৪৫তম বিসিএস, প্রাইমারি শিক্ষক নিয়োগ, সরকারি চাকরি, ব্যাংক প্রস্তুতি

Algebra (87 টি প্রশ্ন )

Q1. For x² − Sx + P = 0 with nonzero roots r1, r2, find Σ 1/r_i and Σ 1/r_i^2.

ক) Σ 1/r_i = S/P; Σ 1/r_i^2 = (S^2 − 2P)/P^2

খ) Σ 1/r_i = P/S; Σ 1/r_i^2 = (P^2 − 2S)/S^2

গ) Σ 1/r_i = −S/P; Σ 1/r_i^2 = (S^2 + 2P)/P^2

ঘ) Σ 1/r_i = S/P; Σ 1/r_i^2 = (S^2 + 2P)/P^2

ব্যাখ্যা (Explanation):

Correct answer: Option 1. Explanation (textbook style).

Let the quadratic be x^2 − Sx + P = 0 with nonzero roots r1, r2. By Viète’s relations
r1 + r2 = S, r1 r2 = P.
Hence the sum of reciprocals is
Σ 1/r_i = 1/r1 + 1/r2 = (r1 + r2)/(r1 r2) = S/P.
To get the sum of reciprocal squares,
Σ 1/r_i^2 = 1/r1^2 + 1/r2^2 = (r1^2 + r2^2)/(r1^2 r2^2).
But r1^2 + r2^2 = (r1 + r2)^2 − 2r1 r2 = S^2 − 2P, and r1^2 r2^2 = (r1 r2)^2 = P^2. Therefore
Σ 1/r_i^2 = (S^2 − 2P)/P^2.
(Equivalently, letting y = 1/x and dividing the original equation by x^2 gives P y^2 − S y + 1 = 0, so the sum of the roots y = S/P and the sum of their squares equals (S^2 − 2P)/P^2.)

Example (to check): x^2 − 2x − 1 = 0 has S = 2, P = −1. Then Σ 1/r_i = 2/(−1) = −2 and Σ 1/r_i^2 = (4 − 2(−1))/1 = 6 (which matches a direct computation of the reciprocals).

Q2. For monic x^3 + ax^2 + bx + c with roots r_i, give p1 = Σ r_i, p2 = Σ r_i^2, p3 = Σ r_i^3.

ক) p1 = −a; p2 = a^2 − 2b; p3 = −a^3 + 3ab − 3c

খ) p1 = a; p2 = a^2 + 2b; p3 = a^3 + 3ab + 3c

গ) p1 = −a; p2 = a^2 + 2b; p3 = −a^3 − 3ab + 3c

ঘ) p1 = −b; p2 = b^2 − 2c; p3 = −b^3 + 3bc − 3a

ব্যাখ্যা (Explanation):

Let the monic cubic be \(f(x)=x^{3}+ax^{2}+bx+c\) with roots \(r_{1},r_{2},r_{3}\). By Viète's relations
\[
e_{1}=r_{1}+r_{2}+r_{3}=-a,\qquad
e_{2}=r_{1}r_{2}+r_{2}r_{3}+r_{3}r_{1}=b,\qquad
e_{3}=r_{1}r_{2}r_{3}=-c.
\]
Denote \(p_{k}=r_{1}^{k}+r_{2}^{k}+r_{3}^{k}\). Then
1) \(p_{1}=e_{1}=-a.\)

2) For \(p_{2}\) use the square of the sum:
\[
p_{2}=(r_{1}+r_{2}+r_{3})^{2}-2\sum_{i\]
Substituting \(e_{1}=-a,\ e_{2}=b\) gives
\[
p_{2}=(-a)^{2}-2b=a^{2}-2b.
\]

3) For \(p_{3}\) use Newton's identity for degree 3:
\[
p_{3}=e_{1}p_{2}-e_{2}p_{1}+3e_{3}.
\]
Substitute the known values:
\[
p_{3}=(-a)(a^{2}-2b)-b(-a)+3(-c)=-a^{3}+2ab+ab-3c=-a^{3}+3ab-3c.
\]

Thus
\[
p_{1}=-a,\qquad p_{2}=a^{2}-2b,\qquad p_{3}=-a^{3}+3ab-3c,
\]
which matches Option 1.

