QBManager

Application of Derivatives

85388 If the tangent to the curve given by $x = t^{2} - 1$ and $y = t^{2} - t$ is parallel to $X$ -axis, then the value of $t$ is

1 0

2

\frac{1}{2}

3

\frac{- 1}{\sqrt{3}}

4

\frac{1}{\sqrt{3}}

Explanation:

(B) : The slope of the tangent to the curve
$x = t^{2} - 1$
$y = t^{2} - t$
On differentiating,
$\frac{dx}{dt} = 2 t$ And, $\frac{dy}{dt} = 2 t - 1$
For the tangent to be parallel to the x-axis, the slope must be 0 . Therefore, we need to solve the equation $\frac{dy}{dx} = 0$ , when $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$
So, we need to solve the equation $2 t - 1 = 0$ $2 t = 1$
So, $t = \frac{1}{2}$

MHT CET-2020

Application of Derivatives

85389 The approximate value of the function $f (x) = x^{3} + 5 x^{2} - 7 x + 10$ at $x = 1.1$ is

1 6.6

2 9.6

3 8.6

4 7.6

Explanation:

(B) : Given, $f (x) = x^{3} + 5 x^{2} - 7 x + 10$
$f^{'} (x) = 3 x^{2} + 10 x - 7$
Here, $x + Δ x = 1.1$
$x = 1, Δ x = 0.1$
We know that,
$f (x + Δ x) = f (x) + Δ x f^{'} (x)$
$f (1.1) = f (1) + 0.1 f^{'} (1)$
$f (1.1) = (1)^{3} + 5 (1)^{2} - 7 (1) + 10 + 0.1 [3 (1)^{2} + 10 (1) - 7]$
$= 1 + 5 - 7 + 10 + 0.1 \times (3 + 10 - 7)$
$= 9 + 0.1 \times 6 = 9 + 0.6 = 9.6$
So, The approximate value of the function $f (1.1) = 9.6$

MHT CET-2020

Application of Derivatives

85390 The equation of normal to the curve $2 x^{2} + 3 y^{2} = 5$ at $P (1, 1)$ is

1

3 x + 2 y + 1 = 0

2

3 x + 2 y - 5 = 0

3

3 x - 2 y + 1 = 0

4

3 x - 2 y - 1 = 0

Explanation:

(D) : Given curve,
$2 x^{2} + 3 y^{2} = 5$
Differentiating w.r.t. $x$ we get -
$4 x + 6 y \cdot \frac{d y}{d x} = 0 \Rightarrow 6 y \frac{d y}{d x} = - 4 x$
$\frac{d y}{d x} = \frac{- 4 x}{6 y}$
$∴$ Slope of the tangent at $(1, 1)$ is
${(\frac{dy}{dx})}_{(1, 1)} = \frac{- 4 (1)}{6 (1)} \Rightarrow {(\frac{dy}{dx})}_{(1, 1)} = \frac{- 4}{6} = \frac{- 2}{3}$
Equation of tangent at $(a, b)$ is
$y - b = {(\frac{d y}{d x})}_{(a, b)} (x - a)$
Tangent at $(1, 1)$ ,
$(y - 1) = \frac{- 2}{3} (x - 1)$
$3 y - 3 = - 2 x + 2$
$3 y + 2 x - 5 = 0$
$2 x + 3 y - 5 = 0$
Slope of the normal at $(1, 1)$ is $\frac{- 1}{{(\frac{dy}{dx})}_{(1, 1)}} = \frac{3}{2}$
Equation of normal at $(a, b)$ is
$y - b = \frac{- 1}{{(\frac{d y}{d x})}_{(a, b)}} (x - a)$
Equation of the normal at $(1, 1)$ is,
$(y - 1) = \frac{3}{2} (x - 1)$
$2 y - 2 = 3 x - 3$
$3 x - 2 y - 1 = 0$

