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TU BCA 2nd Semester — 2026, 2024 BOARD EXAM — Mathematics-II MCQs

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TRIBHUVAN UNIVERSITY

Bachelor of Computer Application

OFFICE OF THE DEAN

2026, 2024 BOARD EXAM

Faculty: Bachelor of Computer Application (BCA)

Course Title: Mathematics-II (MCQs)

Code No: BCA 154

Semester: 2nd Semester

Full Marks: 60

Pass Marks: 24

Time: 3 hours

Group A (Multiple Choice Questions)

Attempt all questions. Correct answers are highlighted in yellow.

1.Length of a continuous curve $y = f(x)$ in the interval [a, b] is calculated by using the formula

\int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx

✓ Correct Answer

\int_a^b \sqrt{1 - \left(\frac{dy}{dx}\right)^2} dx

\int_a^b 1 + \left(\frac{dy}{dx}\right)^2 dx

\int_a^b \sqrt{1 + \frac{dy}{dx}} dx

💡 Complete Solution & Explanation:

The arc length formula for a function y=f(x) i

s \int_a^b \sqrt{1 + (f'(x))^2} dx

2.If the matrix A = [[2, 5], [a, 3]] does not have inverse then the value of a is

5/6

6/5

✓ Correct Answer

💡 Complete Solution & Explanation:

For a matrix to not have an inverse, its determinant must be zero. Det(A) = (2)(3) - (5)(a) = 6 - 5a = 0 =>

a = 6/5

3.The value o $f \lim_{x \to 0} \frac{a^x - 1}{x}$ is equal to

\log_a e

\log_e a

✓ Correct Answer

\log_a 1

\log_a 0

💡 Complete Solution & Explanation:

Using L'Hôpital's rule, the limit i

s \ln(a)

which i

s \log_e a

4.Inclination of the tangent to the curve $2y = 2 - x^2$ at $x = 1$ is

\frac{\pi}{4}

\frac{3\pi}{4}

✓ Correct Answer

\frac{\pi}{2}

\pi

💡 Complete Solution & Explanation:

y = 1 - x^2/2 =>

y' = -x

. At

x=1

, slope

m = -1

. \tan(\theta) = -1 =>

\theta = 3\pi/4

5.Value of the integratio $n \int x e^x dx$ is equal to

x e^x - e^x

✓ Correct Answer

x e^x + e^x

x e^x + x e^x

x^2 e^x + x e^x

💡 Complete Solution & Explanation:

Using integration by parts:

\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + c

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6.The function f(x) defined by $f(x) = x^2 - 4$ is continuous in the interval [-3, 3] then the value of c prescribed by Rolle's theorem is

-3

✓ Correct Answer

💡 Complete Solution & Explanation:

Rolle's theorem states f'(c) = 0.

f'(x) = 2x

, so 2c = 0 implies

c = 0

7.The next iterative value of the root of x^2 - $4 = 0 Newton Raphson method$ if the initial guess 1 is

1.5

2.5

✓ Correct Answer

3.5

3.75

💡 Complete Solution & Explanation:

x_{1} = x_0 - f(x_0)/f'(x_0).

f(1) = -3

, f'(1) = 2. x_1 = 1 - (-3)/2 = 2.5.

8.If a function f(x) = {(x^2-1)/(x-1) for x != 1, k for x = 1} is continuous at $x = 1$ , then value of k is

✓ Correct Answer

💡 Complete Solution & Explanation:

For continuity, lim_{x->1} f(x) must equal f(1). lim_{x->1} (x+1) = 2, so

k = 2

9.Limit of the function lim_{x \to 0} $\frac{\sin x^0}{x}$ is

\frac{\pi}{180}

✓ Correct Answer

\frac{\pi}{90}

\frac{3\pi}{180}

\pi

💡 Complete Solution & Explanation:

x^0 means x degrees. sin(x^\circ) approx x * (

\pi/180)

. The limit i

s \pi/180

10.Which of the following differential equation is not linear?

y' +

y = x

y' + y^2 = 2x

✓ Correct Answer

xy' -

xy = x^2y

y' +

2y = e^x

💡 Complete Solution & Explanation:

A linear differential equation has the dependent variable and its derivatives to the first power only. y^2 makes it non-linear.

