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Mathematics-III BCA Question Paper with Solutions, Book Pdf

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Important Questions For Mathematics-III:
* Important Short Questions
* Solved Question Paper
* Syllabus

Section A: Mathematics-III Very Short Question Solutions

Q1. Show that 

|z1 + z2|2 + |z1 – z2|2 = 2|z1|2 + 2|z2|2

Ans. We have, 

Mathematics-III Show that |z1 + z2|2 + |z1 - z2|2 = 2|z1|2 + 2|z2|2

Q2. Define sequence with example. 

Ans. Let S be any non-empty set. A function whose domain is the set of N natural numbers and whose range is a subset of S, is called a sequence in the set S. 

Define sequence with example. 

Q3. If f(x, y, z) 3x2y – y3z2 then find grad fat point (1, -2, -1). 

Sol. The gradient is a vector:  

If f(x, y, z) 3x2y - y3z2 then find grad fat point (1, -2, -1).

Q4. Solve

Solve important Question

Sol. Let    x + y = z

Solve mathematics Important
Solve mathematics Important
Solve mathematics Important

Q5. Solve

(D3 + 3D2 + 3D + 1)y = ex + e-x 

Sol. Given equation is 

(D3 + 3D2 + 3D + 1)y = ex + e-x 

∴ Auxillary equation is 

D3 + 3D2 + 3D + 1 = 0 

∴ (D+ 1)³ =0  

∴ D = -1, -1, -1 are the roots. 

∴ C.F. = (c1x2 + c2x + c3) (ex + e-x

Solve (D3 + 3D2 + 3D + 1)y = ex + e-x 

Section B: Mathematics-III Short Question Solutions

Q6. Determine the regions defined by 

|z-1| + |z + 1| ≤ 4  

Sol. Here, |z1 -z2| = |1 – (-1)| = 2 

and a = 4 

∴ |z1 – z2| < a

Since |z – z0| + |z – z1| = c (constant) is the equation of ellipse with z0 and z1 as focii, so |z – 1| + |z + 1| ≤ 4 is the equation of all points lying inside and on an ellipse with focii (1, 0) and (-1,0). 

Hence, region represented by |z – 1| + |z + 1| ≤ 4 is an interior and boundary of an ellipse. 


Q7. Solve

Solve Important Mathematical Numerical

Ans. The given equation can be written as  

Solve Important Mathematical Numerical
Solve Important Mathematical Numerical

Q8. Show 

Show Mathematics BCA Important Numerical solution

Sol. Given, 

Show Mathematics BCA Important Numerical solution

Section C: Mathematics-III Detailed Question Solutions

Q9. Show that

Show that Numerical Solution BCA Important

Sol. Let

Show that Numerical Solution BCA Important

Similarly

Show that Numerical Solution BCA Important

Comparing eqs. (1) and (2), we get 

Sn+1 ≥ Sn, ∀n

The sequence 〈Sn〉 is monotonically increasing. 

Now, from eq. (1), we have 

Show that Numerical Solution BCA Important

⇒ The sequence 〈Sn〉 is bounded.  

Thus, the sequence〈Sn〉, being a monotonically increasing sequence bounded above by 3 is convergent and equal to e where 2 < e < 3. 

Show that Numerical Solution BCA Important

⇒ Limit of the sequence 〈Sn〉 lies between 2 and 3, i.e. = e.  


Q10. Test the convergence of following series. 

Test the convergence of following series.

Sol. Here, we have  

Test the convergence of following series.
Test the convergence of following series.

If x< 1,the series is convergent

If x > 1, the series is divergent 

If x = 1, 

Test the convergence of following series.

Therefore, we have to use another test. 

Test the convergence of following series.

Hence, given series is divergent. 


Q11. Obtain the Fourier series of 

Obtain the Fourier series of in the interval (0, 2𝝅) and hence deduce 

in the interval (0, 2𝝅) and hence deduce 

Obtain the Fourier series of in the interval (0, 2𝝅) and hence deduce 

Sol. Fourier series off(x) having period 2𝝅, 0 

The formula of Fourier series is 

Obtain the Fourier series of in the interval (0, 2𝝅) and hence deduce 

a0 = 0 

Obtain the Fourier series of in the interval (0, 2𝝅) and hence deduce 

From Bernoulli rule

Obtain the Fourier series of in the interval (0, 2𝝅) and hence deduce 

Apply Bernoulli rule  

Obtain the Fourier series of in the interval (0, 2𝝅) and hence deduce 

Substitute a0, an, bn in Fourier series, we get

Obtain the Fourier series of in the interval (0, 2𝝅) and hence deduce 
Obtain the Fourier series of in the interval (0, 2𝝅) and hence deduce 

12. Solve

(1 +  y2)dx – (tan-1y – x)dy = 0 

Sol. The given differential equation can be written as

Solve (1 +  y2)dx - (tan-1y - x)dy = 0

Thus, the solution of eq. (1) is  

Solve (1 +  y2)dx - (tan-1y - x)dy = 0

Which is the required solution. 


Q13. Solve

(D2 + 1)y = sin x sin 2x

Sol. Here, the given equation is 

(D2 + 1)y = sin x. sin2x

To find the C.F. of eq. (1), the auxiliary equation is 

(m2 + 1) = 0 ⇒ m ± i

Solve (D2 + 1)y = sin x sin 2x
Solve (D2 + 1)y = sin x sin 2x

Hence, the complete solution of eq. (1) is given by

y = C.F.+ P.I. 

Solve (D2 + 1)y = sin x sin 2x

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