MHT CET-2020

Application of Derivatives

85391 The approximate value of $(66)^{\frac{1}{3}}$ is

1 4.0433

2 4.0416

3 4.0481

4 4.0447

Explanation:

(B) : Consider the function,
$y = f (x)$
$y = (x)^{\frac{1}{3}}$
Then, $x + Δ x = 66$
$x = 64$
$Δ x = 2$
$y = (64)^{\frac{1}{3}}$
$y = 4$
Let,
$dx = Δ x = 2$
Now, $y = (x)^{\frac{1}{3}}$
On differentiating both sides
$\frac{d y}{d x} = \frac{1}{3 (x)^{2 / 3}}$
$\frac{d y}{d x} = \frac{1}{3 (64)^{2 / 3}} [x = 64)$
$\frac{d y}{d x} = \frac{1}{3 (4)^{3 \times \frac{2}{3}}} \Rightarrow \frac{d y}{d x} = \frac{1}{48}$
$Δ y = d y = \frac{d y}{d x} \cdot x$
$Δ y = \frac{1}{48} \times 2 \Rightarrow Δ y = 0.0416$
So, $(66)^{\frac{1}{3}} = y + Δ y$
$[y = 4]$
$= 4 + 0.0416 = 4.0416$

MHT CET-2020

Application of Derivatives

85392 The equation of the normal to the curve $2 x^{2} + y^{2} = 12$ at the point $(2, 2)$ is

1

x - 2 y + 2 = 0

2

2 x - y + 6 = 0

3

x + 2 y + 2 = 0

4

2 x + y - 6 = 0

Explanation:

(A) : Given,
$2 x^{2} + y^{2} = 12$
On differentiating both sides
$4 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{y}$
Slope of tangent at $(2, 2)$ is
${(\frac{dy}{dx})}_{(2, 2)} = \frac{- 2 (2)}{2} \Rightarrow {(\frac{dy}{dx})}_{(2, 2)} = - 2$
$∴$ Slope of normal $= \frac{- dx}{dy} = \frac{1}{2}$
Equation of normal at $(2, 2)$ is,
$(y - b) = \frac{d y}{d x} (x - a) \Rightarrow y - 2 = \frac{1}{2} (x - 2)$
$2 (y - 2) = (x - 2)$
$2 y - 4 = x - 2$
$x - 2 y + 2 = 0$

MHT CET-2020

Application of Derivatives

85388 If the tangent to the curve given by $x = t^{2} - 1$ and $y = t^{2} - t$ is parallel to $X$ -axis, then the value of $t$ is

1 0

2

\frac{1}{2}

3

\frac{- 1}{\sqrt{3}}

4

\frac{1}{\sqrt{3}}

Explanation:

(B) : The slope of the tangent to the curve
$x = t^{2} - 1$
$y = t^{2} - t$
On differentiating,
$\frac{dx}{dt} = 2 t$ And, $\frac{dy}{dt} = 2 t - 1$
For the tangent to be parallel to the x-axis, the slope must be 0 . Therefore, we need to solve the equation $\frac{dy}{dx} = 0$ , when $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$
So, we need to solve the equation $2 t - 1 = 0$ $2 t = 1$
So, $t = \frac{1}{2}$

MHT CET-2020

Application of Derivatives

85389 The approximate value of the function $f (x) = x^{3} + 5 x^{2} - 7 x + 10$ at $x = 1.1$ is

1 6.6

2 9.6

3 8.6

4 7.6

Explanation:

(B) : Given, $f (x) = x^{3} + 5 x^{2} - 7 x + 10$
$f^{'} (x) = 3 x^{2} + 10 x - 7$
Here, $x + Δ x = 1.1$
$x = 1, Δ x = 0.1$
We know that,
$f (x + Δ x) = f (x) + Δ x f^{'} (x)$
$f (1.1) = f (1) + 0.1 f^{'} (1)$
$f (1.1) = (1)^{3} + 5 (1)^{2} - 7 (1) + 10 + 0.1 [3 (1)^{2} + 10 (1) - 7]$
$= 1 + 5 - 7 + 10 + 0.1 \times (3 + 10 - 7)$
$= 9 + 0.1 \times 6 = 9 + 0.6 = 9.6$
So, The approximate value of the function $f (1.1) = 9.6$