Q3. For a cubic ax^3 + bx^2 + cx + d = 0, express Σ r_i^2 via coefficients.

ক) (−b/a)^2 − 2(c/a)

খ) (−b/a)^2 + 2(c/a)

গ) −b/a − 2c/a

ঘ) (c/a)^2 − 2(b/a)

ব্যাখ্যা (Explanation):

Let the roots be \(r_1,r_2,r_3\) of \(ax^3+bx^2+cx+d=0\). By Viète's relations,
\[
r_1+r_2+r_3=-\frac{b}{a},\qquad
r_1r_2+r_2r_3+r_3r_1=\frac{c}{a},\qquad
r_1r_2r_3=-\frac{d}{a}.
\]
Use the identity
\[
r_1^2+r_2^2+r_3^2=(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1).
\]
Substitute the Viète expressions:
\[
r_1^2+r_2^2+r_3^2=\Big(-\frac{b}{a}\Big)^2-2\Big(\frac{c}{a}\Big).
\]
Thus the correct choice is Option 1: \(\displaystyle\big(-\frac{b}{a}\big)^2-2\frac{c}{a}.\)

Q4. Express r1^2 + r2^2 for a quadratic ax^2 + bx + c = 0 in terms of coefficients.

ক) (−b/a)^2 − 2(c/a)

খ) (b/a)^2 + 2(c/a)

গ) 2(−b/a) − (c/a)

ঘ) (−b/a)^2 + 2(c/a)

ব্যাখ্যা (Explanation):

Start with Viète's relations for the quadratic ax^2 + bx + c = 0:
r1 + r2 = −b/a, r1 r2 = c/a.

Compute
\[
r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1r_2
= \Big(\!-\frac{b}{a}\!\Big)^2 - 2\cdot\frac{c}{a}
= \frac{b^2}{a^2} - \frac{2c}{a}
= \frac{b^2 - 2ac}{a^2}.
\]

Thus the correct choice is Option 1: \(\displaystyle \Big(-\frac{b}{a}\Big)^2 - 2\frac{c}{a}.\) (Option 4, which has a plus sign, is incorrect.)

Q5. What are the elementary symmetric polynomials in roots r1,…,rn?

ক) e1 = Σri, e2 = Σ_{i

খ) e1 = Πri, e2 = Σri, …, en = Σ r_i^n

গ) e1 = Σ r_i^2, e2 = Σ r_i^3, …, en = Σ r_i^n

ঘ) e1 = r1 + r2, e2 = r1r2 only for n = 2; undefined for n > 2

ব্যাখ্যা (Explanation):

Correct answer: Option 1.

Explanation:

The elementary symmetric polynomials in the roots \(r_1,\dots,r_n\) are defined for \(k=1,\dots,n\) by
\[
e_k(r_1,\dots,r_n)=\sum_{1\le i_1<\cdots\]
Thus
\[
e_1=\sum_{i=1}^n r_i,\qquad
e_2=\sum_{1\le i\ldots,\qquad
e_n=r_1r_2\cdots r_n.
\]

Examples: for \(n=3\),
\[
e_1=r_1+r_2+r_3,\quad
e_2=r_1r_2+r_1r_3+r_2r_3,\quad
e_3=r_1r_2r_3.
\]

Why this is the standard choice: if \(p(x)\) is the monic polynomial with roots \(r_i\),
\[
p(x)=\prod_{i=1}^n (x-r_i)=x^n-e_1x^{\,n-1}+e_2x^{\,n-2}-\cdots+(-1)^n e_n,
\]
so the coefficients of \(p\) are exactly the elementary symmetric polynomials (Vieta’s formulas).

Why the other options are incorrect:
- Option 2 mixes products and sums incorrectly (it places \(\prod r_i\) as \(e_1\) and uses power sums for later \(e_k\)); that does not match the definition above.
- Option 3 gives power sums \(p_k=\sum_i r_i^k\), which are symmetric but are not the elementary symmetric polynomials.
- Option 4 only describes the \(n=2\) case and does not generalize; elementary symmetric polynomials are defined for every \(n\) as in Option 1.

