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AlgebraQuestion and Answers: Page 357

Question Number 21422 Answers: 1 Comments: 3

Find all integer values of a such that the quadratic expression (x + a)(x + 1991) + 1 can be factored as a product (x + b)(x + c) where b and c are integers.

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{integer}\:\mathrm{values}\:\mathrm{of}\:{a}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{quadratic}\:\mathrm{expression} \\ $$$$\left({x}\:+\:{a}\right)\left({x}\:+\:\mathrm{1991}\right)\:+\:\mathrm{1}\:\mathrm{can}\:\mathrm{be}\:\mathrm{factored} \\ $$$$\mathrm{as}\:\mathrm{a}\:\mathrm{product}\:\left({x}\:+\:{b}\right)\left({x}\:+\:{c}\right)\:\mathrm{where}\:{b}\:\mathrm{and} \\ $$$${c}\:\mathrm{are}\:\mathrm{integers}. \\ $$

Question Number 21357 Answers: 1 Comments: 0

Solve : log_(2x+3) x^2 < 1

$$\mathrm{Solve}\::\:\mathrm{log}_{\mathrm{2}{x}+\mathrm{3}} {x}^{\mathrm{2}} \:<\:\mathrm{1} \\ $$

Question Number 21356 Answers: 1 Comments: 0

Solve : (2^((3x−1)/(x−1)) )^(1/3) < 8^((x−3)/(3x−7))

$$\mathrm{Solve}\::\:\sqrt[{\mathrm{3}}]{\mathrm{2}^{\frac{\mathrm{3}{x}−\mathrm{1}}{{x}−\mathrm{1}}} }\:<\:\mathrm{8}^{\frac{{x}−\mathrm{3}}{\mathrm{3}{x}−\mathrm{7}}} \\ $$

Question Number 21355 Answers: 1 Comments: 0

Solve : ∣x^2 + 3x∣ + x^2 − 2 ≥ 0

$$\mathrm{Solve}\::\:\mid{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\mid\:+\:{x}^{\mathrm{2}} \:−\:\mathrm{2}\:\geqslant\:\mathrm{0} \\ $$

Question Number 21354 Answers: 0 Comments: 4

Solve : (√(2x + 5)) + (√(x − 1)) > 8

$$\mathrm{Solve}\::\:\sqrt{\mathrm{2}{x}\:+\:\mathrm{5}}\:+\:\sqrt{{x}\:−\:\mathrm{1}}\:>\:\mathrm{8} \\ $$

Question Number 21321 Answers: 1 Comments: 0

The number of real solutions of the equation 4x^(99) + 5x^(98) + 4x^(97) + 5x^(96) + ..... + 4x + 5 = 0 is

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{4}{x}^{\mathrm{99}} \:+\:\mathrm{5}{x}^{\mathrm{98}} \:+\:\mathrm{4}{x}^{\mathrm{97}} \:+\:\mathrm{5}{x}^{\mathrm{96}} \:+ \\ $$$$.....\:+\:\mathrm{4}{x}\:+\:\mathrm{5}\:=\:\mathrm{0}\:\mathrm{is} \\ $$

Question Number 21319 Answers: 0 Comments: 0

If x, y, z are three real numbers such that x + y + z = 4 and x^2 + y^2 + z^2 = 6, then (1) (2/3) ≤ x, y, z ≤ 2 (2) 0 ≤ x, y, z ≤ 2 (3) 1 ≤ x, y, z ≤ 3 (4) 2 ≤ x, y, z ≤ 3

$$\mathrm{If}\:{x},\:{y},\:{z}\:\mathrm{are}\:\mathrm{three}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{such} \\ $$$$\mathrm{that}\:{x}\:+\:{y}\:+\:{z}\:=\:\mathrm{4}\:\mathrm{and}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \:=\:\mathrm{6}, \\ $$$$\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{2}}{\mathrm{3}}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{2} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{0}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{1}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{3} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{2}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{3} \\ $$

Question Number 21316 Answers: 0 Comments: 0

Let p = (x_1 − x_2 )^2 + (x_1 − x_3 )^2 + .... + (x_1 − x_6 )^2 + (x_2 − x_3 )^2 + (x_2 − x_4 )^2 + .... + (x_2 − x_6 )^2 + .... + (x_5 − x_6 )^2 = Σ_(1≤i<j≤6) ^6 (x_i − x_j )^2 . Then the maximum value of p if each x_i (i = 1, 2, ....., 6) has the value 0 and 1 is