MHT CET-2020

Application of Derivatives

85390 The equation of normal to the curve $2 x^{2} + 3 y^{2} = 5$ at $P (1, 1)$ is

1

3 x + 2 y + 1 = 0

2

3 x + 2 y - 5 = 0

3

3 x - 2 y + 1 = 0

4

3 x - 2 y - 1 = 0

Explanation:

(D) : Given curve,
$2 x^{2} + 3 y^{2} = 5$
Differentiating w.r.t. $x$ we get -
$4 x + 6 y \cdot \frac{d y}{d x} = 0 \Rightarrow 6 y \frac{d y}{d x} = - 4 x$
$\frac{d y}{d x} = \frac{- 4 x}{6 y}$
$∴$ Slope of the tangent at $(1, 1)$ is
${(\frac{dy}{dx})}_{(1, 1)} = \frac{- 4 (1)}{6 (1)} \Rightarrow {(\frac{dy}{dx})}_{(1, 1)} = \frac{- 4}{6} = \frac{- 2}{3}$
Equation of tangent at $(a, b)$ is
$y - b = {(\frac{d y}{d x})}_{(a, b)} (x - a)$
Tangent at $(1, 1)$ ,
$(y - 1) = \frac{- 2}{3} (x - 1)$
$3 y - 3 = - 2 x + 2$
$3 y + 2 x - 5 = 0$
$2 x + 3 y - 5 = 0$
Slope of the normal at $(1, 1)$ is $\frac{- 1}{{(\frac{dy}{dx})}_{(1, 1)}} = \frac{3}{2}$
Equation of normal at $(a, b)$ is
$y - b = \frac{- 1}{{(\frac{d y}{d x})}_{(a, b)}} (x - a)$
Equation of the normal at $(1, 1)$ is,
$(y - 1) = \frac{3}{2} (x - 1)$
$2 y - 2 = 3 x - 3$
$3 x - 2 y - 1 = 0$

MHT CET-2020

Application of Derivatives

85391 The approximate value of $(66)^{\frac{1}{3}}$ is

1 4.0433

2 4.0416

3 4.0481

4 4.0447

Explanation:

(B) : Consider the function,
$y = f (x)$
$y = (x)^{\frac{1}{3}}$
Then, $x + Δ x = 66$
$x = 64$
$Δ x = 2$
$y = (64)^{\frac{1}{3}}$
$y = 4$
Let,
$dx = Δ x = 2$
Now, $y = (x)^{\frac{1}{3}}$
On differentiating both sides
$\frac{d y}{d x} = \frac{1}{3 (x)^{2 / 3}}$
$\frac{d y}{d x} = \frac{1}{3 (64)^{2 / 3}} [x = 64)$
$\frac{d y}{d x} = \frac{1}{3 (4)^{3 \times \frac{2}{3}}} \Rightarrow \frac{d y}{d x} = \frac{1}{48}$
$Δ y = d y = \frac{d y}{d x} \cdot x$
$Δ y = \frac{1}{48} \times 2 \Rightarrow Δ y = 0.0416$
So, $(66)^{\frac{1}{3}} = y + Δ y$
$[y = 4]$
$= 4 + 0.0416 = 4.0416$

MHT CET-2020

Application of Derivatives

85392 The equation of the normal to the curve $2 x^{2} + y^{2} = 12$ at the point $(2, 2)$ is

1

x - 2 y + 2 = 0

2

2 x - y + 6 = 0

3

x + 2 y + 2 = 0

4

2 x + y - 6 = 0

Explanation:

(A) : Given,
$2 x^{2} + y^{2} = 12$
On differentiating both sides
$4 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{y}$
Slope of tangent at $(2, 2)$ is
${(\frac{dy}{dx})}_{(2, 2)} = \frac{- 2 (2)}{2} \Rightarrow {(\frac{dy}{dx})}_{(2, 2)} = - 2$
$∴$ Slope of normal $= \frac{- dx}{dy} = \frac{1}{2}$
Equation of normal at $(2, 2)$ is,
$(y - b) = \frac{d y}{d x} (x - a) \Rightarrow y - 2 = \frac{1}{2} (x - 2)$
$2 (y - 2) = (x - 2)$
$2 y - 4 = x - 2$
$x - 2 y + 2 = 0$

MHT CET-2020

Application of Derivatives

85388 If the tangent to the curve given by $x = t^{2} - 1$ and $y = t^{2} - t$ is parallel to $X$ -axis, then the value of $t$ is

1 0

2

\frac{1}{2}

3

\frac{- 1}{\sqrt{3}}

4

\frac{1}{\sqrt{3}}

Explanation:

(B) : The slope of the tangent to the curve
$x = t^{2} - 1$
$y = t^{2} - t$
On differentiating,
$\frac{dx}{dt} = 2 t$ And, $\frac{dy}{dt} = 2 t - 1$
For the tangent to be parallel to the x-axis, the slope must be 0 . Therefore, we need to solve the equation $\frac{dy}{dx} = 0$ , when $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$
So, we need to solve the equation $2 t - 1 = 0$ $2 t = 1$
So, $t = \frac{1}{2}$

MHT CET-2020

Application of Derivatives

85389 The approximate value of the function $f (x) = x^{3} + 5 x^{2} - 7 x + 10$ at $x = 1.1$ is

1 6.6

2 9.6

3 8.6

4 7.6

Explanation:

(B) : Given, $f (x) = x^{3} + 5 x^{2} - 7 x + 10$
$f^{'} (x) = 3 x^{2} + 10 x - 7$
Here, $x + Δ x = 1.1$
$x = 1, Δ x = 0.1$
We know that,
$f (x + Δ x) = f (x) + Δ x f^{'} (x)$
$f (1.1) = f (1) + 0.1 f^{'} (1)$
$f (1.1) = (1)^{3} + 5 (1)^{2} - 7 (1) + 10 + 0.1 [3 (1)^{2} + 10 (1) - 7]$
$= 1 + 5 - 7 + 10 + 0.1 \times (3 + 10 - 7)$
$= 9 + 0.1 \times 6 = 9 + 0.6 = 9.6$
So, The approximate value of the function $f (1.1) = 9.6$

MHT CET-2020

Application of Derivatives

85390 The equation of normal to the curve $2 x^{2} + 3 y^{2} = 5$ at $P (1, 1)$ is

1

3 x + 2 y + 1 = 0

2

3 x + 2 y - 5 = 0

3

3 x - 2 y + 1 = 0

4

3 x - 2 y - 1 = 0

Explanation:

(D) : Given curve,
$2 x^{2} + 3 y^{2} = 5$
Differentiating w.r.t. $x$ we get -
$4 x + 6 y \cdot \frac{d y}{d x} = 0 \Rightarrow 6 y \frac{d y}{d x} = - 4 x$
$\frac{d y}{d x} = \frac{- 4 x}{6 y}$
$∴$ Slope of the tangent at $(1, 1)$ is
${(\frac{dy}{dx})}_{(1, 1)} = \frac{- 4 (1)}{6 (1)} \Rightarrow {(\frac{dy}{dx})}_{(1, 1)} = \frac{- 4}{6} = \frac{- 2}{3}$
Equation of tangent at $(a, b)$ is
$y - b = {(\frac{d y}{d x})}_{(a, b)} (x - a)$
Tangent at $(1, 1)$ ,
$(y - 1) = \frac{- 2}{3} (x - 1)$
$3 y - 3 = - 2 x + 2$
$3 y + 2 x - 5 = 0$
$2 x + 3 y - 5 = 0$
Slope of the normal at $(1, 1)$ is $\frac{- 1}{{(\frac{dy}{dx})}_{(1, 1)}} = \frac{3}{2}$
Equation of normal at $(a, b)$ is
$y - b = \frac{- 1}{{(\frac{d y}{d x})}_{(a, b)}} (x - a)$
Equation of the normal at $(1, 1)$ is,
$(y - 1) = \frac{3}{2} (x - 1)$
$2 y - 2 = 3 x - 3$
$3 x - 2 y - 1 = 0$