Thus Option 1 is the correct description.

Q6. Real-coefficient polynomial with roots 2, 1 ± i; write the monic cubic.

ক) x³ − 4x² + 6x − 4

খ) x³ − 4x² + 5x − 2

গ) x³ − 3x² + 4x − 2

ঘ) x³ − 2x² + 2x − 2

ব্যাখ্যা (Explanation):

We are given roots \(2,\;1+i,\;1-i\) and asked for the monic cubic polynomial with real coefficients.

Since complex roots come in conjugate pairs for real-coefficient polynomials, the quadratic factor from \(1\pm i\) is
\[
(x-(1+i))(x-(1-i))=((x-1)-i)((x-1)+i)=(x-1)^2+1 = x^2-2x+2.
\]
Multiplying by the linear factor for the root \(2\) gives the cubic:
\[
(x-2)(x^2-2x+2).
\]
Expand:
\[
\begin{aligned}
(x-2)(x^2-2x+2) &= x^3-2x^2+2x -2x^2+4x-4\\
&= x^3-4x^2+6x-4.
\end{aligned}
\]
So the monic cubic is \(x^3-4x^2+6x-4\), which matches Option 1.

Q7. Given roots 1 and −3 for a monic quadratic, find the polynomial.

ক) x² − 2x − 3

খ) x² − 1x − 3

গ) x² + 2x − 3

ঘ) x² − 4x + 3

ব্যাখ্যা (Explanation):

Since the polynomial is monic and its roots are \(1\) and \(-3\), it factors as
\[
p(x)=(x-1)(x-(-3))=(x-1)(x+3).
\]
Expanding,
\[
p(x)=x^2+3x-x-3=x^2+2x-3.
\]
Thus the correct choice is Option 3: \(x^2+2x-3\).

Q8. From x² − 5x + 6 = 0, find the roots using Vieta.

ক) 2 and 3

খ) −2 and −3

গ) 1 and 6

ঘ) 3 and 5

ব্যাখ্যা (Explanation):

Correct answer: Option 1 (2 and 3). Explanation (textbook style, using Vieta and factoring):

For the quadratic \(x^{2}-5x+6=0\) (here \(a=1,\;b=-5,\;c=6\)), Vieta's formulas state that if the roots are \(r_1,r_2\) then
\(\;r_1+r_2=-\dfrac{b}{a}\) and \(\;r_1r_2=\dfrac{c}{a}\).
Thus \(r_1+r_2=-\dfrac{-5}{1}=5\) and \(r_1r_2=\dfrac{6}{1}=6\). We need two numbers whose sum is \(5\) and product is \(6\); these numbers are \(2\) and \(3\). Therefore the roots are \(2\) and \(3\).

Equivalently, factor the quadratic: \(x^{2}-5x+6=(x-2)(x-3)\). Setting this to zero gives \((x-2)(x-3)=0\), so \(x=2\) or \(x=3\).

Q9. For ax³ + bx² + cx + d with roots r1, r2, r3, state Vieta’s relations.

ক) Σr_i = b/a; Σ r_ir_j = −c/a; r1r2r3 = d/a

খ) Σr_i = −b/a; Σ r_ir_j = c/a; r1r2r3 = −d/a

গ) Σr_i = −c/a; Σ r_ir_j = b/a; r1r2r3 = −d/a

ঘ) Σr_i = −b/a; Σ r_ir_j = −c/a; r1r2r3 = d/a

ব্যাখ্যা (Explanation):

