81279 + Interview Questions in Mathematics QA in Mathematics

1.	\left. \begin{array} { l l } { 2 } & { f ( x ) = \sqrt { x - 1 } + \sqrt { 6 - x } } \\ { 4 } & { f ( x ) = \frac { 1 } { \sqrt { x - [ x ] } } } \end{array} \right.
Answer» f(x) = √x-1 +√6-x the domain of this function should be such that . the root value should not be less than <0so the fomain is [1,6] f(x) = 1/√(x-[x]) = 1/√{x} we know that {x} is fractional part of x , and it is Zero at every integer.. so, the domain should not be any integer... so, domain = R - I.

1.

\left. \begin{array} { l l } { 2 } & { f ( x ) = \sqrt { x - 1 } + \sqrt { 6 - x } } \\ { 4 } & { f ( x ) = \frac { 1 } { \sqrt { x - [ x ] } } } \end{array} \right.

Answer»

f(x) = √x-1 +√6-x

the domain of this function should be such that . the root value should not be less than <0so the fomain is [1,6]

f(x) = 1/√(x-[x]) = 1/√{x}

we know that {x} is fractional part of x , and it is Zero at every integer.. so, the domain should not be any integer...

so, domain = R - I.

Discussion

2.	inorde,2), (9, 4) and (p, 3) are the vertices of parallelogram,find the value of p.
Answer»

Discussion

3.	[ ( \frac { 2 } { 5 } ) ^ { 0 } \times ( \frac { 4 } { 5 } ) ^ { 0 } ] \div ( \frac { 1 } { 7 } ) ^ { 0 }
Answer» Thankyou so much 🤓

Discussion

4.	(3^0 %2B 2^0)/5^0
Answer» (3^0 + 2^0) / 5^0 = ( 1 + 1 ) / 1= 2/1 = 2 2 is the correct answer 2 is correct answer in this question 2 is the correct answer 2 is the correct answer.... 2 is your answer......... (3^0+2^0)/5^0=(1+1)/1=2/1=2

4.

(3^0 %2B 2^0)/5^0

Answer»

(3^0 + 2^0) / 5^0 = ( 1 + 1 ) / 1= 2/1 = 2

2 is the correct answer

2 is correct answer in this question

2 is the correct answer

2 is the correct answer....

2 is your answer.........

(3^0+2^0)/5^0=(1+1)/1=2/1=2

Discussion

5.	5^0*(3^0 %2B 2^0)
Answer» I think it is 1 because the not represents the 1 and 1x1 is 1 (3^0+2^0)×5^0 = 2 is right answer

5.

5^0*(3^0 %2B 2^0)

Answer»

I think it is 1 because the not represents the 1 and 1x1 is 1

(3^0+2^0)×5^0 = 2 is right answer

Discussion

6.	raph for each linear equatioDraw theiven below:(ii) x t 30(iv) 2x -7 0(vi) y + 6 0(viii) 3y +5(x) y 0ii) x-5=0ii) y-2-0
Answer» when x-5=0X=5 X+3=0then X=-3

6.

raph for each linear equatioDraw theiven below:(ii) x t 30(iv) 2x -7 0(vi) y + 6 0(viii) 3y +5(x) y 0ii) x-5=0ii) y-2-0

Answer»

when x-5=0X=5

X+3=0then X=-3

Discussion

7.	Draw the graph of the linear equation x +2 y=8. At which point(s) does the graph cut thex-axis and y-axis.
Answer»

Discussion

8.	Find he CE LCMI for the numben 300and 190.
Answer» From which book have you taken this question? Please tell us so that we can provide you faster answer.

Discussion

9.	8Express the following numbers in the form 2, where p and q are both integers9 0:() 03(i) 52(iii) 0-404040...
Answer» Let x = 0.333333... (1) 10x =3.333333...(2) subtracting (1) from (2) 9x =3.o x=1/3= o.3333333 tell 2nd plzz

9.

8Express the following numbers in the form 2, where p and q are both integers9 0:() 03(i) 52(iii) 0-404040...