MHT CET-2020

Application of Derivatives

85391 The approximate value of $(66)^{\frac{1}{3}}$ is

1 4.0433

2 4.0416

3 4.0481

4 4.0447

Explanation:

(B) : Consider the function,
$y = f (x)$
$y = (x)^{\frac{1}{3}}$
Then, $x + Δ x = 66$
$x = 64$
$Δ x = 2$
$y = (64)^{\frac{1}{3}}$
$y = 4$
Let,
$dx = Δ x = 2$
Now, $y = (x)^{\frac{1}{3}}$
On differentiating both sides
$\frac{d y}{d x} = \frac{1}{3 (x)^{2 / 3}}$
$\frac{d y}{d x} = \frac{1}{3 (64)^{2 / 3}} [x = 64)$
$\frac{d y}{d x} = \frac{1}{3 (4)^{3 \times \frac{2}{3}}} \Rightarrow \frac{d y}{d x} = \frac{1}{48}$
$Δ y = d y = \frac{d y}{d x} \cdot x$
$Δ y = \frac{1}{48} \times 2 \Rightarrow Δ y = 0.0416$
So, $(66)^{\frac{1}{3}} = y + Δ y$
$[y = 4]$
$= 4 + 0.0416 = 4.0416$

MHT CET-2020

Application of Derivatives

85392 The equation of the normal to the curve $2 x^{2} + y^{2} = 12$ at the point $(2, 2)$ is

1

x - 2 y + 2 = 0

2

2 x - y + 6 = 0

3

x + 2 y + 2 = 0

4

2 x + y - 6 = 0

Explanation:

(A) : Given,
$2 x^{2} + y^{2} = 12$
On differentiating both sides
$4 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{y}$
Slope of tangent at $(2, 2)$ is
${(\frac{dy}{dx})}_{(2, 2)} = \frac{- 2 (2)}{2} \Rightarrow {(\frac{dy}{dx})}_{(2, 2)} = - 2$
$∴$ Slope of normal $= \frac{- dx}{dy} = \frac{1}{2}$
Equation of normal at $(2, 2)$ is,
$(y - b) = \frac{d y}{d x} (x - a) \Rightarrow y - 2 = \frac{1}{2} (x - 2)$
$2 (y - 2) = (x - 2)$
$2 y - 4 = x - 2$
$x - 2 y + 2 = 0$

MHT CET-2020

Application of Derivatives

85388 If the tangent to the curve given by $x = t^{2} - 1$ and $y = t^{2} - t$ is parallel to $X$ -axis, then the value of $t$ is

1 0

2

\frac{1}{2}

3

\frac{- 1}{\sqrt{3}}

4

\frac{1}{\sqrt{3}}

Explanation:

(B) : The slope of the tangent to the curve
$x = t^{2} - 1$
$y = t^{2} - t$
On differentiating,
$\frac{dx}{dt} = 2 t$ And, $\frac{dy}{dt} = 2 t - 1$
For the tangent to be parallel to the x-axis, the slope must be 0 . Therefore, we need to solve the equation $\frac{dy}{dx} = 0$ , when $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$
So, we need to solve the equation $2 t - 1 = 0$ $2 t = 1$
So, $t = \frac{1}{2}$