We start from the factorization by the roots. If the cubic
\[
ax^3+bx^2+cx+d
\]
has roots \(r_1,r_2,r_3\), then (up to the leading coefficient)
\[
ax^3+bx^2+cx+d = a(x-r_1)(x-r_2)(x-r_3).
\]
Expand the right-hand side:
\[
(x-r_1)(x-r_2)(x-r_3)=x^3-(r_1+r_2+r_3)x^2+(r_1r_2+r_1r_3+r_2r_3)x-r_1r_2r_3.
\]
Multiplying by \(a\) gives
\[
ax^3 - a(r_1+r_2+r_3)x^2 + a(r_1r_2+r_1r_3+r_2r_3)x - a\,r_1r_2r_3.
\]
Compare coefficients with \(ax^3+bx^2+cx+d\):
- Coefficient of \(x^2\): \(\;b = -a(r_1+r_2+r_3)\;\) so \(\;r_1+r_2+r_3 = -\dfrac{b}{a}.\)
- Coefficient of \(x\): \(\;c = a(r_1r_2+r_1r_3+r_2r_3)\;\) so \(\;r_1r_2+r_1r_3+r_2r_3 = \dfrac{c}{a}.\)
- Constant term: \(\;d = -a\,r_1r_2r_3\;\) so \(\;r_1r_2r_3 = -\dfrac{d}{a}.\)

Therefore the correct Vieta relations are
\[
\boxed{\,r_1+r_2+r_3 = -\dfrac{b}{a},\qquad r_1r_2+r_1r_3+r_2r_3 = \dfrac{c}{a},\qquad r_1r_2r_3 = -\dfrac{d}{a}\,}
\]
This corresponds to Option 2. The alternative list you gave (with the pairwise sum \(-c/a\) and product \(d/a\)) has the signs reversed and is not correct for the polynomial written as \(ax^3+bx^2+cx+d\).

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Q10. For ax² + bx + c = 0 with roots r1, r2, state Vieta’s relations.

ক) r1 + r2 = b/a, r1 r2 = −c/a

খ) r1 + r2 = −b/a, r1 r2 = c/a

গ) r1 + r2 = −c/a, r1 r2 = b/a

ঘ) r1 + r2 = c/a, r1 r2 = −b/a

ব্যাখ্যা (Explanation):

For the quadratic equation \(ax^2+bx+c=0\) (with \(a\neq0\)) let the roots be \(r_1\) and \(r_2\). Then the polynomial can be written in factored form
\[
a(x-r_1)(x-r_2)=ax^2-a(r_1+r_2)x+a\,r_1r_2.
\]
Comparing coefficients with \(ax^2+bx+c\) gives
\[
-a(r_1+r_2)=b\quad\Rightarrow\quad r_1+r_2=-\frac{b}{a},
\]
\[
a\,r_1r_2=c\quad\Rightarrow\quad r_1r_2=\frac{c}{a}.
\]
Thus Vieta’s relations are \(r_1+r_2=-\dfrac{b}{a}\) and \(r_1r_2=\dfrac{c}{a}\). (If \(a=1\) these reduce to \(r_1+r_2=-b\) and \(r_1r_2=c\).)

Q11. If there is exactly one sign change in p(x), what can be concluded?

ক) Exactly one positive real root (counting multiplicity)

খ) At most two positive real roots

গ) No positive real roots

ঘ) Exactly two positive real roots

Q12. If there are no sign changes in p(x), how many positive real roots are there?

ক) Exactly one

খ) At least one

গ) None

ঘ) Infinitely many

Q13. How do you test for negative roots via Descartes’ rule?

ক) Apply the rule to p(x^2)

খ) Count sign changes in the derivative p′(x)

গ) Apply the rule to p(−x)

ঘ) Replace x by 1/x in p(x)

Q14. For p(x) = x^3 − 6x^2 + 11x − 6, how many positive roots at most?

ক) At most 1

খ) At most 2

গ) At most 3 (indeed 1, 2, 3)

ঘ) At most 0

Q15. State Descartes’ rule of signs for positive roots.

ক) Number of positive real roots equals the number of sign changes

খ) Number of positive real roots ≤ number of sign changes and differs by an even integer

গ) Number of positive real roots ≥ number of sign changes

ঘ) Number of positive real roots equals degree minus number of sign changes

Q16. Deflate p(x) = x^3 − 7x + 6 given that x = 1 is a root; what is the quotient?

