# 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,

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.

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

Sol. The gradient is a vector:

Q4. Solve

Sol. Let    x + y = z

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

## 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

Ans. The given equation can be written as

Q8. Show

Sol. Given,

## Section C: Mathematics-III Detailed Question Solutions

Q9. Show that

Sol. Let

Similarly

Comparing eqs. (1) and (2), we get

Sn+1 ≥ Sn, ∀n

The sequence 〈Sn〉 is monotonically increasing.

Now, from eq. (1), we have

⇒ 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.

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

Q10. Test the convergence of following series.

Sol. Here, we have

If x< 1,the series is convergent

If x > 1, the series is divergent

If x = 1,

Therefore, we have to use another test.

Hence, given series is divergent.

Q11. 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

a0 = 0

From Bernoulli rule

Apply Bernoulli rule

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

12. Solve

(1 +  y2)dx – (tan-1y – x)dy = 0

Sol. The given differential equation can be written as

Thus, the solution of eq. (1) is

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

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

y = C.F.+ P.I.