MHT CET-2020

Application of Derivatives

85389 The approximate value of the function $f (x) = x^{3} + 5 x^{2} - 7 x + 10$ at $x = 1.1$ is

1 6.6

2 9.6

3 8.6

4 7.6

Explanation:

(B) : Given, $f (x) = x^{3} + 5 x^{2} - 7 x + 10$
$f^{'} (x) = 3 x^{2} + 10 x - 7$
Here, $x + Δ x = 1.1$
$x = 1, Δ x = 0.1$
We know that,
$f (x + Δ x) = f (x) + Δ x f^{'} (x)$
$f (1.1) = f (1) + 0.1 f^{'} (1)$
$f (1.1) = (1)^{3} + 5 (1)^{2} - 7 (1) + 10 + 0.1 [3 (1)^{2} + 10 (1) - 7]$
$= 1 + 5 - 7 + 10 + 0.1 \times (3 + 10 - 7)$
$= 9 + 0.1 \times 6 = 9 + 0.6 = 9.6$
So, The approximate value of the function $f (1.1) = 9.6$

MHT CET-2020

Application of Derivatives

85390 The equation of normal to the curve $2 x^{2} + 3 y^{2} = 5$ at $P (1, 1)$ is

1

3 x + 2 y + 1 = 0

2

3 x + 2 y - 5 = 0

3

3 x - 2 y + 1 = 0

4

3 x - 2 y - 1 = 0

Explanation:

(D) : Given curve,
$2 x^{2} + 3 y^{2} = 5$
Differentiating w.r.t. $x$ we get -
$4 x + 6 y \cdot \frac{d y}{d x} = 0 \Rightarrow 6 y \frac{d y}{d x} = - 4 x$
$\frac{d y}{d x} = \frac{- 4 x}{6 y}$
$∴$ Slope of the tangent at $(1, 1)$ is
${(\frac{dy}{dx})}_{(1, 1)} = \frac{- 4 (1)}{6 (1)} \Rightarrow {(\frac{dy}{dx})}_{(1, 1)} = \frac{- 4}{6} = \frac{- 2}{3}$
Equation of tangent at $(a, b)$ is
$y - b = {(\frac{d y}{d x})}_{(a, b)} (x - a)$
Tangent at $(1, 1)$ ,
$(y - 1) = \frac{- 2}{3} (x - 1)$
$3 y - 3 = - 2 x + 2$
$3 y + 2 x - 5 = 0$
$2 x + 3 y - 5 = 0$
Slope of the normal at $(1, 1)$ is $\frac{- 1}{{(\frac{dy}{dx})}_{(1, 1)}} = \frac{3}{2}$
Equation of normal at $(a, b)$ is
$y - b = \frac{- 1}{{(\frac{d y}{d x})}_{(a, b)}} (x - a)$
Equation of the normal at $(1, 1)$ is,
$(y - 1) = \frac{3}{2} (x - 1)$
$2 y - 2 = 3 x - 3$
$3 x - 2 y - 1 = 0$

MHT CET-2020

Application of Derivatives

85391 The approximate value of $(66)^{\frac{1}{3}}$ is

1 4.0433

2 4.0416

3 4.0481

4 4.0447

Explanation:

(B) : Consider the function,
$y = f (x)$
$y = (x)^{\frac{1}{3}}$
Then, $x + Δ x = 66$
$x = 64$
$Δ x = 2$
$y = (64)^{\frac{1}{3}}$
$y = 4$
Let,
$dx = Δ x = 2$
Now, $y = (x)^{\frac{1}{3}}$
On differentiating both sides
$\frac{d y}{d x} = \frac{1}{3 (x)^{2 / 3}}$
$\frac{d y}{d x} = \frac{1}{3 (64)^{2 / 3}} [x = 64)$
$\frac{d y}{d x} = \frac{1}{3 (4)^{3 \times \frac{2}{3}}} \Rightarrow \frac{d y}{d x} = \frac{1}{48}$
$Δ y = d y = \frac{d y}{d x} \cdot x$
$Δ y = \frac{1}{48} \times 2 \Rightarrow Δ y = 0.0416$
So, $(66)^{\frac{1}{3}} = y + Δ y$
$[y = 4]$
$= 4 + 0.0416 = 4.0416$