Question Number 21315 Answers: 0 Comments: 0

The number of real solutions of the equation ((97 − x))^(1/4) + (x)^(1/4) = 5

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\sqrt[{\mathrm{4}}]{\mathrm{97}\:−\:{x}}\:+\:\sqrt[{\mathrm{4}}]{{x}}\:=\:\mathrm{5} \\ $$

Question Number 21314 Answers: 1 Comments: 0

Let α and β be the root of x^2 + px − (1/(2p^2 )) = 0, p ∈ R. The minimum value of α^4 + β^4 is

$$\mathrm{Let}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{be}\:\mathrm{the}\:\mathrm{root}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{px}\:−\:\frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }\:=\:\mathrm{0}, \\ $$$${p}\:\in\:{R}.\:\mathrm{The}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\alpha^{\mathrm{4}} \:+\:\beta^{\mathrm{4}} \:\mathrm{is} \\ $$

Question Number 21313 Answers: 0 Comments: 4

Let k be a real number such that the inequality (√(x − 3)) + (√(6 − x)) ≥ k has a solution then the maximum value of k is

$$\mathrm{Let}\:{k}\:\mathrm{be}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{inequality}\:\sqrt{{x}\:−\:\mathrm{3}}\:+\:\sqrt{\mathrm{6}\:−\:{x}}\:\geqslant\:{k}\:\mathrm{has}\:\mathrm{a} \\ $$$$\mathrm{solution}\:\mathrm{then}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:{k} \\ $$$$\mathrm{is} \\ $$

Question Number 21311 Answers: 0 Comments: 0

Let a and b be positive real numbers with a^3 + b^3 = a − b, and k = a^2 + 4b^2 , then (1) k < 1 (2) k >1 (3) k = 1 (4) k > 2

$$\mathrm{Let}\:{a}\:\mathrm{and}\:{b}\:\mathrm{be}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers} \\ $$$$\mathrm{with}\:{a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} \:=\:{a}\:−\:{b},\:\mathrm{and}\:{k}\:=\:{a}^{\mathrm{2}} \:+\:\mathrm{4}{b}^{\mathrm{2}} , \\ $$$$\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:{k}\:<\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:{k}\:>\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:{k}\:=\:\mathrm{1} \\ $$$$\left(\mathrm{4}\right)\:{k}\:>\:\mathrm{2} \\ $$

Question Number 21309 Answers: 0 Comments: 0

Suppose p is a polynomial with complex coefficients and an even degree. If all the roots of p are complex non-real numbers with modulus 1, prove that p(1) ∈ R iff p(−1) ∈ R.

$$\mathrm{Suppose}\:{p}\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{with}\:\mathrm{complex} \\ $$$$\mathrm{coefficients}\:\mathrm{and}\:\mathrm{an}\:\mathrm{even}\:\mathrm{degree}.\:\mathrm{If}\:\mathrm{all} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{p}\:\mathrm{are}\:\mathrm{complex}\:\mathrm{non}-\mathrm{real} \\ $$$$\mathrm{numbers}\:\mathrm{with}\:\mathrm{modulus}\:\mathrm{1},\:\mathrm{prove}\:\mathrm{that} \\ $$$${p}\left(\mathrm{1}\right)\:\in\:{R}\:\mathrm{iff}\:{p}\left(−\mathrm{1}\right)\:\in\:{R}. \\ $$

Question Number 21308 Answers: 0 Comments: 0

Find all complex numbers z such that ∣z − ∣z + 1∣∣ = ∣z + ∣z − 1∣∣

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{complex}\:\mathrm{numbers}\:{z}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mid{z}\:−\:\mid{z}\:+\:\mathrm{1}\mid\mid\:=\:\mid{z}\:+\:\mid{z}\:−\:\mathrm{1}\mid\mid \\ $$

Question Number 21307 Answers: 0 Comments: 5

Let z_1 , z_2 , z_3 be complex numbers such that (i) ∣z_1 ∣ = ∣z_2 ∣ = ∣z_3 ∣ = 1 (ii) z_1 + z_2 + z_3 ≠ 0 (iii) z_1 ^2 + z_2 ^2 + z_3 ^2 = 0 Prove that for all n ≥ 2, ∣z_1 ^n + z_2 ^n + z_3 ^n ∣ ∈ {0, 1, 2, 3}.

