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(i) If x<y and b<0 , then xb>yb.

(ii) If xy>0 , then x>0 and y<0 .

(iii) If xy>0 , then x<0 and y<0 .

(iv) If xy<0 , then x<0 and y<0 .

(v) If x<−5 and x<−2 , then x∈(−∞,−5) .

(vi) If x<−5 and x>2 , then x∈(−5,2) .

(vii) If x>−2 and x<9 , then x∈(−2,9) .

(viii) If ?x?>5 , then x∈(−∞,−5)∪(5,∞).

(ix) If ?x?≤4 , then x∈[−4,4].

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

(xi) Graph of x≥0 corresponds to Fig 6.13.

(xii) Graph of y≤0 corresponds to Fig 6.14.

(xiii) Solution set of x≥0 and y≤0 corresponds to Fig 6.15.

(xiv) Solution set of x≥0 and y≤1 corresponds to Fig 6.16.

(xv) Solution set of x+y≥0 corresponds to Fig 6.17.

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    Payal Gupta | Contributor-Level 10

    2 months ago

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

    ( i ) I f     x < y     a n d     b < 0           ⇒             x b > 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     ⇒ x ∈ ( − ∞ , − 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     ⇒ x ∈ ( − 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                     ⇒             x ∈ ( − ∞ , − 5 ) ∪ ( 5 , ∞ )                     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 | ≤ 4     t h e n     − 4 ≤ x ≤ 4                 ⇒             x ∈ [ − 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 ≤ 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 ≥ 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 ≥ 0                   H e n c e ,     s t a t e m e n t     ( x i i )     i s &thins

Similar Questions for you

If one of the diameters of the circle x 2 + y 2 − 2 2 x − 6 2 y + 1 4 = 0  is a chord of circle. ( x − 2 2 ) 2 + ( y − 2 2 ) 2 = r 2 ,  then the value of r2 is equal to…………….
A
alok kumar singh

  x 2 + y 2 − 2 2 x − 6 2 y + 1 4 = 0

 centre ( 2 , 3 2 )

radius  ( ( 2 ) 2 + ( 3 2 ) 2 − 1 4 ) 1 / 2

= ( 2 + 1 8 − 1 4 ) 1 / 2 = ( 6 )

( x − 2 2 ) 2 + ( y − 2 2 ) 2 = r 2

centre  ( 2 2 , 2 2 )

OA = ( 2 2 − 2 ) 2 + ( 2 2 − 3 2 ) 2 = 2 + 2 = 2                  

r2 = ( 6 ) 2 + ( 2 ) 2 = 6 + 4 = 1 0  

 

LE the common tangents to the curves 4(x2+y2)=9 and y2 = 4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and l respectively denote the eccentricity and the length of the latus rectum of this ellipse, then le2 is equal to
P
Payal Gupta

Let y = mx + c is the common tangent

so  c=1m=±321+m2⇒m2=13

so equation of common tangents will be

y=±13x±3

which intersects at Q (3, 0)

Major axis and minor axis of ellipse are 12 and 6. So eccentricity

e2=1−14=34

and length of latus rectum

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Hence

le2=33/4=4

if z 2 + z + 1 = 0 ,     z ∈ C ,     t h e n     | ∑ n = 1 1 5 ( z n + ( − 1 ) n 1 z n ) 2 |  is equal to……….
V
Vishal Baghel

z 2 + z + 1 = 0 ⇒ z = w , w 2

| ∑ n = 1 1 5 ( z n + ( − 1 ) n 1 z n ) 2 | = | ∑ n = 1 1 5 ( z 2 n + 1 z 2 n + 2 ( − 1 ) n ) | = | ∑ n = 1 1 5 w 2 n + 1 w 2 n + 2 ( − 1 ) n |

= | w 2 ( 1 − 1 ) 1 − w 2 + 1 w 2 ( 1 − 1 ) 1 − 1 w 2 − 2 | = 2

(i) If −4x≥12 , then x…−3 .

(ii) If 34−x≤−3, then x…4 .

(iii) If 2x+2>0, then x…−2 .

(iv) If x>−5 , then 4x…−20 .

(v) If x>y and z<0 , then −xz…−yz .

(vi) If p>0 and q<0 , then p−q…p .

(vii) If x+2>5 , then x…−7 or x…3 .

(viii)If −2x+1≥9 , then x…−4 .
P
Payal Gupta

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

( i ) − 4 x ≥ 1 2         ⇒ x ≤ − 3             H e n c e ,     t h e     f i l l e r     i s     ( ≤ ) ( i i ) I f     − 3 4 x ≤ − 3         ⇒ x ≥ 3 x × 4 3         ⇒ x ≥ 4             H e n c e ,     t h e     f i l l e r     i s     ( ≥ ) ( i i i ) I f     2 x + 2 > 0         ⇒ x > − 2               H e n c e ,     t h e     f i l l e r     i s     ( > ) ( i v ) I f     x > − 5         ⇒ 4 x > − 2 0             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 ⇒ − 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 ⇒           x < − 5 − 2     o r     x > 5 − 2 ⇒           x < − 7     o r     x > 3                 S o ,     x ∈ ( − ∞ , − 7 ) ∪ ( 3 , ∞ )             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 ≥ 9     t h e n ⇒           − 2 x ≥ 9 − 1         ⇒ − 2 x ≥ 8           ⇒ 2 x ≤ − 8           ⇒ x ≤ − 4             H e n c e ,     t h e     f i l l e r     i s     ( ≤ )

30. Kindly consider the following y + 8 ≥ 2x
P
Payal Gupta

30. 

30. For inequality y+8 ≤ 2x, the equation of the line is y+8=2x. We consider the table below to plot of y+8=2x.

xy |   0−8 | 40 |

The line devides the xy-plane into half planer I and II. We select a point (0,0) and check the correctness of the inequality.

i.e., 0+8 ≤ 2 × 0

0 ≤ 0 which is true.

So, the solution region is I which includes the rigin (0,0). The continuous line also indicates that any points on the line also satisfy the given inequality.

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