Answer»

Let x = 0.333333... (1)

10x =3.333333...(2)

subtracting (1) from (2)

9x =3.o

x=1/3= o.3333333

tell

2nd plzz

Discussion

10.	(1) JUL09)9. Find the value of the following:() - 12 + (-7) + 36 + (-32) + (- 49)D. The sum of two integers is -23. If one of the integers is 145, find the other.. If one of the integers is -43, find the other
Answer» -145 x =-23 x -145 -23 = +168

Discussion

11.	and OD is the diameter of the silancrOA-7 cm, find the area of the shaded region.The area of an equilateral triangle ABC is 17320.5cm2. With each vertex of the triangle as centre, acircle is drawn with radius equal to half the lengthof the side of the triangle (see Fig. 12.28). Find thearea of the shaded region. (Use Tt 3.14 andV3 1.73205)Fig. 12.28
Answer»

Discussion

12.	15. Draw the graph of the linear equation x +2 y-8. At which point(s) does the graph cut thex-axis and y-axis.
Answer» Given is equation of line x + 2y = 8 At x = 8 At y = 4

12.

15. Draw the graph of the linear equation x +2 y-8. At which point(s) does the graph cut thex-axis and y-axis.

Answer»

Given is equation of line x + 2y = 8

At x = 8 At y = 4

Discussion

13.	15. Draw the graph of the linear equation x +2 y -8. At which point(s) does the graph cut thex-axis and y-axis.
Answer» It is clear from the table as well as from the graph that the equation will cut X-axis at 8 and Y-axis at 4.

Discussion

14.	f(l(i*m))
Answer» 17. ( 1 + x)ⁿ C(0) + C(1)x + C(2) x² +.....C(n) x^10 Put x = 1 and n = 10 Therefore, 2^ 10 = C(0) + C(1) +.... C(10)

14.

f(l(i*m))

Answer»

17. ( 1 + x)ⁿ C(0) + C(1)x + C(2) x² +.....C(n) x^10

Put x = 1 and n = 10

Therefore,

2^ 10 = C(0) + C(1) +.... C(10)

Discussion

15.	ptg 드 2them. tool the Valuep^{3}+q^{3}+6 p^{2}+5 q^{2}+6 p q
Answer» Your answer is not in the options given in the question.

Discussion

16.	0 2f \ L -
Answer» lcm of 15 n 20 is 60 24= 815 4= 60 130 =302030=60 ans ....30-8=22/60 1/12is the right answer

16.

0 2f \ L -

Answer»

lcm of 15 n 20 is 60 2*4= 815 *4= 60

1*30 =3020*30=60

ans ....30-8=22/60

1/12is the right answer

Discussion

17.	1.Draw the graph of a linear equation + y = 7. At what point does the graphCut the x--axis and y--axis.
Answer»

Discussion

18.	Draw the graph of the linear equation x +2 y 8. At which point(s) does thexis and y-axis.graph cut the
Answer» Given line is x + 2y = 8 At x = 8 At y = 4

Discussion

19.	Write a polynomial function with the real zeros-1.5,-0.3, and 1.75.
Answer» the function with zeros as -1.5 , -0.3 and 1.75 are = (x+3/2)(x+3/10)(x-7/4)= (x²+3x/10+3x/2 +9/20)(x-7/4)= x³-7x²/4+3x²/10-21x/40+3x²/2-21x/8+9x/20-63/80= x³ +(3/2-7/4+3/10)x²+(9/20-21/8-21/40)x-63/80

19.

Write a polynomial function with the real zeros-1.5,-0.3, and 1.75.

Answer»

the function with zeros as -1.5 , -0.3 and 1.75 are

= (x+3/2)(x+3/10)(x-7/4)= (x²+3x/10+3x/2 +9/20)(x-7/4)= x³-7x²/4+3x²/10-21x/40+3x²/2-21x/8+9x/20-63/80= x³ +(3/2-7/4+3/10)x²+(9/20-21/8-21/40)x-63/80

Discussion

20.	6. Prove that. points A (3, 4), B (0,1) andC(2,3)are counle cs (a, al, (p,6) and D (3,5) are the vertices of a paraitelogram AdCo,taken in orde Ftuq), C (p.6) and D (3,5) are the vertices of a parallelogram ABCD, taken in order. Find ouFHnvalue of p and q.
Answer»