$$\mathrm{Let}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{be}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{such} \\ $$$$\mathrm{that} \\ $$$$\left(\mathrm{i}\right)\:\mid{z}_{\mathrm{1}} \mid\:=\:\mid{z}_{\mathrm{2}} \mid\:=\:\mid{z}_{\mathrm{3}} \mid\:=\:\mathrm{1} \\ $$$$\left(\mathrm{ii}\right)\:{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} \:\neq\:\mathrm{0} \\ $$$$\left(\mathrm{iii}\right)\:{z}_{\mathrm{1}} ^{\mathrm{2}} \:+\:{z}_{\mathrm{2}} ^{\mathrm{2}} \:+\:{z}_{\mathrm{3}} ^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{all}\:{n}\:\geqslant\:\mathrm{2}, \\ $$$$\mid{z}_{\mathrm{1}} ^{{n}} \:+\:{z}_{\mathrm{2}} ^{{n}} \:+\:{z}_{\mathrm{3}} ^{{n}} \mid\:\in\:\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3}\right\}. \\ $$

Question Number 21294 Answers: 1 Comments: 0

Let z_1 , z_2 , z_3 be complex numbers, not all real, such that ∣z_1 ∣ = ∣z_2 ∣ = ∣z_3 ∣ = 1 and 2(z_1 + z_2 + z_3 ) − 3z_1 z_2 z_3 ∈ R. Prove that max(arg z_1 , arg z_2 , arg z_3 ) ≥ (π/6) . Where 0 < arg(z_1 ), arg(z_2 ), arg(z_3 ) < 2π.

$$\mathrm{Let}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{be}\:\mathrm{complex}\:\mathrm{numbers},\:\mathrm{not} \\ $$$$\mathrm{all}\:\mathrm{real},\:\mathrm{such}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \mid\:=\:\mid{z}_{\mathrm{2}} \mid\:=\:\mid{z}_{\mathrm{3}} \mid\:=\:\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{2}\left({z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} \right)\:−\:\mathrm{3}{z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}} \:\in\:{R}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{max}\left(\mathrm{arg}\:{z}_{\mathrm{1}} ,\:\mathrm{arg}\:{z}_{\mathrm{2}} ,\:\mathrm{arg}\:{z}_{\mathrm{3}} \right)\:\geqslant \\ $$$$\frac{\pi}{\mathrm{6}}\:.\:\mathrm{Where}\:\mathrm{0}\:<\:\mathrm{arg}\left({z}_{\mathrm{1}} \right),\:\mathrm{arg}\left({z}_{\mathrm{2}} \right),\:\mathrm{arg}\left({z}_{\mathrm{3}} \right) \\ $$$$<\:\mathrm{2}\pi. \\ $$

Question Number 21293 Answers: 1 Comments: 0

Let n be an even positive integer such that (n/2) is odd and let α_0 , α_1 , ...., α_(n−1) be the complex roots of unity of order n. Prove that Π_(k=0) ^(n−1) (a + bα_k ^2 ) = (a^(n/2) + b^(n/2) )^2 for any complex numbers a and b.

$$\mathrm{Let}\:{n}\:\mathrm{be}\:\mathrm{an}\:\mathrm{even}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{such} \\ $$$$\mathrm{that}\:\frac{{n}}{\mathrm{2}}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{and}\:\mathrm{let}\:\alpha_{\mathrm{0}} ,\:\alpha_{\mathrm{1}} ,\:....,\:\alpha_{{n}−\mathrm{1}} \:\mathrm{be} \\ $$$$\mathrm{the}\:\mathrm{complex}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{of}\:\mathrm{order}\:{n}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({a}\:+\:{b}\alpha_{{k}} ^{\mathrm{2}} \right)\:=\:\left({a}^{\frac{{n}}{\mathrm{2}}} \:+\:{b}^{\frac{{n}}{\mathrm{2}}} \right)^{\mathrm{2}} \\ $$$$\mathrm{for}\:\mathrm{any}\:\mathrm{complex}\:\mathrm{numbers}\:{a}\:\mathrm{and}\:{b}. \\ $$

Question Number 21248 Answers: 0 Comments: 0

The locus of the centre of a circle which touches the given circles ∣z − z_1 ∣ = ∣3 + 4i∣ and ∣z − z_2 ∣ = ∣1 + i(√3)∣ is a hyperbola, then the length of its transverse axis is

$$\mathrm{The}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{which} \\ $$$$\mathrm{touches}\:\mathrm{the}\:\mathrm{given}\:\mathrm{circles}\:\mid{z}\:−\:{z}_{\mathrm{1}} \mid\:= \\ $$$$\mid\mathrm{3}\:+\:\mathrm{4}{i}\mid\:\mathrm{and}\:\mid{z}\:−\:{z}_{\mathrm{2}} \mid\:=\:\mid\mathrm{1}\:+\:{i}\sqrt{\mathrm{3}}\mid\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{hyperbola},\:\mathrm{then}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{its} \\ $$$$\mathrm{transverse}\:\mathrm{axis}\:\mathrm{is} \\ $$