ক) x^2 − x − 6

খ) x^2 + x − 6

গ) x^2 + 2x − 7

ঘ) x^2 − 1x + 6

Q17. Is x − 2 a factor of x^3 − 3x^2 − 4x + 12?

ক) No, remainder −2 by synthetic division

খ) No, remainder 2 by synthetic division

গ) Yes, remainder 0 by synthetic division

ঘ) Yes, remainder 1 by synthetic division

Q18. In Horner’s scheme, when dividing by (x − α), what does the algorithm return?

ক) Remainder p(α) and coefficients of the quotient

খ) Only the remainder p(α)

গ) Only the leading coefficient of the quotient

ঘ) Coefficients of the original polynomial

Q19. Evaluate p(2) for p(x) = 2x^3 − x^2 + 3x − 5 using Horner’s method.

ক) 15

খ) 9

গ) 11

ঘ) 13

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Q20. Divide x^3 − 6x^2 + 11x − 6 by (x − 1) via synthetic division; what is the remainder and quotient?

ক) Remainder −1; quotient x^2 − 5x + 5

খ) Remainder 0; quotient x^2 − 5x + 6

গ) Remainder 1; quotient x^2 − 7x + 6

ঘ) Remainder 0; quotient x^2 − 6x + 11

Q21. Factor x^4 + 1 over ℝ.

ক) (x^2 + 1)^2

খ) (x^2 + √2 x + 1)(x^2 − √2 x + 1)

গ) (x^2 − 1)(x^2 + 2)

ঘ) (x^2 + x + 1)(x^2 − x + 1)

Q22. If x = 2 is a double root of p(x), what does that imply?

ক) (x − 2) divides p(x) but p′(2) ≠ 0

খ) (x − 2)^3 divides p(x) and p(2) = 0

গ) (x − 2)^2 divides p(x) and p(2) = p′(2) = 0

ঘ) p(2) ≠ 0 but p′(2) = 0

Q23. What is the remainder when dividing p(x) by (x − r)?

ক) p(r)

খ) p′(r)

গ) 0 iff r is not a root

ঘ) p(r)/(x − r)

Q24. If a + bi is a root of a real-coefficient polynomial, what other number must also be a root?

ক) −a + bi

খ) a + b

গ) a − bi

ঘ) −a − bi

Q25. State the Fundamental Theorem of Algebra.

ক) Every real polynomial factors into linear real factors

খ) Every non-constant complex polynomial has a complex root

গ) Every polynomial has only real roots

ঘ) Every constant polynomial has a complex root

Q26. In ℝ^k, if ∑ ||x_n|| converges absolutely, does ∑ x_n converge?

ক) Yes, absolute convergence implies convergence

খ) No, absolute convergence does not imply convergence

গ) Only if x_n ≥ 0 componentwise

ঘ) Only if the terms are monotone

Q27. State the Cauchy criterion for convergence of a series in a metric space.

ক) The sequence of terms x_n is Cauchy

খ) The partial sums S_N are bounded

গ) The partial sums S_N form a Cauchy sequence

ঘ) The remainders R_N → 0 implies convergence

Q28. Does ∑ 1/n converge? What about ∑ 1/n^p?

ক) ∑ 1/n diverges; ∑ 1/n^p converges iff p > 1

খ) Both converge for all p > 0

গ) Both diverge for all p

ঘ) ∑ 1/n converges; ∑ 1/n^p diverges if p > 1

Q29. What does it mean for a metric space to be complete?

ক) Every bounded sequence has a convergent subsequence

খ) Every Cauchy sequence has a convergent subsequence

গ) Every Cauchy sequence converges to a limit in the space

ঘ) Every sequence converges

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Q30. Define a Cauchy sequence in a metric space (X, d).

ক) For every ε > 0, ∃N s.t. n ≥ N ⇒ d(x_n, x_1) < ε

খ) For every ε > 0, ∃N s.t. m, n ≥ N ⇒ d(x_m, x_n) < ε

গ) For every ε > 0, ∃N s.t. d(x_n, x) < ε for all n ≥ N

ঘ) There exists ε0 > 0 with d(x_m, x_n) < ε0 for all m, n

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Algebra (87 টি প্রশ্ন )