Discussion

21.	Find the distance of the point P(6,-6) from the origin.
Answer»

Discussion

22.	A numbeu Cenkikts of tuue digit whose omnumben the diguintee Change theieFihd the humber
Answer» Thanks correct answer

Discussion

23.	humber? Can you write it in the form P, where p and gare integers9and q#0?
Answer»

Discussion

24.	ations for 0° < 0 < 90१ ;24/3¢0s20 = 3W 9/\ ८056 -- 5४00 _ 1-5cosf+sin® 1.3
Answer»

Discussion

25.	9. In Fig. 12.27, AB and CD are two diameters of acircle (with centre O) perpendicular to each otherand OD is the diameter of the smaller circle. IfOA 7 cm, find the area of the shaded region.
Answer»

Discussion

26.	Find the Distance between the pair of points(-5, 7) and (-1, 3).
Answer» distance= √(x2-x1)^2+(y2-y1)^2=√{(-1-(-7)}^2+(3-7)^2=√{(6)^2+(-4)^2=√36+16=√52 unit 16√2 is correct answer correct answer is 4√2

26.

Find the Distance between the pair of points(-5, 7) and (-1, 3).

Answer»

distance= √(x2-x1)^2+(y2-y1)^2=√{(-1-(-7)}^2+(3-7)^2=√{(6)^2+(-4)^2=√36+16=√52 unit

16√2 is correct answer

correct answer is 4√2

Discussion

27.	-varnABC isc
Answer» tqsm bro

Discussion

28.	ii) 331 % of 372
Answer» 133 ------ % 3 100/3 × ₹72 ₹2400

28.

ii) 331 % of 372

Answer»

133 ------ % 3

100/3 × ₹72

₹2400

Discussion

29.	If (x, 3) and (3, 5) are the extremities of a diameter ofa circle with centre at (2, y). Then the value of x and yare-(A) x 1, y 4 (B) x 4, y 1(C) x 8, y 2 (D) of these
Answer» Thanks for the suggestion

Discussion

30.	In the given figure, BE 1 AC. AD is any line from A toBC intersecting BE at H. P, Q and R are mid-points ofAH,AB and BC respectively, then prove that
Answer» Let ABC be a triangle.We have BE⊥AC and AD is any line from A intersecting BE at H.P, Q and R are the mid points of AH, AB and BC respectively. Given Q and R are the mid points of sides AB and BC of ∆ABC respectively∴ QR\|\|AC [Mid-point theorem]Given BE⊥AC∴ BE⊥QR In ∆ABH and Q and P are the mid points of sides AB and AH respectively.∴ QP\|\|BH [Mid-point theorem]∴QP\|\|BE⇒QP⊥QR (BE⊥QR)⇒∠PQR = 90°

30.

In the given figure, BE 1 AC. AD is any line from A toBC intersecting BE at H. P, Q and R are mid-points ofAH,AB and BC respectively, then prove that

Answer»

Let ABC be a triangle.We have BE⊥AC and AD is any line from A intersecting BE at H.P, Q and R are the mid points of AH, AB and BC respectively.

Given Q and R are the mid points of sides AB and BC of ∆ABC respectively∴ QR||AC [Mid-point theorem]Given BE⊥AC∴ BE⊥QR In ∆ABH and Q and P are the mid points of sides AB and AH respectively.∴ QP||BH [Mid-point theorem]∴QP||BE⇒QP⊥QR (BE⊥QR)⇒∠PQR = 90°

Discussion

31.	I6. AB is a diameter of a circle. AH and BKAB.are perpendiculars from A and B respectively to the tangent at P. Provethat AH + BK
Answer» SOLUTION:[FIGURE IS IN THE ATTACHMENT]Given:AH and BK are perpendiculars drawn from A and B on tangent at P.To Prove:AH + BK = ABPROOF:ABKH is a rectangle.(AH & BK are Perpendiculars.AB = HK …..…….....(1)[Opposite sides of rectangle are equal]We know that, the lengths of tangents drawn from an external point to the circle are equal.AH = HP ……............(2)BK = PK ..….............(3)Adding eq. (2) and (3), AH + BK = HP + PK = HKAH + BK = AB [from eq. (1)]

