Question

8
In the given figure, PQRT is a cyclic quadrilateral
in which the side PT is produced to such that
TS = PQ. If angle QPR = angle TPR, prove that PR = SR.​

Answer 1

Step-by-step explanation:

Evaluate →0

4

|2|

a) 2 b) -2 c) does not exist. d) 0

2)

→2+

2+

4−

2

=

a) 1 b) −∞ c) ∞ d) none of these.

3) The value of the limit

→0+

7

5

3

is

a) -7/5 b) +∞ c) −∞ d) 7/5

4) →2−

2−4

|−2|

=

a) -4 b) does not exist. c) 0 d) 4

5) If [x] is the bracket/greatest integer function, then value of the limit →3−

[]

2

is

a) 1/2 b) 1/3 c) 4/3 d) -1

6) Given that 9 − 6

2 ≤ () ≤ −5 + 8 then the value of lim→1

() is

a) 1 b) 3 c) -1 d) not unique

7) The value of derivative of f (x) = |x –1| + |x –3| at x = 2 is

a) 2 b) -2 c) -1 d) 0

8) →2 −

[]

||

=

a) ∞ b) +1 c) -1 d) 1/2

9) Functions defined by () = √ and () = √1 − . The Domain of f.g is

a) [0, ∞) b) (−∞, 1] c) [0,1] d) (0,1)

10) The range of the function √ is

a) [0, ∞) b) (0, ∞] c) (0, ∞) d) (−∞, ∞)

11) =

+2

−2

− 3 is continuous ∀ ∈

a) ] − ∞, ∞[ b) ]−∞, 2[ ∪ ]2, ∞[ c) ]−∞, 2] ∪ [2, ∞[ d) ]−∞, −

Answer 2

In △TPS and △TRQ

∠PST=∠RQT

∠SPQ=∠QRT

[∵ exterior angle of cyclic quadrilateral is equal to the interior opposite angle]

∠T=∠T (common)

∴△TPS−△TRQ (By AA)

By CPCT , PR = SR.

Hope it helps you..