MHT CET-2020

Application of Derivatives

85392 The equation of the normal to the curve $2 x^{2} + y^{2} = 12$ at the point $(2, 2)$ is

1

x - 2 y + 2 = 0

2

2 x - y + 6 = 0

3

x + 2 y + 2 = 0

4

2 x + y - 6 = 0

Explanation:

(A) : Given,
$2 x^{2} + y^{2} = 12$
On differentiating both sides
$4 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{y}$
Slope of tangent at $(2, 2)$ is
${(\frac{dy}{dx})}_{(2, 2)} = \frac{- 2 (2)}{2} \Rightarrow {(\frac{dy}{dx})}_{(2, 2)} = - 2$
$∴$ Slope of normal $= \frac{- dx}{dy} = \frac{1}{2}$
Equation of normal at $(2, 2)$ is,
$(y - b) = \frac{d y}{d x} (x - a) \Rightarrow y - 2 = \frac{1}{2} (x - 2)$
$2 (y - 2) = (x - 2)$
$2 y - 4 = x - 2$
$x - 2 y + 2 = 0$

MHT CET-2020

Application of Derivatives

85388 If the tangent to the curve given by $x = t^{2} - 1$ and $y = t^{2} - t$ is parallel to $X$ -axis, then the value of $t$ is

1 0

2

\frac{1}{2}

3

\frac{- 1}{\sqrt{3}}

4

\frac{1}{\sqrt{3}}

Explanation:

(B) : The slope of the tangent to the curve
$x = t^{2} - 1$
$y = t^{2} - t$
On differentiating,
$\frac{dx}{dt} = 2 t$ And, $\frac{dy}{dt} = 2 t - 1$
For the tangent to be parallel to the x-axis, the slope must be 0 . Therefore, we need to solve the equation $\frac{dy}{dx} = 0$ , when $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$
So, we need to solve the equation $2 t - 1 = 0$ $2 t = 1$
So, $t = \frac{1}{2}$

MHT CET-2020

Application of Derivatives

85389 The approximate value of the function $f (x) = x^{3} + 5 x^{2} - 7 x + 10$ at $x = 1.1$ is

1 6.6

2 9.6

3 8.6

4 7.6

Explanation:

(B) : Given, $f (x) = x^{3} + 5 x^{2} - 7 x + 10$
$f^{'} (x) = 3 x^{2} + 10 x - 7$
Here, $x + Δ x = 1.1$
$x = 1, Δ x = 0.1$
We know that,
$f (x + Δ x) = f (x) + Δ x f^{'} (x)$
$f (1.1) = f (1) + 0.1 f^{'} (1)$
$f (1.1) = (1)^{3} + 5 (1)^{2} - 7 (1) + 10 + 0.1 [3 (1)^{2} + 10 (1) - 7]$
$= 1 + 5 - 7 + 10 + 0.1 \times (3 + 10 - 7)$
$= 9 + 0.1 \times 6 = 9 + 0.6 = 9.6$
So, The approximate value of the function $f (1.1) = 9.6$

MHT CET-2020

Application of Derivatives

85390 The equation of normal to the curve $2 x^{2} + 3 y^{2} = 5$ at $P (1, 1)$ is

1

3 x + 2 y + 1 = 0

2

3 x + 2 y - 5 = 0

3

3 x - 2 y + 1 = 0

4

3 x - 2 y - 1 = 0

Explanation:

