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New answer posted

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Maths Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

What is the length of the foot of the perpendicular drawn from the point $P (3, 4, 5)$ on the $y$ -axis?

(a) $\sqrt[]{41}$

(b) $\sqrt[]{34}$

(c) 5

(d) None of these

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} O n y - a x i s, x = 0 a n d z = 0 g i v e n point i s P (3, 4, 5) \\ ∴ T h e point A i s (0, 4, 0) \\ ∴ P A = \sqrt[]{{(0 - 3)}^{2} + {(4 - 4)}^{2} + {(0 - 5)}^{2}} = \sqrt[]{9 + 0 + 25} = \sqrt[]{34} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

The distance of point $P (3, 4, 5)$ from the $y z$ -plane is

(a) 3 units

(b) 4 units

(c) 5 units

(d) 550

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n point i s P (3, 4, 5) \\ ∴ Distance o f P f r o m y z p l a n e = \sqrt[]{{(0 - 3)}^{2} + {(4 - 4)}^{2} + {(5 - 5)}^{2}} = \sqrt[]{9} = 3 u n i t s \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Six

(i) If $- 4 x \geq 12$ , then $x \dots - 3$ .

(ii) If $\frac{3}{4} - x \leq - 3,$ then $x \dots 4$ .

(iii) If $\frac{2}{x + 2} > 0,$ then $x \dots - 2$ .

(iv) If $x > - 5$ , then $4 x \dots - 20$ .

(v) If $x > y$ and $z < 0$ , then $- x z \dots - y z$ .

(vi) If $p > 0$ and $q < 0$ , then $p - q \dots p$ .

(vii) If $x + 2 > 5$ , then $x \dots - 7$ or $x \dots 3$ .

(viii)If $- 2 x + 1 \geq 9$ , then $x \dots - 4$ .

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} (i) - 4 x \geq 12 \Rightarrow x \leq - 3 \\ H e n c e, t h e f i l l e r i s (\leq) \\ (i i) I f - \frac{3}{4} x \leq - 3 \Rightarrow x \geq 3 x * \frac{4}{3} \Rightarrow x \geq 4 \\ H e n c e, t h e f i l l e r i s (\geq) \\ (i i i) I f \frac{2}{x + 2} > 0 \Rightarrow x > - 2 \\ H e n c e, t h e f i l l e r i s (>) \\ (i v) I f x > - 5 \Rightarrow 4 x > - 20 \\ H e n c e, t h e f i l l e r i s (>) \\ (v) I f x > y a n d z < 0, t h e n \\ x z < y z \Rightarrow - x z > - y z \\ H e n c e, t h e f i l l e r i s (>) \\ (v i) I f p > 0 a n d q < 0, t h e n \\ p - q > p \\ H e n c e, t h e f i l l e r i s (>) \\ (v i i) I f | x + 2 | > 5 t h e n \\ x + 2 < - 5 o r x + 2 > 5 \\ \Rightarrow x < - 5 - 2 o r x > 5 - 2 \\ \Rightarrow x < - 7 o r x > 3 \\ S o, x \in (- \infty, - 7) \cup (3, \infty) \\ H e n c e, t h e f i l l e r i s (<) o r (>) \\ (v i i i) I f - 2 x + 1 \geq 9 t h e n \\ \Rightarrow - 2 x \geq 9 - 1 \Rightarrow - 2 x \geq 8 \Rightarrow 2 x \leq - 8 \Rightarrow x \leq - 4 \\ H e n c e, t h e f i l l e r i s (\leq) \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Six

(i) If $x < y$ and $b < 0$ , then $\frac{x}{b} > \frac{y}{b} .$

(ii) If $x y > 0$ , then $x > 0$ and $y < 0$ .

(iii) If $x y > 0$ , then $x < 0$ and $y < 0$ .

(iv) If $x y < 0$ , then $x < 0$ and $y < 0$ .

(v) If $x < - 5$ and $x < - 2$ , then $x \in (- \infty, - 5)$ .

(vi) If $x < - 5$ and $x > 2$ , then $x \in (- 5, 2)$ .