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11.The solution of $\int \frac{dx}{x^2 + 4}$ is:

\frac{1}{2} \sin^{-1} \frac{x}{2} + c

\tan^{-1} \frac{x}{2} + c

\frac{1}{2} \tan^{-1} \frac{x}{2} + c

✓ Correct Answer

\sin^{-1} \frac{x}{2} + c

💡 Complete Solution & Explanation:

Using the standard integral formula

\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c

. Here

a=2

, so

\int \frac{dx}{x^2 + 2^2} = \frac{1}{2} \tan^{-1}(\frac{x}{2}) + c

12.The limit of the function $f(x) = \frac{e^x - 1}{x}$ as $x \to 0$ is:

-1

1

✓ Correct Answer

0

does not exist

💡 Complete Solution & Explanation:

Using L'Hôpital's rule for the indeterminate form

\frac{0}{0}

\lim_{x \to 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{e^x}{1} = e^0 = 1

13.The function $f(x) = x^3 - 2x$ is increasing at the interval:

[0, \infty)

[2, \infty)

(0, \infty)

(1, \infty)

✓ Correct Answer

💡 Complete Solution & Explanation:

Find

f'(x) = 3x^2 - 2

. For the function to be increasing,

f'(x) > 0 \implies 3x^2 > 2 \implies x^2 > \frac{2}{3} \implies x > \sqrt{\frac{2}{3}} \approx 0.816

. Among choices,

(1, \infty)

fits the criteria.

14.The stationary points of $y = x^3 - 6x$ are:

0, \pm\sqrt{6}

✓ Correct Answer

0, 1

0, \pm4

\pm\sqrt{2}

💡 Complete Solution & Explanation:

Set the first derivative to zero:

y' = 3x^2 - 6 = 0

. This gives

x^2 = 2

, so

x = \pm\sqrt{2}

. However, the stationary points are values where the slope is zero, hence

3x^2 - 6 = 0 \implies x = \pm\sqrt{2}

15.The area below the curve $y = x^3$ , x-axis and ordinates at $x = 0$ and $x = 2$ is:

4

✓ Correct Answer

32

0

20

💡 Complete Solution & Explanation:

Area =$$$\int_{0}^{2} x^3 dx =

\frac{x^4}{4}

_{0}^{2} = \frac{2^4}{4} - \frac{0^4}{4} = \frac{16}{4} = 4$.

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16.The maximum value of $Z = 4x + y$ at $(0,0), (30,0), (20,30)$ and $(0,50)$ is:

510

120

✓ Correct Answer

110

50

💡 Complete Solution & Explanation:

Evaluate

Z = 4x + y

at each point:

Z(0,0)=0, Z(30,0)=4(30)+0=120, Z(20,30)=4(20)+30=110, Z(0,50)=4(0)+50=50

. The maximum value is

120

17.If the initial value to approximate the root of $x - 13 = 0$ is $3.5$ , then the value of one iteration using Newton-Raphson method is:

2.687

2.678

✓ Correct Answer

3.607

3.597

💡 Complete Solution & Explanation:

Newton-Raphson formula:

x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

. Here

f(x) = x - 13

f'(x) = 1

x_1 = 3.5 - \frac{3.5 - 13}{1} = 3.5 - (-9.5) = 13

. Correction: Given provided options, if

x

was meant as

x^2-13=0

, then

x_1 = 3.5 - \frac{3.5^2 - 13}{2(3.5)} = 3.5 - \frac{12.25 - 13}{7} = 3.5 - \frac{-0.75}{7} = 3.5 + 0.107 = 3.607

18.The solution of $\frac{dy}{dx} = e^{-x}$ when $y(0) = 0$ is:

y = -e^{-x} + 1

✓ Correct Answer

y = e^x + c

y = -e^x + e^{-c}

y = e^x

💡 Complete Solution & Explanation:

Integrating both sides:

\int dy = \int e^{-x} dx \implies y = -e^{-x} + c

. Applying the condition

y(0) = 0

0 = -e^0 + c \implies c = 1

. Thus,

y = -e^{-x} + 1

19.What is the value of $\frac{dy}{dx}$ if $x = e^t$ and $y = 2t$ ?

\frac{1}{e^t}

t e^t

\frac{2}{e^t}

✓ Correct Answer

e^t

💡 Complete Solution & Explanation:

By chain rule,

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

. Here,

\frac{dy}{dt} = 2

and

\frac{dx}{dt} = e^t

. Therefore,

\frac{dy}{dx} = \frac{2}{e^t}

20.The differential equation $\frac{d^2y}{dx^2} = \sqrt{1 + (\frac{dy}{dx})^2}$ is of:

Second degree, second order

Second degree, first order

First degree, second order

✓ Correct Answer

First degree, first order

💡 Complete Solution & Explanation:

The order is the highest derivative present, which is 2. The degree is the power of the highest derivative when the equation is free from radicals. Squaring both sides gives

(\frac{d^2y}{dx^2})^2 = 1 + (\frac{dy}{dx})^2

. The degree is 2.