(D) : Given curve,
$2 x^{2} + 3 y^{2} = 5$
Differentiating w.r.t. $x$ we get -
$4 x + 6 y \cdot \frac{d y}{d x} = 0 \Rightarrow 6 y \frac{d y}{d x} = - 4 x$
$\frac{d y}{d x} = \frac{- 4 x}{6 y}$
$∴$ Slope of the tangent at $(1, 1)$ is
${(\frac{dy}{dx})}_{(1, 1)} = \frac{- 4 (1)}{6 (1)} \Rightarrow {(\frac{dy}{dx})}_{(1, 1)} = \frac{- 4}{6} = \frac{- 2}{3}$
Equation of tangent at $(a, b)$ is
$y - b = {(\frac{d y}{d x})}_{(a, b)} (x - a)$
Tangent at $(1, 1)$ ,
$(y - 1) = \frac{- 2}{3} (x - 1)$
$3 y - 3 = - 2 x + 2$
$3 y + 2 x - 5 = 0$
$2 x + 3 y - 5 = 0$
Slope of the normal at $(1, 1)$ is $\frac{- 1}{{(\frac{dy}{dx})}_{(1, 1)}} = \frac{3}{2}$
Equation of normal at $(a, b)$ is
$y - b = \frac{- 1}{{(\frac{d y}{d x})}_{(a, b)}} (x - a)$
Equation of the normal at $(1, 1)$ is,
$(y - 1) = \frac{3}{2} (x - 1)$
$2 y - 2 = 3 x - 3$
$3 x - 2 y - 1 = 0$

MHT CET-2020

Application of Derivatives

85391 The approximate value of $(66)^{\frac{1}{3}}$ is

1 4.0433

2 4.0416

3 4.0481

4 4.0447

Explanation:

(B) : Consider the function,
$y = f (x)$
$y = (x)^{\frac{1}{3}}$
Then, $x + Δ x = 66$
$x = 64$
$Δ x = 2$
$y = (64)^{\frac{1}{3}}$
$y = 4$
Let,
$dx = Δ x = 2$
Now, $y = (x)^{\frac{1}{3}}$
On differentiating both sides
$\frac{d y}{d x} = \frac{1}{3 (x)^{2 / 3}}$
$\frac{d y}{d x} = \frac{1}{3 (64)^{2 / 3}} [x = 64)$
$\frac{d y}{d x} = \frac{1}{3 (4)^{3 \times \frac{2}{3}}} \Rightarrow \frac{d y}{d x} = \frac{1}{48}$
$Δ y = d y = \frac{d y}{d x} \cdot x$
$Δ y = \frac{1}{48} \times 2 \Rightarrow Δ y = 0.0416$
So, $(66)^{\frac{1}{3}} = y + Δ y$
$[y = 4]$
$= 4 + 0.0416 = 4.0416$

MHT CET-2020

Application of Derivatives

85392 The equation of the normal to the curve $2 x^{2} + y^{2} = 12$ at the point $(2, 2)$ is

1

x - 2 y + 2 = 0

2

2 x - y + 6 = 0

3

x + 2 y + 2 = 0

4

2 x + y - 6 = 0

Explanation:

(A) : Given,
$2 x^{2} + y^{2} = 12$
On differentiating both sides
$4 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{y}$
Slope of tangent at $(2, 2)$ is
${(\frac{dy}{dx})}_{(2, 2)} = \frac{- 2 (2)}{2} \Rightarrow {(\frac{dy}{dx})}_{(2, 2)} = - 2$
$∴$ Slope of normal $= \frac{- dx}{dy} = \frac{1}{2}$
Equation of normal at $(2, 2)$ is,
$(y - b) = \frac{d y}{d x} (x - a) \Rightarrow y - 2 = \frac{1}{2} (x - 2)$
$2 (y - 2) = (x - 2)$
$2 y - 4 = x - 2$
$x - 2 y + 2 = 0$

MHT CET-2020