QBManager

Application of Derivatives

85393 The approximate value of $(1.002)^{300}$ using differentiation is

1 1.8

2 1.6

3 1.2

4 1.4

Explanation:

(B) : Given,
$x + Δ x = 1.002$
$x = 1$
$Δ x = 0.002$
Let,
$f (x) = x^{300}$
$f^{'} (x) = 300 x^{299}$
We know that,
$f (x + Δ x) = f (x) + Δ x . f^{'} (x)$
$f (1.002) = f (1) + 0.002 \times 300 (1)^{299}$
$f (1.002) = 1 + 0.002 \times 300 = 1 + 0.6$
$f (1.002) = 1.6$

AP EAMCET-2002

Application of Derivatives

85394 Using differentiation, the approximate value of $\sin 46^{\circ}$ ,
Given that $1^{0} = {0.0175}^{\circ}$ , is

1 0.07194

2

\frac{1.0175}{\sqrt{2}}

3 0.7194

4

\frac{0.0175}{\sqrt{2}}

Explanation:

(B) : Given
$f (x) = \sin x$
$f^{'} (x) = \cos x$
$46^{\circ} = 45^{\circ} + 1^{\circ}$
$1^{\circ} = 0.0175$
$(x + Δ x) = 46^{\circ}$
$x = 45^{\circ}$
$Δ x = 1^{\circ} = 0.0175$
We know that,
$f (x + Δ x) = f^{'} (x) Δ x + f (x)$
$= \cos 45^{\circ} \times (0.0175) + \sin 45^{\circ}$
$= \frac{1}{\sqrt{2}} (0.0175) + \frac{1}{\sqrt{2}}$
$= \frac{1}{\sqrt{2}} [0.0175 + 1] = \frac{1.0175}{\sqrt{2}}$
So, the approximate value of $\sin 46^{\circ} = \frac{1.0175}{\sqrt{2}}$

[ -2019]

Application of Derivatives

85395 The slope of normal to the curve $x = \sqrt{t}$ and
$y = t - \frac{1}{\sqrt{t}} at t = 4 is$
$\begin{array}{llll} (a) \frac{17}{4} (b) \frac{- 17}{4} (c) \frac{4}{17} (d) \frac{- 4}{17} \end{array}$

1

\frac{17}{4}

2

\frac{- 17}{4}

3

\frac{4}{17}

4

\frac{- 4}{17}

Explanation:

(D) : Given,
$x = \sqrt{t}, y = t - \frac{1}{\sqrt{t}} at t = 4$
Now, $\frac{dx}{dt} = \frac{1}{2 \sqrt{t}}$
Similarly $\frac{dy}{dt} = 1 + \frac{1}{2} t^{- 3 / 2}$
$∴ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$= \frac{(1 + \frac{1}{2} t^{- 3 / 2})}{\frac{1}{2 \sqrt{t}}} = 2 \sqrt{t} (1 + \frac{1}{2} t^{- 3 / 2}) = 2 \sqrt{t} (1 + \frac{1}{2 t^{3 / 2}})$
$\frac{dy}{dx} = \frac{2 t^{3 / 2} + 1}{t}$
At, on putting $t = 4$
$\frac{dy}{dx} = \frac{2 (4)^{3 / 2} + 1}{4} = \frac{17}{4}$
$∴$ Slope of normal at $(t = 4) = - \frac{\frac{1}{d y}}{d x} = \frac{- 4}{17}$

MHT CET-2019

Application of Derivatives

85396 The equation of normal to the curve $y = \log_{e} x$ at the point $P (1, 0)$ is

1

x + y = 1

2

2 x + y = 2

3

x - y = 1

4

x - 2 y = 1

Explanation:

(A) : Given,
Curve, $y = \log_{e} x$
On differentiating w.r.t $x$ ,
$\frac{dy}{dx} = \frac{1}{x}$
$∴$ Slope of normal at $P = (1, 0)$
$= - \frac{1}{{(\frac{dy}{dx})}_{(1, 0)}} = \frac{- 1}{1} = - 1$
$∴$ Equation of normal at $P = (1, 0)$
$y - 0 = (- 1) (x - 1)$
$y = - x + 1$
$x + y = 1$