TU BCA 2nd Semester — 2026, 2024 BOARD EXAM — Mathematics-II MCQs

TRIBHUVAN UNIVERSITY

OFFICE OF THE DEAN

Group A (Multiple Choice Questions)

1.Length of a continuous curve y=f(x)y = f(x)y=f(x) in the interval [a, b] is calculated by using the formula

2.If the matrix A = [[2, 5], [a, 3]] does not have inverse then the value of a is

3.The value oflim⁡x→0ax−1xf \lim_{x \to 0} \frac{a^x - 1}{x}flimx→0​xax−1​ is equal to

4.Inclination of the tangent to the curve 2y=2−x22y = 2 - x^22y=2−x2 at x=1x = 1x=1 is

5.Value of the integration∫xexdxn \int x e^x dxn∫xexdx is equal to

6.The function f(x) defined by f(x)=x2−4f(x) = x^2 - 4f(x)=x2−4 is continuous in the interval [-3, 3] then the value of c prescribed by Rolle's theorem is

7.The next iterative value of the root of x^2 - 4=0NewtonRaphsonmethod4 = 0 Newton Raphson method4=0NewtonRaphsonmethod if the initial guess 1 is

8.If a function f(x) = {(x^2-1)/(x-1) for x != 1, k for x = 1} is continuous at x=1x = 1x=1, then value of k is

9.Limit of the function lim_{x \to 0} sin⁡x0x\frac{\sin x^0}{x}xsinx0​ is

10.Which of the following differential equation is not linear?

11.The solution of ∫dxx2+4\int \frac{dx}{x^2 + 4}∫x2+4dx​ is:

12.The limit of the function f(x)=ex−1xf(x) = \frac{e^x - 1}{x}f(x)=xex−1​ as x→0x \to 0x→0 is:

13.The function f(x)=x3−2xf(x) = x^3 - 2xf(x)=x3−2x is increasing at the interval:

14.The stationary points of y=x3−6xy = x^3 - 6xy=x3−6x are:

15.The area below the curve y=x3y = x^3y=x3, x-axis and ordinates at x=0x = 0x=0 and x=2x = 2x=2 is:

16.The maximum value of Z=4x+yZ = 4x + yZ=4x+y at (0,0),(30,0),(20,30)(0,0), (30,0), (20,30)(0,0),(30,0),(20,30) and (0,50)(0,50)(0,50) is:

17.If the initial value to approximate the root of x−13=0x - 13 = 0x−13=0 is 3.53.53.5, then the value of one iteration using Newton-Raphson method is:

18.The solution of dydx=e−x\frac{dy}{dx} = e^{-x}dxdy​=e−x when y(0)=0y(0) = 0y(0)=0 is:

19.What is the value of dydx\frac{dy}{dx}dxdy​ if x=etx = e^tx=et and y=2ty = 2ty=2t?

20.The differential equation d2ydx2=1+(dydx)2\frac{d^2y}{dx^2} = \sqrt{1 + (\frac{dy}{dx})^2}dx2d2y​=1+(dxdy​)2​ is of:

💎 Upgrade to Elite Pass

1.Length of a continuous curve $y = f(x)$ in the interval [a, b] is calculated by using the formula

3.The value o $f \lim_{x \to 0} \frac{a^x - 1}{x}$ is equal to

4.Inclination of the tangent to the curve $2y = 2 - x^2$ at $x = 1$ is

5.Value of the integratio $n \int x e^x dx$ is equal to

6.The function f(x) defined by $f(x) = x^2 - 4$ is continuous in the interval [-3, 3] then the value of c prescribed by Rolle's theorem is

7.The next iterative value of the root of x^2 - $4 = 0 Newton Raphson method$ if the initial guess 1 is

8.If a function f(x) = {(x^2-1)/(x-1) for x != 1, k for x = 1} is continuous at $x = 1$ , then value of k is

9.Limit of the function lim_{x \to 0} $\frac{\sin x^0}{x}$ is

11.The solution of $\int \frac{dx}{x^2 + 4}$ is:

12.The limit of the function $f(x) = \frac{e^x - 1}{x}$ as $x \to 0$ is:

13.The function $f(x) = x^3 - 2x$ is increasing at the interval:

14.The stationary points of $y = x^3 - 6x$ are:

15.The area below the curve $y = x^3$ , x-axis and ordinates at $x = 0$ and $x = 2$ is:

16.The maximum value of $Z = 4x + y$ at $(0,0), (30,0), (20,30)$ and $(0,50)$ is:

17.If the initial value to approximate the root of $x - 13 = 0$ is $3.5$ , then the value of one iteration using Newton-Raphson method is:

18.The solution of $\frac{dy}{dx} = e^{-x}$ when $y(0) = 0$ is:

19.What is the value of $\frac{dy}{dx}$ if $x = e^t$ and $y = 2t$ ?

20.The differential equation $\frac{d^2y}{dx^2} = \sqrt{1 + (\frac{dy}{dx})^2}$ is of: