CUET Mathematics - Calculus - Application Of Derivatives | Free Questions - Page 11

Q1:

2022: 17 Aug Shift 2

Calculus

>

Application of Derivatives

Easy

Core

The slope of the tangent to the curve $y = 3x^2 + 2kx - 5$ at $x=1$ is $9$ . The value of $k$ is :

$-\frac{3}{2}$

$\frac{1}{2}$

$\frac{5}{2}$

$\frac{3}{2}$

Correct Answer

Option 4

Correct Answer

Explanation

Q2:

2022: 17 Aug Shift 2

Calculus

>

Application of Derivatives

Easy

Core

An architect designs a garden in a circular shape with centre E and radius 50 m. There is a flower bed ABCD of rectangular shape inside the garden as shown in the figure. Suppose length and width of the flower bed are $2x$ and $2y$ meters respectively.

Based on above information answer the following question :

$x$ and $y$ will satisfy :

$x^2 + y^2 = 50$

$x^2 + y^2 = 100$

$x^2 + y^2 = 2500$

$x + y = 50$

Correct Answer

Option 3

Correct Answer

Explanation

Q3:

2022: 17 Aug Shift 2

Calculus

>

Application of Derivatives

Easy

Core

An architect designs a garden in a circular shape with centre E and radius 50 m. There is a flower bed ABCD of rectangular shape inside the garden as shown in the figure. Suppose length and width of the flower bed are $2x$ and $2y$ meters respectively.

Based on above information answer the following question :

The area of the flower bed $(A(x))$ is given by :

$x\sqrt{50 - x^2}$

$4x\sqrt{2500 - x^2}$

$4x\sqrt{2500 + x^2}$

$x\sqrt{50 + x^2}$

Correct Answer

Option 2

Correct Answer

Explanation

Q4:

2022: 17 Aug Shift 2

Calculus

>

Application of Derivatives

Medium

Core

An architect designs a garden in a circular shape with centre E and radius 50 m. There is a flower bed ABCD of rectangular shape inside the garden as shown in the figure. Suppose length and width of the flower bed are $2x$ and $2y$ meters respectively.

Based on above information answer the following question :

$\frac{dA(x)}{dx} =$

$8\frac{[1250 - x^2]}{\sqrt{2500 - x^2}}$

$4\left[\frac{2500 - x^2}{\sqrt{2500 - x^2}}\right]$

$8\frac{[2500 + x^2]}{\sqrt{2500 - x^2}}$

$4\frac{[1250 + x^2]}{\sqrt{2500 - x^2}}$

Correct Answer

Option 1

Correct Answer

Explanation

Q5:

2022: 17 Aug Shift 2

Calculus

>

Application of Derivatives

Medium

Core

An architect designs a garden in a circular shape with centre E and radius 50 m. There is a flower bed ABCD of rectangular shape inside the garden as shown in the figure. Suppose length and width of the flower bed are $2x$ and $2y$ meters respectively.

Based on above information answer the following question :

The maximum area (in $m^2$ ) of the flower bed is :

1250

2500

5000

625

Correct Answer

Option 3

Correct Answer

Explanation

Q6:

2022: 17 Aug Shift 2

Calculus

>

Application of Derivatives

Medium

Core

An architect designs a garden in a circular shape with centre E and radius 50 m. There is a flower bed ABCD of rectangular shape inside the garden as shown in the figure. Suppose length and width of the flower bed are $2x$ and $2y$ meters respectively.

Based on above information answer the following question :

If area of the flower bed is maximum, then area (in $m^2$ ) of the garden, which is outside the flower bed is :

$1250(\pi - 2)\,m^2$

$1250(\pi + 2)\,m^2$

$2500(\pi - 2)\,m^2$

$2500(\pi + 2)\,m^2$

Correct Answer

Option 3

Correct Answer

Explanation

Q7:

2022: 6 Aug Shift 2

Calculus

>

Application of Derivatives

Medium

Core

The function $f(x) = 6(2x^4 - x^2)$ is strictly increasing in

$(-\infty, -\frac{1}{2}) \cup (\frac{1}{2}, \infty)$

$(-\frac{1}{2}, 0) \cup (\frac{1}{2}, \infty)$

$(-\frac{1}{2}, \frac{1}{2})$

$(-\infty, -\frac{1}{2}) \cup (0, \frac{1}{2})$

Correct Answer

Option 2

Correct Answer

Explanation

Q8:

2022: 6 Aug Shift 2

Calculus

>

Application of Derivatives

Easy

Core

The maximum slope of the tangents to the curve $y(x) = -x^3 + 3x^2 + 9x - 30$ is

$0$

$12$

$16$

$18$

Correct Answer

Option 2

Correct Answer

Explanation

Q9:

2022: 6 Aug Shift 2

Calculus

>

Application of Derivatives

Hard

Core

The line $ax + by = 7$ is a tangent to the curve $y = \frac{x-7}{(x-2)(x-3)}$ at the point where it cuts the x-axis

A. The y-intercept of the line is $-0.7$

B. $b = -7$

C. $a = 1$

D. the line passes through the point $(-13, -1)$

E. $b = -20$

Choose the correct answer from the options given below:

A, B, D Only

C, D, E Only

A, D, E Only

B, C, D Only

Correct Answer

Option 2

Correct Answer

Explanation

Q10:

2022: 6 Aug Shift 2

Calculus

>

Application of Derivatives

Easy

Core

The slope of normal to the curve $y = 3x^2 + 3 \sin 3x$ , at $x = 0$ is:

$9$

$\frac{1}{9}$

$-\frac{1}{9}$

$-9$

Correct Answer

Option 3

Correct Answer

Explanation

Q11:

2022: 4 Aug Shift 1

Calculus

>

Application of Derivatives

Easy

Core

The interval in which the function $f(x) = 2x^3 + 3x^2 - 12x + 1$ is strictly increasing is -

$[-2, 1]$

$(-\infty, 1]$

$(-\infty, -2) \cup (1, \infty)$

$(-\infty, -1] \cup [2, \infty)$

Correct Answer

Option 3

Correct Answer

Explanation

Q12:

2022: 4 Aug Shift 1

Calculus

>

Application of Derivatives

Easy

Core

If x is real, then minimum value of $x^2 - 8x + 17$ is :

-1

0

1

2

Correct Answer

Option 3

Correct Answer

Explanation

Q13:

2022: 4 Aug Shift 1

Calculus

>

Application of Derivatives

Easy

Core

The tangent to the curve $y^2 + 2x - 5 = 0$ at the point (h, k) is parallel to the line $x + 2y = 4$ , then the value of 'k' is:

$-\frac{1}{2}$

-2

2

$\frac{1}{2}$

Correct Answer

Option 3

Correct Answer

Explanation

Q14:

2022: 4 Aug Shift 1

Calculus

>

Application of Derivatives

Easy

Core

The slope of normal to the curve $y = kx^2 - 3x + 2$ at $x = \frac{1}{2}$ is 5. The value of 'k' is

8

$\frac{16}{5}$

$\frac{14}{5}$

$-\frac{1}{8}$

Correct Answer

Option 3

Correct Answer

Explanation

Q15:

2022: 10 Aug Shift 1

Calculus

>

Application of Derivatives

Easy

Core

The point(s) on the curve $\frac{x^2}{9} + \frac{y^2}{64} = 1$ , at which the tangents are parallel to x-axis are

$(0, \pm 3)$

$(\pm 8, 0)$

$(0, \pm 8)$

$(\pm 3, 0)$

Correct Answer

Option 3

Correct Answer

Explanation

Q16:

2022: 10 Aug Shift 1

Calculus

>

Application of Derivatives

Medium

Core

An energy DRONE is flying along the curve $y = x^2 + 7$ . A soldier is placed at $(3, 7)$ . The nearest distance of the DRONE from soldier's position is

2

3

$\sqrt{5}$

$\sqrt{7}$

Correct Answer

Option 3

Correct Answer

Explanation

Q17:

2022: 10 Aug Shift 1

Calculus

>

Application of Derivatives

Medium

Core

The maximum value of $x^{-x}$ is

$e^{e-1}$

$e^{\frac{1}{e}}$

$e^{\frac{-1}{e}}$

$e^{\frac{-4}{e}}$

Correct Answer

Option 2

Correct Answer

Explanation

Q18:

2022: 10 Aug Shift 1

Calculus

>

Application of Derivatives

Easy

Core

The absolute maximum value of $y = x^3 - 3x + 2$ , $0 \leq x \leq 2$ , is

4

6

2

0

Correct Answer

Option 1

Correct Answer

Explanation

Q19:

2022: 10 Aug Shift 1

Calculus

>

Application of Derivatives

Easy

Core

The interval in which the function, $f(x) = 7 - 4x - x^2$ is strictly increasing is

$(-\infty, \infty)$

$(-2, \infty)$

$(-\infty, -2)$

$(-\infty, -2]$

Correct Answer

Option 3

Correct Answer

Explanation

Q20:

2022: 30 Aug Shift 1

Calculus

>

Application of Derivatives

Medium

Core

Consider the function $f(x) = x^{\frac{1}{x}}$ . Its

minimum value is $e^{\frac{1}{e}}$

maximum value is $e^{\frac{1}{e}}$

minimum value is $e^e$

maximum value is $\left(\frac{1}{e}\right)^e$

Correct Answer

Option 2

Correct Answer

Explanation

Q21:

2022: 30 Aug Shift 1

Calculus

>

Application of Derivatives

Medium

Core

The given function $f(x) = x^5 - 5x^4 + 5x^3 - 1$ ; has/have

(a) local maxima at $x = 1$

(b) local maximum value is 0

(c) local minimum at $x = 3$

(d) local minimum value is $-28$

(e) The point of inflexion is $x = 1$

Choose the correct answer from the options given below

(a), (b) only

(a), (b), (c) only

(a), (b), (c), (d) only

(a), (c), (e) only

Correct Answer

Option 3

Correct Answer

Explanation

Q22:

2022: 30 Aug Shift 1

Calculus

>

Application of Derivatives

Medium

Core

Match List-I with List-II

List-I	List-II
(a) If $x = t^2$ and $y = t^3$ , then $\frac{d^2y}{dx^2}$ at $t = 1$	(i) $-2$
(b) If $f(x) = \sqrt{x} + 1$ , then $f''(1)$	(ii) $-1$
(c) The minimum value of $f(x) = 9x^2 + 12x + 2$ is	(iii) $\frac{3}{4}$
(d) The point of inflexion of the function $f(x) = (x-2)^4 (x+1)^3$ is	(iv) $-\frac{1}{4}$

Choose the correct answer from the options given below

(a) - (i), (b) - (iii), (c) - (ii), (d) - (iv)

(a) - (ii), (b) - (iii), (c) - (i), (d) - (iv)

(a) - (iii), (b) - (iv), (c) - (i), (d) - (ii)

(a) - (iv), (b) - (i), (c) - (iii), (d) - (ii)

Correct Answer

Option 3

Correct Answer

Explanation

Q23:

2022: 16 July Shift 2

Calculus

>

Application of Derivatives

Easy

Core

The equation of the normal to the curve $y = x - \frac{1}{x}$ at $(1,0)$ is:

$x + 2y = 3$

$-x + 2y + 1 = 0$

$x + 2y = 1$

$x - 2y = 1$

Correct Answer

Option 3

Correct Answer

Explanation

Q24:

2022: 16 July Shift 2

Calculus

>

Application of Derivatives

Easy

Core

The minimum value of $x^2 - 8x + 17$ on the set $\mathbb{R}$ of all real numbers is:

-1

0

1

2

Correct Answer

Option 3

Correct Answer

Explanation

Q25:

2022: 16 July Shift 2

Calculus

>

Application of Derivatives

Easy

Core

The interval in which the function given by $f(x) = x^2 e^{-x}$ is strictly increasing is:

$(-\infty, \infty)$

$(-\infty, 0)$

$(2, \infty)$

$(0, 2)$

Correct Answer

Option 4

Correct Answer

Explanation

Q26:

2022: 23 Aug Shift 1

Calculus

>

Application of Derivatives

Medium

Core

The interval in which $f(x) = \frac{x}{2} + \frac{2}{x}$ is a decreasing function of $x$ is :

$[-2, 2]$

$(-\infty, -2) \cup (2, \infty)$

$(-2, 0) \cup (0, 2)$

$(-2, 2)$

Correct Answer

Option 3

Correct Answer

Explanation

Q27:

2022: 23 Aug Shift 1

Calculus

>

Application of Derivatives

Easy

Core

The tangent to the curve $y = e^{3x}$ at the point $(0, 1)$ , meets the x-axis at :

$(0, 1)$

$(0, 3)$

$\left(-\frac{1}{3}, 0\right)$

$\left(\frac{1}{3}, 0\right)$

Correct Answer

Option 3

Correct Answer

Explanation

Q28:

2022: 23 Aug Shift 1

Calculus

>

Application of Derivatives

Medium

Core

Match list I with list II

List - I	List - II
A. Slope of tangent to the curve $y = x^3 - x$ at $x = 2$	I. $-2$
B. Slope of tangent to the curve $y = 3x^3 - 4x$ at $x = 0$	II. $11$
C. Slope of normal to the curve $y = \sin\theta$ at $\theta = \frac{\pi}{3}$	III. $2$
D. Slope of normal to the curve $y = \cos\theta$ at $\theta = \frac{\pi}{6}$	IV. $-4$

Choose the correct answer from the option given below :

A-II, B-IV, C-I, D-III

A-II, B-III, C-I, D-IV

A-I, B-IV, C-II, D-III

A-I, B-III, C-II, D-IV

Correct Answer

Option 1

Correct Answer

Explanation

Q29:

2022: 23 Aug Shift 1

Calculus

>

Application of Derivatives

Easy

Core

A student Sumit is running on a playground along the curve given by $y = x^2 + 7$ . Another student Amit standing at a point with co-ordinates $(3, 7)$ on the playground wants to hit Sumit by paper ball, when Sumit is nearest to him.

Based on the above information answer the following.

Sumit's position, when $x = 10$ is :

$(10, 93)$

$(10, 17)$

$(10, 107)$

$(10, 3)$

Correct Answer

Option 3

Correct Answer

Explanation

Q30:

2022: 23 Aug Shift 1

Calculus

>

Application of Derivatives

Easy

Core

A student Sumit is running on a playground along the curve given by $y = x^2 + 7$ . Another student Amit standing at a point with co-ordinates $(3, 7)$ on the playground wants to hit Sumit by paper ball, when Sumit is nearest to him.

Based on the above information answer the following.

Distance between Sumit and Amit in terms of $x$ is :

$\sqrt{2(x-1)(2x^2 + 2x + 3)}$

$(x-3)^2 + x^4$

$\sqrt{(x-3)^2 + x^4}$

$\sqrt{2(x-1)(2x^2 + 2x + 3)}$

Correct Answer

Option 3

Correct Answer

Explanation

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