Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 3 Theory of Equations Ex 3.7 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.7

Choose the most suitable answer.

Question 1.

A zero of x³ + 64 is:

(a) 0

(b) 4

(c) 4i

(d) -4

Solution:

(d) -4

Hint:

x³ = -64

x³ = -4 × -4 × -4 = (-4)³

x = -4

Question 2.

If f and g are polynomials of degrees m and n respectively, and if h(x) = (f o g) (x), then the degree of h is:

(a) mn

(b) m + n

(c) m^{n}

(d) n^{m}

Solution:

(a) mn

Hint:

Let f(x) = x^{m} and g(x) = x^{n}

degree = m, degree = n

(f o g) (x) = f(g(x)) = f(x^{n}) = (x^{n})^{m} =x^{mn}

degree = mn

Question 3.

A polynomial equation in x of degree n always has:

(a) n distinct roots

(b) n real roots

(c) n imaginary roots

(d) at most one root.

Solution:

(a) n distinct roots

Question 4.

If α, β and γ are the roots of x³ + px² + qx + r, then Σ \(\frac{1}{α}\) is:

(a) –\(\frac{q}{r}\)

(b) –\(\frac{p}{r}\)

(c) \(\frac{q}{r}\)

(d) –\(\frac{q}{p}\)

Solution:

(a) –\(\frac{q}{r}\)

Hint:

x³ + px² + qx + r = 0

α, β, γ are the roots

α + β + γ = -p

αβ + βγ + γα = q

αβγ = -r

Σ \(\frac{1}{α}\) = \(\frac{1}{α}\) + \(\frac{1}{β}\) + \(\frac{1}{γ}\) = \(\frac{βγ+αβ+αγ}{αβγ}\) = –\(\frac{q}{r}\)

Question 5.

According to the rational root theorem, which number is not possible rational root of 4x^{7} + 2x^{7} – 10x³ – 5?

(a) -1

(b) \(\frac{5}{4}\)

(c) \(\frac{4}{5}\)

(d) 5

Solution:

(c) \(\frac{4}{5}\)

Hint:

By rational root of theorem,

\(\frac{p}{q}\) is a root of a polynomial a_{0} = -5, a_{n} = 4 and (4, -5) = 1, then p must divide 5 and q must divide 4.

Possible values of p are +1, -1, +5, -5

Possible values of q are +1, -1, 4, -4

∴ \(\frac{4}{5}\) is not possible rational roots.

Question 6.

The polynomial x³ – kx² + 9x has three real roots if and only if, k satisfies:

(a) |k | ≤ 6

(b) k = 0

(c) |k| > 6

(d) |k| ≥ 6

Solution:

(d) |k| ≥ 6

Hint:

Δ ≥ 0

k² – 4(1)(9) ≥ 0

k² ≥ 36

|k| ≥ 6

Question 7.

The number of real numbers in [0, 2π] satisfying sin^{4}x – 2 sin²x + 1 is:

(a) 2

(b) 4

(c) 1

(d) ∞

Solution:

(a) 2

Hint:

sin^{4}x – 2sin²x + 1 = 0

Put sin²x = t [t² – 2t + 1 = 0 ⇒ (t – 1)² = 0]

t = 1, t = 1

sin²x = 1

sin x = ± 1

∴ sin x = 1, sin x = -1

sin x = sin \(\frac{π}{2}\), sin x = sin (π + \(\frac{π}{2}\))

x = \(\frac{π}{2}\), x = \(\frac{3π}{2}\)

Number of real numbers in [0, 2, π] is 2.

Question 8.

If x³ + 12x² + 10ax + 1999 definitely has a positive root, if and only if:

(a) a ≥ 0

(b) a > 0

(c) a < 0

(d) a ≤ 0

Solution:

(c) a < 0

Hint:

∴ If a < 0, then only we can get one sign change.

Question 9.

The polynomial x³ + 2x + 3 has:

(a) one negative and two real roots

(b) one positive and two imaginary roots

(c) three real roots

(d) no solution

Solution:

(a) one negative and two real roots

Hint:

p(x) = x³ + 2x + 3, no sign changes ⇒ no positive real roots.

p(-x) = -x³ – 2x + 3, one sign changes ⇒ one negative real roots.

∴ It has 1 negative root and 2 imaginary roots.

Question 10.

The number of positive roots of the polynomials \(\sum_{r=0}^{n}\) ^{n}C_{r} (-1)^{r} x^{r} is

(a) 0

(b) n

(c) <n

(d) r

Solution:

(b) n

Hint:

\(\sum_{r=0}^{n}\) ^{n}C_{r} (-1)^{r} x^{r} = x^{n} – ^{n}C_{1} x^{n-1} + ^{n}C_{2} x^{n-2} + …. + (-1)^{n}

P(x) has n changes

∴ It has n positive changes

P(-x) has no changes

∴ no negative roots.