Calculus MCQs Question and Answer in Engineering Maths Page-2 section-1

11.
The function f(x) = 2x³ - 3x² - 36x + 2 has its maxima at

A. x = -2 only

B. x = 0 only

C. x = 3 only

D. both x = -2 and x = 3

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12.
The minimum value of the function $${\text{f}}\left( {\text{x}} \right) = \left( {\frac{{{{\text{x}}^3}}}{3}} \right) - {\text{x}}$$ occurs at

A. x = 1

B. x = -1

C. x = 0

D. x = $$\frac{1}{{\sqrt 3 }}$$

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13.
The function f(x) = 2x - x² + 3 has

A. a maxima at x = 1 and a minima at x = 5

B. a maxima at x = 1 and a minima at x = -5

C. only a maxima at x = 1

D. only a minima at x = 1

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14.
The function y = |2 - 3x|

A. is continuous $$\forall {\text{x}} \in {\text{R}}$$ and differentiable $$\forall {\text{x}} \in {\text{R}}$$

B. is continuous $$\forall {\text{x}} \in {\text{R}}$$ and differentiable $$\forall {\text{x}} \in {\text{R}}$$ except at x = $$\frac{3}{2}$$

C. is continuous $$\forall {\text{x}} \in {\text{R}}$$ and differentiable $$\forall {\text{x}} \in {\text{R}}$$ except at x = $$\frac{2}{3}$$

D. is continuous $$\forall {\text{x}} \in {\text{R}}$$ except x = 3 and differentiable $$\forall {\text{x}} \in {\text{R}}$$

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15.
Which of the following is correct?

A. $$\mathop {\lim }\limits_{{\text{x}} \to 0} \left( {\frac{{\sin \,4{\text{x}}}}{{\sin \,2{\text{x}}}}} \right) = 1{\text{ and }}\mathop {\lim }\limits_{{\text{x}} \to 0} \left( {\frac{{\tan {\text{x}}}}{{\text{x}}}} \right) = 1$$

B. $$\mathop {\lim }\limits_{{\text{x}} \to 0} \left( {\frac{{\sin \,4{\text{x}}}}{{\sin \,2{\text{x}}}}} \right) = \infty {\text{ and }}\mathop {\lim }\limits_{{\text{x}} \to 0} \left( {\frac{{\tan {\text{x}}}}{{\text{x}}}} \right) = 1$$

C. $$\mathop {\lim }\limits_{{\text{x}} \to 0} \left( {\frac{{\sin \,4{\text{x}}}}{{\sin \,2{\text{x}}}}} \right) = 2{\text{ and }}\mathop {\lim }\limits_{{\text{x}} \to 0} \left( {\frac{{\tan {\text{x}}}}{{\text{x}}}} \right) = \infty $$

D. $$\mathop {\lim }\limits_{{\text{x}} \to 0} \left( {\frac{{\sin \,4{\text{x}}}}{{\sin \,2{\text{x}}}}} \right) = 2{\text{ and }}\mathop {\lim }\limits_{{\text{x}} \to 0} \left( {\frac{{\tan {\text{x}}}}{{\text{x}}}} \right) = 1$$

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16.
The double integral $$\int\limits_0^{\text{a}} {\int\limits_0^{\text{y}} {{\text{f}}\left( {{\text{x, y}}} \right){\text{dx dy}}} } $$ is equivalent to

A. $$\int\limits_0^{\text{x}} {\int\limits_0^{\text{y}} {{\text{f}}\left( {{\text{x, y}}} \right){\text{dx dy}}} } $$

B. $$\int\limits_0^{\text{a}} {\int\limits_{\text{x}}^{\text{y}} {{\text{f}}\left( {{\text{x, y}}} \right){\text{dx dy}}} } $$

C. $$\int\limits_0^{\text{a}} {\int\limits_{\text{x}}^{\text{a}} {{\text{f}}\left( {{\text{x, y}}} \right){\text{dy dx}}} } $$

D. $$\int\limits_0^{\text{a}} {\int\limits_0^{\text{a}} {{\text{f}}\left( {{\text{x, y}}} \right){\text{dx dy}}} } $$

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17.
The values of x for which the function $${\text{f}}\left( {\text{x}} \right) = \frac{{{{\text{x}}^2} - 3{\text{x}} - 4}}{{{{\text{x}}^2} + 3{\text{x}} - 4}}$$ is NOT continuous are

A. 4 and -1

B. 4 and 1

C. -4 and 1

D. -4 and -1

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18.
The expression $${\text{V}} = \int_0^{\text{H}} {\pi {{\text{R}}^2}{{\left( {1 - \frac{{\text{h}}}{{\text{H}}}} \right)}^2}} {\text{dh}}$$ for the volume of a cone is equal to

