MCQ and Introduction on Calculus

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Calculus is a branch of mathematics that focuses on the investigation of continuous transition. Calculus is also referred to as "infinite calculus" or "infinitesimal calculus." Calculus in the classical sense analyzes functions that change continuously. The two most crucial concepts in calculus are derivatives and integrals. The derivative measures the rate of change of a function, whereas the integral measures the area under the curve. While the derivative defines the function at a certain point, the integral adds together the discrete values of a function over a range of values.

Differential calculus and integral calculus are the two main categories in calculus mathematics. In both differential and integral calculus, the impact of a slight change in the independent variable as the equation gets closer to zero is taken into account. The higher discipline of mathematics known as Analysis is based on both discrete and integral calculus.

Calculus Overview

Calculus Overview

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Ques 1. What is y's derivative in relation to x if y = cos (x²)?

sin (x²)
- sin x²
2x sin (x²)
- 2x sin (x²)

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Ans. (d) - 2x sin (x²)

Explanation: y = cos (x²)

= dx / dy = (- sin x²) × (2x)

= dx / dy = - 2x sin (x²)

Ques 2. If the curve ay + x² = 7 and x³ = y, sliced orthogonally at (1,1), then the value of a is:

1
0
6
-6

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Ans. (c ) 6

Explanation: Given in the question:

ay + x² = 7 and x³ = y

So, ay + x² = 7

Differentiation with respect to x, we get

= ady / dx + 2x = 0

= dy / dx = -2x / a = 1st curve slope

Now, x³ = y

Differentiation with respect to x, we get

= 3x² = dx / dy = 2nd curve slope

So, the curve cuts at (1,1)

Thus, (-2x / a) (1,1) × (3x²) (1,1) = -1

So, a = 6.

Ques 3. A formula for the tangent to a curve is y = x³ at (1,1):

x – 10y + 50 = 0
3x – y – 2 = 0
x – 3y – 4 = 0
x + 2y – 7 = 0

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Ans. (b) 3x – y – 2 = 0

Explanation: y = f(x) = x³

= y’ = f’(x) = 3x²

= m = f’(x) = 3 × 1² = 3

Formula for tangent at (1,1) will be:

(y – b) = m(x – a)

= (y – 1) = 3(x – 1)

= y – 1 = 3x – 3

So, 3x – y – 2 = 0.

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Ques 4. If f(x) = x - 1/x then, f’(-1) will be

0
2
1
-2

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Ans. (b) 2

Explanation: Given in the question:

f(x) – x 1/x

Differentiation with respect to x, we get

= f’(x) = 1 – (-1/x)

= 1 + 1/x²

Put x = -1

= f’(-1) = 1 + 1/(-1)² = 1+1 = 2

Therefore, f’(-1) = 2

Ques 5. What will be the minimum value of f(x) = x² - x + 2

1/2
3/4
7/4
1/4

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Ans. (c ) 7/4

Explanation: Given in the question:

f(x) = x² - x + 2

f’(x) = 2x – 1

Let the derivative be equal to 0, we get

f’(x) = 2x – 1 = 0

x = 1 /2

Now, f’(x) = 2 > 0

Thus, option (c ) is correct answer.

Ques 6. How would the slope of the curve change if y = 2x – x² and x grows at a pace of 3 units/sec?

Increases at 6 units/sec
Decreases at 6 units/sec
Increases at 3 units/sec
Decreases at 3 units/sec

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Ans. (b) Decreases at 6 units/sec

Explanation: Given in the question, y = 2x – x² and dx / dt = 3 units/sec

Curve slope will be, dy / dx = 2 – 2x = m

= dm / dt = 0 – 2 × dx / dt

= -2 × 3

= -6 units per second

Thus, option (b) is the correct answer.