Question Number 21247 Answers: 1 Comments: 0

If [ ] represents the greatest integer function and f(x) = x − [x] then number of real roots of the equation f(x) + f((1/x)) = 1 are infinite. True/False

$$\mathrm{If}\:\left[\:\right]\:\mathrm{represents}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer} \\ $$$$\mathrm{function}\:\mathrm{and}\:{f}\left({x}\right)\:=\:{x}\:−\:\left[{x}\right]\:\mathrm{then} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$${f}\left({x}\right)\:+\:{f}\left(\frac{\mathrm{1}}{{x}}\right)\:=\:\mathrm{1}\:\mathrm{are}\:\mathrm{infinite}. \\ $$$$\boldsymbol{\mathrm{True}}/\boldsymbol{\mathrm{False}} \\ $$

Question Number 21236 Answers: 1 Comments: 0

If (1/a) + (1/(2a)) + (1/(3a)) = (1/(b^2 − 2b)) a and b are positive integers Find minimum value of a + b

$$\mathrm{If}\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{a}}\:+\:\frac{\mathrm{1}}{\mathrm{3}{a}}\:=\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} \:−\:\mathrm{2}{b}} \\ $$$${a}\:\mathrm{and}\:{b}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{integers} \\ $$$$\mathrm{Find}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:{a}\:+\:{b} \\ $$

Question Number 21235 Answers: 1 Comments: 0

For any integer k, let α_k = cos (((kπ)/7)) + i sin (((kπ)/7)), where i = (√(−1)). The value of the expression ((Σ_(k=1) ^(12) ∣α_(k+1) − α_k ∣)/(Σ_(k=1) ^3 ∣α_(4k−1) − α_(4k−2) ∣)) is

$$\mathrm{For}\:\mathrm{any}\:\mathrm{integer}\:{k},\:\mathrm{let}\:\alpha_{{k}} \:=\:\mathrm{cos}\:\left(\frac{{k}\pi}{\mathrm{7}}\right)\:+ \\ $$$${i}\:\mathrm{sin}\:\left(\frac{{k}\pi}{\mathrm{7}}\right),\:\mathrm{where}\:{i}\:=\:\sqrt{−\mathrm{1}}.\:\mathrm{The}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{expression}\:\frac{\underset{{k}=\mathrm{1}} {\overset{\mathrm{12}} {\sum}}\mid\alpha_{{k}+\mathrm{1}} \:−\:\alpha_{{k}} \mid}{\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\mid\alpha_{\mathrm{4}{k}−\mathrm{1}} \:−\:\alpha_{\mathrm{4}{k}−\mathrm{2}} \mid}\:\mathrm{is} \\ $$

Question Number 21234 Answers: 1 Comments: 0

Let f(x) = ax^2 + bx + c, where a, b, c are real numbers. If the numbers 2a, a + b, and c are all integers, then the number of integral values between 1 and 5 that f(x) can take is

$$\mathrm{Let}\:{f}\left({x}\right)\:=\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c},\:\mathrm{where}\:{a},\:{b},\:{c} \\ $$$$\mathrm{are}\:\mathrm{real}\:\mathrm{numbers}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{numbers}\:\mathrm{2}{a}, \\ $$$${a}\:+\:{b},\:\mathrm{and}\:{c}\:\mathrm{are}\:\mathrm{all}\:\mathrm{integers},\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{integral}\:\mathrm{values}\:\mathrm{between}\:\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{5}\:\mathrm{that}\:{f}\left({x}\right)\:\mathrm{can}\:\mathrm{take}\:\mathrm{is} \\ $$

Question Number 21230 Answers: 1 Comments: 0

For each positive integer n, consider the highest common factor h_n of the two numbers n! + 1 and (n + 1)!. For n < 100, find the largest value of h_n .

$$\mathrm{For}\:\mathrm{each}\:\mathrm{positive}\:\mathrm{integer}\:{n},\:\mathrm{consider} \\ $$$$\mathrm{the}\:\mathrm{highest}\:\mathrm{common}\:\mathrm{factor}\:{h}_{{n}} \:\mathrm{of}\:\mathrm{the}\:\mathrm{two} \\ $$$$\mathrm{numbers}\:{n}!\:+\:\mathrm{1}\:\mathrm{and}\:\left({n}\:+\:\mathrm{1}\right)!.\:\mathrm{For}\:{n}\:<\:\mathrm{100}, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{value}\:\mathrm{of}\:{h}_{{n}} . \\ $$