A. $$\int_0^{\text{R}} {\pi {{\text{R}}^2}{{\left( {1 - \frac{{\text{h}}}{{\text{H}}}} \right)}^2}} {\text{dr}}$$

B. $$\int_0^{\text{R}} {\pi {{\text{R}}^2}{{\left( {1 - \frac{{\text{h}}}{{\text{H}}}} \right)}^2}} {\text{dh}}$$

C. $$\int_0^{\text{H}} {2\pi {\text{rH}}{{\left( {1 - \frac{{\text{r}}}{{\text{R}}}} \right)}^2}} {\text{dh}}$$

D. $$4\int_0^{\text{R}} {\pi {\text{rH}}{{\left( {1 - \frac{{\text{r}}}{{\text{R}}}} \right)}^2}} {\text{dr}}$$

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19.
Consider the function f(x) = |x| in the interval -1 < x ≤ 1. At the point x = 0, f(x) is

A. continuous and differentiable

B. non continuous and differentiable

C. continuous and non-differentiable

D. neither continuous nor differentiable

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20.
For a scalar function f(x, y, z) = x² + 3y² + 2z², the gradient at the point P(1, 2, -1) is

A. $$2\overrightarrow {\text{i}} + 6\overrightarrow {\text{j}} + 4\overrightarrow {\text{k}} $$

B. $$2\overrightarrow {\text{i}} + 12\overrightarrow {\text{j}} - 4\overrightarrow {\text{k}} $$

C. $$2\overrightarrow {\text{i}} + 12\overrightarrow {\text{j}} + 4\overrightarrow {\text{k}} $$

D. $$\sqrt {56} $$

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Calculus MCQs Question and Answer in Engineering Maths

11. The function f(x) = 2x3 - 3x2 - 36x + 2 has its maxima at

Answer & Solution

12. The minimum value of the function $${\text{f}}\left( {\text{x}} \right) = \left( {\frac{{{{\text{x}}^3}}}{3}} \right) - {\text{x}}$$ occurs at

Answer & Solution

13. The function f(x) = 2x - x2 + 3 has

Answer & Solution

14. The function y = |2 - 3x|

Answer & Solution

15. Which of the following is correct?

Answer & Solution

16. The double integral $$\int\limits_0^{\text{a}} {\int\limits_0^{\text{y}} {{\text{f}}\left( {{\text{x, y}}} \right){\text{dx dy}}} } $$ is equivalent to

Answer & Solution

17. The values of x for which the function $${\text{f}}\left( {\text{x}} \right) = \frac{{{{\text{x}}^2} - 3{\text{x}} - 4}}{{{{\text{x}}^2} + 3{\text{x}} - 4}}$$ is NOT continuous are

Answer & Solution

18. The expression $${\text{V}} = \int_0^{\text{H}} {\pi {{\text{R}}^2}{{\left( {1 - \frac{{\text{h}}}{{\text{H}}}} \right)}^2}} {\text{dh}}$$ for the volume of a cone is equal to

Answer & Solution

19. Consider the function f(x) = |x| in the interval -1 < x ≤ 1. At the point x = 0, f(x) is

Answer & Solution

20. For a scalar function f(x, y, z) = x2 + 3y2 + 2z2, the gradient at the point P(1, 2, -1) is

Answer & Solution

Read More Section(Calculus)

11.
The function f(x) = 2x³ - 3x² - 36x + 2 has its maxima at

12.
The minimum value of the function $${\text{f}}\left( {\text{x}} \right) = \left( {\frac{{{{\text{x}}^3}}}{3}} \right) - {\text{x}}$$ occurs at

13.
The function f(x) = 2x - x² + 3 has

14.
The function y = |2 - 3x|

15.
Which of the following is correct?

16.
The double integral $$\int\limits_0^{\text{a}} {\int\limits_0^{\text{y}} {{\text{f}}\left( {{\text{x, y}}} \right){\text{dx dy}}} } $$ is equivalent to

17.
The values of x for which the function $${\text{f}}\left( {\text{x}} \right) = \frac{{{{\text{x}}^2} - 3{\text{x}} - 4}}{{{{\text{x}}^2} + 3{\text{x}} - 4}}$$ is NOT continuous are

18.
The expression $${\text{V}} = \int_0^{\text{H}} {\pi {{\text{R}}^2}{{\left( {1 - \frac{{\text{h}}}{{\text{H}}}} \right)}^2}} {\text{dh}}$$ for the volume of a cone is equal to

19.
Consider the function f(x) = |x| in the interval -1 < x ≤ 1. At the point x = 0, f(x) is

20.
For a scalar function f(x, y, z) = x² + 3y² + 2z², the gradient at the point P(1, 2, -1) is