Ques 7. Limx→0 (e^x2 – cos x) / x² is equals to

0
1
2/3
3/2

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Ans. (d) 3/2

Explanation: Given in the question: Lim_x→0 (e^x² – cos x) / x²

= Lim_x→0 (e^x² – cos x -1 + 1) / x²

= Lim_x→0 {(e^x² – 1) / x² + (1 – cos x)} / x²

= Lim_x→0 {(e^x² – 1) / x² + Lim_x→0 (1 – cos x)} / x²

= 1 + ½

= (2 + 1) / 2

= 3/2

Ques 8. If f(x) = (x + 1) / x then df(x) / dx is

1/x
-1/x
-1/x²
1/x²

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Ans. (c) -1/x²

Explanation: Given in the question: f(x) = (x + 1) / x

So, df(x) / dx = d {(x + 1) / x} / dx

= {1 × x – (x + 1) × 1} / x²

= -1/x²

Ques 9. Find dy / dx, if y = e^{log (log x)}

1/x
1/log x
e^{log (log x)}
None of the above

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Ans. (a) 1/x

Explanation: Given in the question: y = e^{log (log x)}

As we know that, e log x = x

Therefore, e^{log (log x)} = log x

Now, y = log x

Differentiation with respect to x, we get

dy / dx = d(log x) / dx = 1/x

Ques 10. If y = sin x° then what is the value of dy / dx?

Cos x
0
- cos x
None of the above

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Ans. (d) none of the above

Explanation: Given in the question: y = sin x°

We know that, 180° = π radian

Therefore, 1° = π / 180 radian

Now, x° = πx / 180 radian

= y = sin (π•x / 180)

Differentiation with respect to x, we get

dy / dx = π / 180 × cos (π•x / 180)

Ques 11. If y = xx, then what is the value of dy / dx at x = 1.

0
1
-1
2

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Ans. (b) 1

Given in the question: y = x^x

By taking log both the sides

= log y = log x^x

= log y = x log x

Differentiation with respect to x, we get

= 1 / y × dy / dx = x × log x / dx + log x

= 1 / y × dy / dx = x × 1 / x + log x

= dy / dx = y × (1 + log x)

= dy / dx = x^x × (1 + log x)

By putting x = 1

= dy / dx = 1¹ × (1 + log 1)

= dy / dx = 1 × (1 + 0)

Here, log 1 = 0

Thus, dy / dx = 1.

Ques 12. What is the value of lim _x_→0 (x sin 1/x) ?

Infinity
-1
0
1

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Ans. (c ) 0

Explanation: suppose, f(x) = lim _x→0 (x sin 1/x)

When the value of x = 0

Then, f(x) = 0 × sin ∞

f(x) = 0 × finite value

Thus, f(x) = lim _x→0 (x sin 1/x) = 0.

Ques 13. The function f(x) = x|x| is ………..

not constant at the beginning
Clearly distinct at the beginning
Limit is lacking at origin
not distinguishable at origin

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Ans. (b) clearly distinct at the beginning

Explanation: Given in the question: f(x) = x|x|

f(x) = { x², x ≥ 0, – x², x < 0

f(0+) = 0

f(0-) = 0

f(0) = 0

Left limit = right limit= function value = 0

Hence, f(x) is constant at x = 0

Now, f’(x) = x•|x|/ x + 1•|x| = |x| + |x| = 2|x|

f'(0+) = 0

f'(0-) = 0

Left derivative= right derivative = 0 (well defined)

Hence, f(x) is distinguished at x = 0.

Ques 14. If y = logyx, then what will be the value of dy / dx.

1 / x + log y
1 / log x (1 + y)
1 / x (1 + log y)
1 / y + log x

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Ans. (c ) 1 / x (1 + log y)

Explanation: Given in the question: y = logyx

= y = log x / log y

= y log y = log x

Differentiation with respect to x, we get

dy / dx• log y + y • 1 / y • dy / dx = 1 / x

log y dy / dx + dy / dx = 1 / x

=dy / dx (1 + log y) = 1/x

Therefore, dy / dx = 1 / x (1 + log y)

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MCQ and Introduction on Calculus

CBSE CLASS XII Related Questions

1.
Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).

2.
Find:
Let \(A=[a_{ij}]\) be a \(2\times2\) matrix whose elements are given by \[ a_{ij}=\frac{(2i-j)^2}{3} \] Find the transpose matrix \(A'\).

3.
Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

4.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

5.
Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

6.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

Similar Mathematics Concepts

Comments

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MCQ and Introduction on Calculus

CBSE CLASS XII Related Questions

1.Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).

2.Find:Let \(A=[a_{ij}]\) be a \(2\times2\) matrix whose elements are given by \[ a_{ij}=\frac{(2i-j)^2}{3} \] Find the transpose matrix \(A'\).

3.Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

4.Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

5. Find: If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

6. Find: The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).

2.
Find:
Let \(A=[a_{ij}]\) be a \(2\times2\) matrix whose elements are given by \[ a_{ij}=\frac{(2i-j)^2}{3} \] Find the transpose matrix \(A'\).

3.
Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

4.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

5.
Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

6.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]