MHT CET-2019

Application of Derivatives

85393 The approximate value of $(1.002)^{300}$ using differentiation is

1 1.8

2 1.6

3 1.2

4 1.4

Explanation:

(B) : Given,
$x + Δ x = 1.002$
$x = 1$
$Δ x = 0.002$
Let,
$f (x) = x^{300}$
$f^{'} (x) = 300 x^{299}$
We know that,
$f (x + Δ x) = f (x) + Δ x . f^{'} (x)$
$f (1.002) = f (1) + 0.002 \times 300 (1)^{299}$
$f (1.002) = 1 + 0.002 \times 300 = 1 + 0.6$
$f (1.002) = 1.6$

AP EAMCET-2002

Application of Derivatives

85394 Using differentiation, the approximate value of $\sin 46^{\circ}$ ,
Given that $1^{0} = {0.0175}^{\circ}$ , is

1 0.07194

2

\frac{1.0175}{\sqrt{2}}

3 0.7194

4

\frac{0.0175}{\sqrt{2}}

Explanation:

(B) : Given
$f (x) = \sin x$
$f^{'} (x) = \cos x$
$46^{\circ} = 45^{\circ} + 1^{\circ}$
$1^{\circ} = 0.0175$
$(x + Δ x) = 46^{\circ}$
$x = 45^{\circ}$
$Δ x = 1^{\circ} = 0.0175$
We know that,
$f (x + Δ x) = f^{'} (x) Δ x + f (x)$
$= \cos 45^{\circ} \times (0.0175) + \sin 45^{\circ}$
$= \frac{1}{\sqrt{2}} (0.0175) + \frac{1}{\sqrt{2}}$
$= \frac{1}{\sqrt{2}} [0.0175 + 1] = \frac{1.0175}{\sqrt{2}}$
So, the approximate value of $\sin 46^{\circ} = \frac{1.0175}{\sqrt{2}}$

[ -2019]

Application of Derivatives

85395 The slope of normal to the curve $x = \sqrt{t}$ and
$y = t - \frac{1}{\sqrt{t}} at t = 4 is$
$\begin{array}{llll} (a) \frac{17}{4} (b) \frac{- 17}{4} (c) \frac{4}{17} (d) \frac{- 4}{17} \end{array}$

1

\frac{17}{4}

2

\frac{- 17}{4}

3

\frac{4}{17}

4

\frac{- 4}{17}

Explanation:

(D) : Given,
$x = \sqrt{t}, y = t - \frac{1}{\sqrt{t}} at t = 4$
Now, $\frac{dx}{dt} = \frac{1}{2 \sqrt{t}}$
Similarly $\frac{dy}{dt} = 1 + \frac{1}{2} t^{- 3 / 2}$
$∴ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$= \frac{(1 + \frac{1}{2} t^{- 3 / 2})}{\frac{1}{2 \sqrt{t}}} = 2 \sqrt{t} (1 + \frac{1}{2} t^{- 3 / 2}) = 2 \sqrt{t} (1 + \frac{1}{2 t^{3 / 2}})$
$\frac{dy}{dx} = \frac{2 t^{3 / 2} + 1}{t}$
At, on putting $t = 4$
$\frac{dy}{dx} = \frac{2 (4)^{3 / 2} + 1}{4} = \frac{17}{4}$
$∴$ Slope of normal at $(t = 4) = - \frac{\frac{1}{d y}}{d x} = \frac{- 4}{17}$

MHT CET-2019

Application of Derivatives

85396 The equation of normal to the curve $y = \log_{e} x$ at the point $P (1, 0)$ is

1

x + y = 1

2

2 x + y = 2

3

x - y = 1

4

x - 2 y = 1

Explanation:

(A) : Given,
Curve, $y = \log_{e} x$
On differentiating w.r.t $x$ ,
$\frac{dy}{dx} = \frac{1}{x}$
$∴$ Slope of normal at $P = (1, 0)$
$= - \frac{1}{{(\frac{dy}{dx})}_{(1, 0)}} = \frac{- 1}{1} = - 1$
$∴$ Equation of normal at $P = (1, 0)$
$y - 0 = (- 1) (x - 1)$
$y = - x + 1$
$x + y = 1$