(vii) If $x > - 2$ and $x < 9$ , then $x \in (- 2, 9)$ .

(viii) If $? x ? > 5$ , then $x \in (- \infty, - 5) \cup (5, \infty) .$

(ix) If $? x ? \leq 4$ , then $x \in [- 4, 4] .$

(x) Graph of $x < 3$ corresponds to Fig 6.12.

(xi) Graph of $x \geq 0$ corresponds to Fig 6.13.

(xii) Graph of $y \leq 0$ corresponds to Fig 6.14.

(xiii) Solution set of $x \geq 0$ and $y \leq 0$ &nbs

...more

(i) If $x < y$ and $b < 0$ , then $\frac{x}{b} > \frac{y}{b} .$

(ii) If $x y > 0$ , then $x > 0$ and $y < 0$ .

(iii) If $x y > 0$ , then $x < 0$ and $y < 0$ .

(iv) If $x y < 0$ , then $x < 0$ and $y < 0$ .

(v) If $x < - 5$ and $x < - 2$ , then $x \in (- \infty, - 5)$ .

(vi) If $x < - 5$ and $x > 2$ , then $x \in (- 5, 2)$ .

(vii) If $x > - 2$ and $x < 9$ , then $x \in (- 2, 9)$ .

(viii) If $? x ? > 5$ , then $x \in (- \infty, - 5) \cup (5, \infty) .$

(ix) If $? x ? \leq 4$ , then $x \in [- 4, 4] .$

(x) Graph of $x < 3$ corresponds to Fig 6.12.

(xi) Graph of $x \geq 0$ corresponds to Fig 6.13.

(xii) Graph of $y \leq 0$ corresponds to Fig 6.14.

(xiii) Solution set of $x \geq 0$ and $y \leq 0$ corresponds to Fig 6.15.

(xiv) Solution set of $x \geq 0$ and $y \leq 1$ corresponds to Fig 6.16.

(xv) Solution set of $x + y \geq 0$ corresponds to Fig 6.17.

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0 Follower 9 Views

Payal Gupta

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} (i) I f x < y a n d b < 0 \\ \Rightarrow \frac{x}{b} > \frac{y}{b} \\ H e n c e, s t a t e m e n t (i) i s F a l s e . \\ (i i) I f x, y > 0 t h e n x > 0, y > 0 o r x < 0, y < 0 \\ H e n c e, s t a t e m e n t (i i) i s F a l s e . \\ (i i i) I f x y > 0 t h e n x < 0, a n d y < 0 \\ H e n c e, s t a t e m e n t (i i i) i s T r u e . \\ (i v) I f x y < 0 t h e n x < 0, y > 0 o r x > 0, y < 0 \\ H e n c e, s t a t e m e n t (i v) i s F a l s e . \\ (v) I f x < - 5 a n d x < - 2 \Rightarrow x \in (- \infty, - 5) \\ H e n c e, s t a t e m e n t (v) i s T r u e . \\ (v i) I f x < - 5 a n d x > 2 t h e n x h a s n o v a l u e . \\ H e n c e, s t a t e m e n t (v i) i s F a l s e . \\ (v i i) I f x > - 2 a n d x < 9 \Rightarrow x \in (- 2, 9) \\ H e n c e, s t a t e m e n t (v i i) i s T r u e . \\ (v i i i) I f | x | > 5 t h e n x < - 5 o r x > 5 \\ \Rightarrow x \in (- \infty, - 5) \cup (5, \infty) \\ H e n c e, s t a t e m e n t (v i i i) i s F a l s e . \\ (i x) I f | x | \leq 4 t h e n - 4 \leq x \leq 4 \\ \Rightarrow x \in [- 4, 4] \\ H e n c e, s t a t e m e n t (i x) i s T r u e . \\ (x) T h e g i v e n g r a p h r e p r e s e n t s x \leq 3 \\ H e n c e, s t a t e m e n t (x) i s F a l s e . \\ (x i) T h e g i v e n g r a p h r e p r e s e n t s x \geq 0 \\ H e n c e, s t a t e m e n t (x i) i s T r u e . \\ (x i i) T h e g i v e n g r a p h r e p r e s e n t s y \geq 0 \\ H e n c e, s t a t e m e n t (x i i) i s &thins \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Five

