Match the following -Vbeginarray -c- c- Whine- Column-I - - Column -IIMhlineB - - z -2-4 - Q - Part of - a - circlell hlineC - z -frac pi 4- R - Full circle lllhlineD - z - t - i - 2 t lin R - S - Straight Line lllhlinelendarray A- A - R - B - q - C - s - D - PB- A - q - B - r - C - s - D - PC- A - p - B - s - C - r - D - qD- A - q - B - s - C - r - D - P

The correct option is B A → q, B → r, C → s , D → P(A)→ (Q), arg z+1/z 1=π/4 will be half the arc of the circle (B)→ (R),|z 2|=4 will be a full circle (C) ...

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Byju's Answer

Standard XII

Mathematics

Parametric Equation of a Circle

Match the fol...

Question

Match the following :
$\begin{matrix} C o l u m n - I & C o l u m n - I I (A) & a r g \frac{z + 1}{z - 1} = \frac{π}{4} & (P) & P a r a b o l a (B) & | z - 2 | = 4 & (Q) & P a r t o f a c i r c l e (C) & a r g z = \frac{π}{4} & (R) & F u l l c i r c l e (D) & z = t + i t^{2} (t \in R) & (S) & S t r a i g h t L i n e \end{matrix}$

A \to R, B \to q, C \to s, D \to P

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A \to q, B \to r, C \to s, D \to P

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A \to q, B \to s, C \to r, D \to P

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A \to p, B \to s, C \to r, D \to q

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Solution

The correct option is B $A \to q, B \to r, C \to s, D \to P$
$(A) \to (Q), a r g \frac{z + 1}{z - 1} = \frac{π}{4}$ will be half the arc of the circle
$(B) \to (R), | z - 2 | = 4$ will be a full circle
$(C) \to (S), a r g z = \frac{π}{4}$ will be a line x =y (such that x > 0, y > 0)
$(D) \to (P), x + i y = t + i t^{2}$
$\Rightarrow x = t, y = t^{2}$
On eliminating t, we get
$y = x^{2}$ which is a parabola

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Similar questions

$\begin{matrix} Column I & Column II (a) & a r g (\frac{z + 1}{z - 1}) = \frac{π}{4} & (p) & Parabola (b) & z = \frac{3 i - 1}{2 + i t} (t ϵ R) & (q) & Part of a circle (c) & a r g z = \frac{π}{4} & (r) & Full circle (d) & z = t + i t^{2} (t ϵ R) & (s) & Line \end{matrix}$

Which of the following is correct?

Q. Answer the following by appropriately matching the lists based on the information in Column I and Column II

\begin{matrix} C o l u m n I & C o l u m n I I a . y = f (x) is given by x = t^{5} - 5 t^{3} - 20 t + 7 and y = 4 t^{3} - 3 t^{2} - 18 t + 3. Then - 5 \times \frac{d y}{d x} at t = 1 & p. 0 b . Let P (x) be a polynomial of degree 4, with P (2) = - 1, P^{'} (2) = 0, P^{''} (2) = 2, P^{'''} (2) = - 12 and P^{i v} (2) = 24, then P^{''} (3) is & q. - 2 c . y = \frac{1}{x}, then \frac{\frac{d y}{\sqrt{1 + y^{4}}}}{\frac{d x}{\sqrt{1 + x^{4}}}} & r . 2 d . f (\frac{2 x + 3 y}{5}) = \frac{2 f (x) + 3 f (y)}{5} and f^{'} (0) = p and f (0) = q . Then, f^{''} (0) is & s. - 1 \end{matrix}

Q. Match the Column-I with Column-II:

Column-I (Compound)	Column-II (Oxidation state of nitrogen)
(a) $M g_{3} N_{2}$	(P) -1
(b) $N O$	(q) +2
(c) $(N_{2} H_{5})_{2} S O_{4}$	(r) -2
(d) $N H_{2} O H$	(s) -3

Q. Match the Column-I with Column-II. Choose the correct option among the following
Column – I Column – II
(A) Borax (p)

s p^{3}

hybridization of B
(B)

B F_{3}

(q)

s p^{2}

hybridization of B
(C) Diborane (r) Triangular unit
(D) Boric acid (s) Tetrahedral unit

Q. Match the following by appropriately matching the lists based on the information given in

Column I

and

Column II

\begin{matrix} C o l u m n I & C o l u m n I I (T y p e o f △ A B C) a. cot \frac{A}{2} = \frac{b + c}{a} & p. always right angled b. a tan A + b tan B = (a + b) tan \frac{A + B}{2} & q. always isosceles c. a cos A = b cos B & r. may be right angled d. cos A = \frac{sin B}{2 sin C} & s. may be right angled isosceles \end{matrix}

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Match the following : Column−IColumn−II(A)argz+1z−1=π4(P)Parabola(B)|z−2|=4(Q)Part of a circle(C)argz=π4(R)Full circle(D)z=t+it2(t∈R)(S)Straight Line

The correct option is B A→q,B→r,C→s,D→P(A)→(Q),argz+1z−1=π4 will be half the arc of the circle (B)→(R),|z−2|=4 will be a full circle (C)→(S),argz=π4 will be a line x =y (such that x > 0, y > 0) (D)→(P),x+iy=t+it2 ⇒x=t,y=t2 On eliminating t, we get y=x2 which is a parabola