $\frac{dy}{dx} = 0$
$\sec θ \tan θ - cosec θ \cot θ = 0$
$\sec θ \tan θ = cosec θ \cot θ$
$\sin^{3} θ = \cos^{3} θ$
$\Rightarrow \tan^{3} θ = 1$
$\Rightarrow \tan θ = 1$
$θ = \tan^{- 1} (π / 4)$
$θ = (π / 4)$
Minimum value of $\sec θ + cosec θ$
$= \sec π / 4 + cosec π / 4$
$= \sqrt{2} + \sqrt{2} = 2 \sqrt{2}$

AMU-2006

Application of Derivatives

85700 Maximum value of $f (x) = \sin x + \cos x$ is

1

\sqrt{2}

2 2

3

1 / \sqrt{2}

4 1

Explanation:

(A) : Find maximum value of
$f (x) = \sin x + \cos x$
We know that,
$- \sqrt{a^{2} + b^{2}} \leq a \sin x + b \cos x \leq \sqrt{a^{2} + b^{2}}$
$\sin x + \cos x \leq \sqrt{1^{2} + 1^{2}}$
$\sin x + \cos x \leq \sqrt{2}$
Maximum value of $\sin x + \cos x$ is $\sqrt{2}$

AMU-2002

Application of Derivatives

85701 If the minimum value of $f (x) = \frac{5 x^{2}}{2} + \frac{α}{x^{5}}, x > 0$ , is 14 then the value of $α$ is equal to :

1 32

2 64

3 128

4 256

Explanation:

(C) : Given,
$f (x) = \frac{5 x^{2}}{2} + \frac{α}{x^{5}}, x > 0$
$f^{'} (x) = \frac{2 \times 5 \times x}{2} - \frac{5 α}{x^{6}}$
$f^{'} (x) = 5 x - \frac{5 α}{x^{6}}$
For maxima $/$ minima, $f^{'} (x) = 0$
$5 x - \frac{5 α}{x^{6}} = 0 \Rightarrow 5 x = \frac{5 α}{x^{6}}$
$5 x^{7} = 5 α$
$x^{7} = α$
$x = α^{1 / 7}$
And, also given, $f (α^{1 / 7}) = 14$
$\frac{5 α^{2 / 7}}{2} + \frac{α}{α^{5 / 7}} = 14$
* $\frac{7 α^{2 / 7}}{α^{2 / 7} = 4} = 14$
$α = 4^{7 / 2} = 2^{7} = 128$

JEE Main-2022-28.07.2022

Application of Derivatives

85698 Let $f (x) = {a x}^{3} + {b x}^{2} + c x + 1$ have extrema at $x$ $= α, β$ such that $α β < 0$ and $f (α) f (β) < 0$ . Then the equation $f (x) = 0$ has

1 three equal roots

2 one negative root if

f (α) < 0

and

f (β) > 0

3 one positive root if

f (α) > 0

and

f (β) < 0

4 None of these

Explanation:

(D) : Given,
$f (x) = a x^{3} + b x^{2} + c x + 1$
Then, $f (x) = 3 a x^{2} + 2 b x + c$
It have extreme at $x = α, β$ such that $α β < 0$
Then,
$f^{'} (α) = 3 a α^{2} + 2 b α + c$
and, $f^{'} (β) = 3 α β^{2} + 2 b β + c$
And also $f (α) f (β) < 0$
$(3 α^{3} + b α^{2} + c α + 1) (a β^{3} + b β^{2} + c β + 1) < 0$
As $α β < 0$ , there will be a root $γ$ between $α$ and $β$ .
So, $γ$ is the third root and may be positive or negative.
Hence, none of these is the correct answer.

AMU-2019

Application of Derivatives

85699 The minimum value of $\sec θ + cosec θ$ is

1 2

2

2 \sqrt{2}

3 4

4

4 \sqrt{2}

Explanation:

(B) : Find, minimum value of $\sec θ + cosec θ$
Let, $y = \sec θ + cosec θ$
$\frac{d y}{d x} = \sec θ \cdot \tan θ - cosec θ \cdot \cot θ$
For minimum $\frac{dy}{dx} = 0$
$\sec θ \tan θ - cosec θ \cot θ = 0$
$\sec θ \tan θ = cosec θ \cot θ$
$\sin^{3} θ = \cos^{3} θ$
$\Rightarrow \tan^{3} θ = 1$
$\Rightarrow \tan θ = 1$
$θ = \tan^{- 1} (π / 4)$
$θ = (π / 4)$
Minimum value of $\sec θ + cosec θ$
$= \sec π / 4 + cosec π / 4$
$= \sqrt{2} + \sqrt{2} = 2 \sqrt{2}$

AMU-2006

Application of Derivatives

85700 Maximum value of $f (x) = \sin x + \cos x$ is

1

\sqrt{2}

2 2

3

1 / \sqrt{2}

4 1

Explanation:

(A) : Find maximum value of
$f (x) = \sin x + \cos x$
We know that,
$- \sqrt{a^{2} + b^{2}} \leq a \sin x + b \cos x \leq \sqrt{a^{2} + b^{2}}$
$\sin x + \cos x \leq \sqrt{1^{2} + 1^{2}}$
$\sin x + \cos x \leq \sqrt{2}$
Maximum value of $\sin x + \cos x$ is $\sqrt{2}$

AMU-2002

Application of Derivatives

85701 If the minimum value of $f (x) = \frac{5 x^{2}}{2} + \frac{α}{x^{5}}, x > 0$ , is 14 then the value of $α$ is equal to :

1 32

2 64

3 128

4 256

Explanation:

(C) : Given,
$f (x) = \frac{5 x^{2}}{2} + \frac{α}{x^{5}}, x > 0$
$f^{'} (x) = \frac{2 \times 5 \times x}{2} - \frac{5 α}{x^{6}}$
$f^{'} (x) = 5 x - \frac{5 α}{x^{6}}$
For maxima $/$ minima, $f^{'} (x) = 0$
$5 x - \frac{5 α}{x^{6}} = 0 \Rightarrow 5 x = \frac{5 α}{x^{6}}$
$5 x^{7} = 5 α$
$x^{7} = α$
$x = α^{1 / 7}$
And, also given, $f (α^{1 / 7}) = 14$
$\frac{5 α^{2 / 7}}{2} + \frac{α}{α^{5 / 7}} = 14$
* $\frac{7 α^{2 / 7}}{α^{2 / 7} = 4} = 14$
$α = 4^{7 / 2} = 2^{7} = 128$

JEE Main-2022-28.07.2022

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