If $f (z) = \frac{2 z - 7}{1 - z}$ , where $z = 1 + 2 i$ , then $f (z)$ is:

(a) $\frac{2}{z}$

(b) $z$

(c) $2 z$

(d) None of these

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z = 1 + 2 i \\ ∴ | z | = \sqrt[]{{(1)}^{2} + {(2)}^{2}} = \sqrt[]{5} \\ N o w f (z) = \frac{7 - z}{1 - z^{2}} \\ = \frac{7 - (1 + 2 i)}{1 - {(1 + 2 i)}^{2}} = \frac{7 - 1 - 2 i}{1 - 1 - 4 i^{2} - 4 i} = \frac{6 - 2 i}{4 - 4 i} \\ = \frac{3 - i}{2 - 2 i} = \frac{3 - i}{2 - 2 i} * \frac{2 + 2 i}{2 + 2 i} = \frac{6 + 6 i - 2 i - 2 i^{2}}{4 - 4 i^{2}} \\ = \frac{6 + 4 i + 2}{4 + 4} = \frac{8 + 4 i}{8} = 1 + \frac{1}{2} i \\ S o, | f (z) | = \sqrt[]{{(1)}^{2} + {(\frac{1}{2})}^{2}} = \sqrt[]{1 + \frac{1}{4}} = \frac{\sqrt[]{5}}{2} = \frac{| z |}{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Five

The value of $arg (x)$ when $x < 0$ is:

(a) $0$

(b) $\frac{π}{2}$

(c) $π$

(d) None of these

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z = - x + 0 i a n d x < 0 \\ ∴ | z | = \sqrt[]{{(- 1)}^{2} + {(0)}^{2}} = 1, x < 0 \\ Since, t h e point (- x, 0) l i e s o n t h e n e g a t i v e s i d e o f t h e r e a l a x i s (? x < 0) \\ ∴ Principle argument (z) = π \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Five

The real value of $θ$ for which the expression $\frac{1 + i c o s θ}{1 - i c o s θ}$ is a real number is:

(a) $\frac{π}{4} + n π$

(b) $n π + \frac{(2 n + 1) π}{4}$

(c) $\pm \frac{π}{2} + n π$

(d) None of these

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z = \frac{1 + i c o s θ}{1 - 2 i c o s θ} = \frac{1 + i c o s θ}{1 - 2 i c o s θ} * \frac{1 + 2 i c o s θ}{1 + 2 i c o s θ} \\ = \frac{1 + 2 i c o s θ + i c o s θ + 2 i^{2} c o s^{2} θ}{1 - 4 i^{2} c o s^{2} θ} \\ = \frac{1 + 3 i c o s θ - 2 c o s^{2} θ}{1 + 4 c o s^{2} θ} = \frac{1 - 2 c o s^{2} θ}{1 + 4 c o s^{2} θ} + \frac{3 c o s θ}{1 + 4 c o s^{2} θ} i \\ I f z i s a r e a l n u m b e r, t h e n \\ \Rightarrow \frac{3 c o s θ}{1 + 4 c o s^{2} θ} = 0 \\ \Rightarrow 3 c o s θ = 0 \Rightarrow c o s θ = 0 \\ ∴ θ = (2 n + 1) \frac{π}{2}, n \in N . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

a year ago

Maths Exemplar Solution

$? z_{1} + z_{2} ? = ? z_{1} ? + ? z_{2} ?$ is possible if:

(a) $z_{2} = \frac{1}{z_{1}}$

(b) $z_{2} = \frac{1}{{\overset{?}{z}}_{1}}$

(c) $arg (z_{1}) = arg (z_{2})$

(d) $z_{1} = z_{2}$ $(d)$

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z_{1} = r_{1} (c o s θ_{1} + i s i n θ_{1}) a n d z_{2} = r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ Since | z_{1} + z_{2} | = | z_{1} | + | z_{2} | \\ z_{1} + z_{2} = r_{1} c o s θ_{1} + i r_{1} s i n θ_{1} + r_{2} c o s θ_{2} + i r_{2} s i n θ_{2} \\ | z_{1} + z_{2} | = \sqrt[]{\begin{array}{l} r_{1}^{2} c o s^{2} θ_{1} + r_{2}^{2} c o s^{2} θ_{2} + 2 r_{1} r_{2} c o s θ_{1} c o s θ_{2} \\ + r_{1}^{2} s i n^{2} θ_{1} + r_{2}^{2} s i n^{2} θ_{2} + 2 r_{1} r_{2} s i n θ_{1} s i n θ_{2} \end{array}} \\ = \sqrt[]{r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} c o s (θ_{1} - θ_{2})} \\ B u t | z_{1} + z_{2} | = | z_{1} | + | z_{2} | \\ S o \sqrt[]{r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} c o s (θ_{1} - θ_{2})} = r_{1} + r_{2} \\ S q u a r i n g b o t h s i d e s, w e g e t \\ r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} c o s (θ_{1} - θ_{2}) = r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} \\ \Rightarrow 2 r_{1} r_{2} - 2 r_{1} r_{2} c o s (θ_{1} - θ_{2}) = 0 \\ \Rightarrow 1 - c o s (θ_{1} - θ_{2}) = 0 \Rightarrow c o s (θ_{1} - θ_{2}) = 1 \\ \Rightarrow θ_{1} - θ_{2} = 0 \Rightarrow θ_{1} = θ_{2} \\ S o, a r g (z_{1}) = a r g (z_{2}) \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Five

If $z$ is a complex number, then

(a) $∣ z ∣^{2} > ∣ z ∣^{2}$

(b) $∣ z ∣^{2} = ∣ z ∣^{2}$

(c) $∣ z ∣^{2} < ∣ z ∣^{2}$

(d) $∣ z ∣^{2} \geq ∣ z ∣^{2}$

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z = x + y i \\ | z | = | x + y i | a n d {| z |}^{2} = {| x + y i |}^{2} \\ \Rightarrow {| z |}^{2} = x^{2} + y^{2} \dots (i) \\ N o w z^{2} = x^{2} + y^{2} i^{2} + 2 x y i \\ \Rightarrow z^{2} = x^{2} - y^{2} + 2 x y i \\ | z^{2} | = \sqrt[]{{(x^{2} - y^{2})}^{2} + {(2 x y)}^{2}} = \sqrt[]{x^{4} + y^{4} - 2 x^{2} y^{2} + 4 x^{2} y^{2}} \\ = \sqrt[]{x^{4} + y^{4} + 2 x^{2} y^{2}} = \sqrt[]{{(x^{2} + y^{2})}^{2}} \\ S o | z^{2} | = x^{2} + y^{2} = {| z |}^{2} \\ S o | z^{2} | = {| z |}^{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Five

The complex number $z$ which satisfies the condition $\frac{1 + i z}{1 - i z} = 1$ lies on Circle

(a) Circle $x^{2} + y^{2} = 1$

(b) The x-axis

(c) The y-axis

(d) The line $x + y = 1$ .

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | \frac{i + z}{i - z} | = 1 \\ L e t z = x + y i \\ ∴ | \frac{i + x + y i}{i - x - y i} | = 1 \Rightarrow | \frac{x + (y + 1) i}{- x - (y - 1) i} | = 1 \\ \Rightarrow | x + (y + 1) i | = | - x - (y - 1) i | \\ \Rightarrow \sqrt[]{x^{2} + {(y + 1)}^{2}} = \sqrt[]{x^{2} + {(y - 1)}^{2}} \\ S q u a r i n g b o t h s i d e s, w e g e t \\ \Rightarrow x^{2} + {(y + 1)}^{2} = x^{2} + {(y - 1)}^{2} \\ \Rightarrow {(y + 1)}^{2} = {(y - 1)}^{2} \\ \Rightarrow y^{2} + 2 y + 1 = y^{2} - 2 y + 1 \Rightarrow 2 y = - 2 y \\ \Rightarrow 4 y = 0 \Rightarrow y = 0 \Rightarrow x - a x i s \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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