MHT CET-2019

Application of Derivatives

85393 The approximate value of $(1.002)^{300}$ using differentiation is

1 1.8

2 1.6

3 1.2

4 1.4

Explanation:

(B) : Given,
$x + Δ x = 1.002$
$x = 1$
$Δ x = 0.002$
Let,
$f (x) = x^{300}$
$f^{'} (x) = 300 x^{299}$
We know that,
$f (x + Δ x) = f (x) + Δ x . f^{'} (x)$
$f (1.002) = f (1) + 0.002 \times 300 (1)^{299}$
$f (1.002) = 1 + 0.002 \times 300 = 1 + 0.6$
$f (1.002) = 1.6$

AP EAMCET-2002

Application of Derivatives

85394 Using differentiation, the approximate value of $\sin 46^{\circ}$ ,
Given that $1^{0} = {0.0175}^{\circ}$ , is

1 0.07194

2

\frac{1.0175}{\sqrt{2}}

3 0.7194

4

\frac{0.0175}{\sqrt{2}}

Explanation:

(B) : Given
$f (x) = \sin x$
$f^{'} (x) = \cos x$
$46^{\circ} = 45^{\circ} + 1^{\circ}$
$1^{\circ} = 0.0175$
$(x + Δ x) = 46^{\circ}$
$x = 45^{\circ}$
$Δ x = 1^{\circ} = 0.0175$
We know that,
$f (x + Δ x) = f^{'} (x) Δ x + f (x)$
$= \cos 45^{\circ} \times (0.0175) + \sin 45^{\circ}$
$= \frac{1}{\sqrt{2}} (0.0175) + \frac{1}{\sqrt{2}}$
$= \frac{1}{\sqrt{2}} [0.0175 + 1] = \frac{1.0175}{\sqrt{2}}$
So, the approximate value of $\sin 46^{\circ} = \frac{1.0175}{\sqrt{2}}$

[ -2019]

Application of Derivatives

85395 The slope of normal to the curve $x = \sqrt{t}$ and
$y = t - \frac{1}{\sqrt{t}} at t = 4 is$
$\begin{array}{llll} (a) \frac{17}{4} (b) \frac{- 17}{4} (c) \frac{4}{17} (d) \frac{- 4}{17} \end{array}$

1

\frac{17}{4}

2

\frac{- 17}{4}

3

\frac{4}{17}

4

\frac{- 4}{17}

Explanation:

(D) : Given,
$x = \sqrt{t}, y = t - \frac{1}{\sqrt{t}} at t = 4$
Now, $\frac{dx}{dt} = \frac{1}{2 \sqrt{t}}$
Similarly $\frac{dy}{dt} = 1 + \frac{1}{2} t^{- 3 / 2}$
$∴ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$= \frac{(1 + \frac{1}{2} t^{- 3 / 2})}{\frac{1}{2 \sqrt{t}}} = 2 \sqrt{t} (1 + \frac{1}{2} t^{- 3 / 2}) = 2 \sqrt{t} (1 + \frac{1}{2 t^{3 / 2}})$
$\frac{dy}{dx} = \frac{2 t^{3 / 2} + 1}{t}$
At, on putting $t = 4$
$\frac{dy}{dx} = \frac{2 (4)^{3 / 2} + 1}{4} = \frac{17}{4}$
$∴$ Slope of normal at $(t = 4) = - \frac{\frac{1}{d y}}{d x} = \frac{- 4}{17}$

MHT CET-2019

Application of Derivatives

85396 The equation of normal to the curve $y = \log_{e} x$ at the point $P (1, 0)$ is

1

x + y = 1

2

2 x + y = 2

3

x - y = 1

4

x - 2 y = 1

Explanation:

(A) : Given,
Curve, $y = \log_{e} x$
On differentiating w.r.t $x$ ,
$\frac{dy}{dx} = \frac{1}{x}$
$∴$ Slope of normal at $P = (1, 0)$
$= - \frac{1}{{(\frac{dy}{dx})}_{(1, 0)}} = \frac{- 1}{1} = - 1$
$∴$ Equation of normal at $P = (1, 0)$
$y - 0 = (- 1) (x - 1)$
$y = - x + 1$
$x + y = 1$