Question Number 21219 Answers: 1 Comments: 0

(A) If ∣w∣ = 2, then the set of points z = w − (1/w) is contained in or equal to (B) If ∣w∣ = 1, then the set of points z = w + (1/w) is contained in or equal to Options for both A and B: (p) An ellipse with eccentricity (4/5) (q) The set of points z satisfying Im z = 0 (r) The set of points z satisfying ∣Im z∣ ≤ 1 (s) The set of points z satisfying ∣Re z∣ ≤ 2 (t) The set of points z satisfying ∣z∣ ≤ 3

$$\left(\mathrm{A}\right)\:\mathrm{If}\:\mid{w}\mid\:=\:\mathrm{2},\:\mathrm{then}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{points} \\ $$$${z}\:=\:{w}\:−\:\frac{\mathrm{1}}{{w}}\:\mathrm{is}\:\mathrm{contained}\:\mathrm{in}\:\mathrm{or}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left(\mathrm{B}\right)\:\mathrm{If}\:\mid{w}\mid\:=\:\mathrm{1},\:\mathrm{then}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{points} \\ $$$${z}\:=\:{w}\:+\:\frac{\mathrm{1}}{{w}}\:\mathrm{is}\:\mathrm{contained}\:\mathrm{in}\:\mathrm{or}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\mathrm{Options}\:\mathrm{for}\:\mathrm{both}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}: \\ $$$$\left(\mathrm{p}\right)\:\mathrm{An}\:\mathrm{ellipse}\:\mathrm{with}\:\mathrm{eccentricity}\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\left(\mathrm{q}\right)\:\mathrm{The}\:\mathrm{set}\:\mathrm{of}\:\mathrm{points}\:{z}\:\mathrm{satisfying}\:\mathrm{Im}\:{z} \\ $$$$=\:\mathrm{0} \\ $$$$\left(\mathrm{r}\right)\:\mathrm{The}\:\mathrm{set}\:\mathrm{of}\:\mathrm{points}\:{z}\:\mathrm{satisfying}\:\mid\mathrm{Im}\:{z}\mid \\ $$$$\leqslant\:\mathrm{1} \\ $$$$\left(\mathrm{s}\right)\:\mathrm{The}\:\mathrm{set}\:\mathrm{of}\:\mathrm{points}\:{z}\:\mathrm{satisfying}\:\mid\mathrm{Re}\:{z}\mid \\ $$$$\leqslant\:\mathrm{2} \\ $$$$\left(\mathrm{t}\right)\:\mathrm{The}\:\mathrm{set}\:\mathrm{of}\:\mathrm{points}\:{z}\:\mathrm{satisfying}\:\mid{z}\mid\:\leqslant\:\mathrm{3} \\ $$

Question Number 21168 Answers: 1 Comments: 0

Let f(x) = ∣x − 1∣ + ∣x − 2∣ + ∣x − 3∣, then find the value of k for which f(x) = k has 1. no solution 2. only one solution 3. two solutions of same sign 4. two solutions of opposite sign

$$\mathrm{Let}\:{f}\left({x}\right)\:=\:\mid{x}\:−\:\mathrm{1}\mid\:+\:\mid{x}\:−\:\mathrm{2}\mid\:+\:\mid{x}\:−\:\mathrm{3}\mid, \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{k}\:\mathrm{for}\:\mathrm{which}\:{f}\left({x}\right) \\ $$$$=\:{k}\:\mathrm{has} \\ $$$$\mathrm{1}.\:\mathrm{no}\:\mathrm{solution} \\ $$$$\mathrm{2}.\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution} \\ $$$$\mathrm{3}.\:\mathrm{two}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{same}\:\mathrm{sign} \\ $$$$\mathrm{4}.\:\mathrm{two}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{opposite}\:\mathrm{sign} \\ $$

Question Number 21137 Answers: 1 Comments: 0

If the minimum value of ∣z+1+i∣ + ∣z−1−i∣ + ∣2 − z∣ + ∣3 − z∣ is k then (k − 8) equals

$$\mathrm{If}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mid{z}+\mathrm{1}+{i}\mid\:+\:\mid{z}−\mathrm{1}−{i}\mid\:+\:\mid\mathrm{2}\:−\:{z}\mid\:+\:\mid\mathrm{3}\:−\:{z}\mid\:\mathrm{is} \\ $$$${k}\:\mathrm{then}\:\left({k}\:−\:\mathrm{8}\right)\:\mathrm{equals} \\ $$

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