MHT CET-2019

Application of Derivatives

85393 The approximate value of $(1.002)^{300}$ using differentiation is

1 1.8

2 1.6

3 1.2

4 1.4

Explanation:

(B) : Given,
$x + Δ x = 1.002$
$x = 1$
$Δ x = 0.002$
Let,
$f (x) = x^{300}$
$f^{'} (x) = 300 x^{299}$
We know that,
$f (x + Δ x) = f (x) + Δ x . f^{'} (x)$
$f (1.002) = f (1) + 0.002 \times 300 (1)^{299}$
$f (1.002) = 1 + 0.002 \times 300 = 1 + 0.6$
$f (1.002) = 1.6$

AP EAMCET-2002

Application of Derivatives

85394 Using differentiation, the approximate value of $\sin 46^{\circ}$ ,
Given that $1^{0} = {0.0175}^{\circ}$ , is

1 0.07194

2

\frac{1.0175}{\sqrt{2}}

3 0.7194

4

\frac{0.0175}{\sqrt{2}}

Explanation:

(B) : Given
$f (x) = \sin x$
$f^{'} (x) = \cos x$
$46^{\circ} = 45^{\circ} + 1^{\circ}$
$1^{\circ} = 0.0175$
$(x + Δ x) = 46^{\circ}$
$x = 45^{\circ}$
$Δ x = 1^{\circ} = 0.0175$
We know that,
$f (x + Δ x) = f^{'} (x) Δ x + f (x)$
$= \cos 45^{\circ} \times (0.0175) + \sin 45^{\circ}$
$= \frac{1}{\sqrt{2}} (0.0175) + \frac{1}{\sqrt{2}}$
$= \frac{1}{\sqrt{2}} [0.0175 + 1] = \frac{1.0175}{\sqrt{2}}$
So, the approximate value of $\sin 46^{\circ} = \frac{1.0175}{\sqrt{2}}$

[ -2019]

Application of Derivatives

85395 The slope of normal to the curve $x = \sqrt{t}$ and
$y = t - \frac{1}{\sqrt{t}} at t = 4 is$
$\begin{array}{llll} (a) \frac{17}{4} (b) \frac{- 17}{4} (c) \frac{4}{17} (d) \frac{- 4}{17} \end{array}$

1

\frac{17}{4}

2

\frac{- 17}{4}

3

\frac{4}{17}

4

\frac{- 4}{17}

Explanation:

(D) : Given,
$x = \sqrt{t}, y = t - \frac{1}{\sqrt{t}} at t = 4$
Now, $\frac{dx}{dt} = \frac{1}{2 \sqrt{t}}$
Similarly $\frac{dy}{dt} = 1 + \frac{1}{2} t^{- 3 / 2}$
$∴ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$= \frac{(1 + \frac{1}{2} t^{- 3 / 2})}{\frac{1}{2 \sqrt{t}}} = 2 \sqrt{t} (1 + \frac{1}{2} t^{- 3 / 2}) = 2 \sqrt{t} (1 + \frac{1}{2 t^{3 / 2}})$
$\frac{dy}{dx} = \frac{2 t^{3 / 2} + 1}{t}$
At, on putting $t = 4$
$\frac{dy}{dx} = \frac{2 (4)^{3 / 2} + 1}{4} = \frac{17}{4}$
$∴$ Slope of normal at $(t = 4) = - \frac{\frac{1}{d y}}{d x} = \frac{- 4}{17}$

MHT CET-2019

Application of Derivatives

85396 The equation of normal to the curve $y = \log_{e} x$ at the point $P (1, 0)$ is

1

x + y = 1

2

2 x + y = 2

3

x - y = 1

4

x - 2 y = 1

Explanation:

(A) : Given,
Curve, $y = \log_{e} x$
On differentiating w.r.t $x$ ,
$\frac{dy}{dx} = \frac{1}{x}$
$∴$ Slope of normal at $P = (1, 0)$
$= - \frac{1}{{(\frac{dy}{dx})}_{(1, 0)}} = \frac{- 1}{1} = - 1$
$∴$ Equation of normal at $P = (1, 0)$
$y - 0 = (- 1) (x - 1)$
$y = - x + 1$
$x + y = 1$

MHT CET-2019

Question Bank Series

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