Discussion

32.	ABCD is a square. E, FL ABCDisas0G = DH and BE = CF = DG = AH. Prove that EFGH is a square.INCERT]R G and H are points on AB, BC, CD and DA respectively such thatH anE CF14.
Answer»

Discussion

33.	Ifa graph y-pix), cut x axis on two points, then find the number of zeros
Answer» If it cuts on two points then the equation has two roots has a cut means that the point satisfies the given equation

Discussion

34.	If p(x) = ax - 3x +76 and 1, -2 are zeroes of pix),then find the value of a and b.
Answer» 1-2=3/a=>a=-3✓and 7b/a=-2=>b=6/7✓ a= - 3 ,b=6/7 is the correct answer of the given question

34.

If p(x) = ax - 3x +76 and 1, -2 are zeroes of pix),then find the value of a and b.

Answer»

1-2=3/a=>a=-3✓and 7b/a=-2=>b=6/7✓

a= - 3 ,b=6/7 is the correct answer of the given question

Discussion

35.	1 y=pix)5. The number of real zeros of y=p (x) is............. in the given figure.A. 1B.2C.3D. OXX'
Answer» A.1 i hope it will hep you D is a right answer and this is nice question the answer could be option c A=1 is correct answer

35.

1 y=pix)5. The number of real zeros of y=p (x) is............. in the given figure.A. 1B.2C.3D. OXX'

Answer»

A.1

i hope it will hep you

D is a right answer and this is nice question

the answer could be option c

A=1 is correct answer

Discussion

36.	Level-IIrf the point Pix y) be equidistant from the points A(e+ b, b- a) and B(a -b, a+ b), thenprove that bx -ay.1.
Answer»

Discussion

37.	is a diameter of a circle. AH and BK are perpendiculars from A and B respectively to the tangent at P6. AB is athat AH + BKAB.Prove
Answer» AH and BK are perpendiculars drawn from A and B on tangent at P.To Prove:AH + BK = ABPROOF:ABKH is a rectangle.(AH & BK are Perpendiculars.AB = HK …..…….....(1)[Opposite sides of rectangle are equal]We know that, tangents drawn from an external point to the circle are equal.AH = HP ……............(2)BK = PK ..….............(3)Adding eq. (2) and (3), AH + BK = HP + PK = HKAH + BK = AB [from eq. (1)]

Discussion

38.	6. AB is diam16.a diameter of a circle. AH and BK are perpendiculars from A and B respectively to the tangent at P: Provethat AH + BKAB.tnnother triangle whose
Answer» SOLUTION:[FIGURE IS IN THE ATTACHMENT]Given:AH and BK are perpendiculars drawn from A and B on tangent at P.To Prove:AH + BK = ABPROOF:ABKH is a rectangle.(AH & BK are Perpendiculars.AB = HK …..…….....(1)[Opposite sides of rectangle are equal]We know that, the lengths of tangents drawn from an external point to the circle are equal.AH = HP ……............(2)BK = PK ..….............(3)Adding eq. (2) and (3), AH + BK = HP + PK = HKAH + BK = AB [from eq. (1)]

Discussion

39.	The tuo numbease in thahumber Js 89 hih themum
Answer»

Discussion

40.	(2) 1 () 0 (0) 1 (d) 14. Let P(O) and () be extremities of two perpendiculardiameters of the ellipse r? + 4 y2 = 4. Then distance ofcentre O from the chord PQ is :V1+3 cose(a) Acos cos o(b)To(c) V5(d) of these
Answer»

Discussion

41.	32+48The value ot + 12s equal to
Answer» thanks yaaar ek sum or bhej raha hu

41.

32+48The value ot + 12s equal to

Answer»

thanks

yaaar ek sum or bhej raha hu

Discussion

42.	Find the value of p for which the points (-5, 1), (1, p) and (4, -2) arecollinear.
Answer» x₁ = -5 x₂ = 1x₃ = 4 y₁ = 1y₂ = py₃ = -2 For the points to be collinear :- x₁(y₂ -y₃) + x₂(y₃ - y₂) + x₃(y₁ - y₂) = 0 -5{ p -(-2)} + 1 ( - 2 - 1) + 4( 1 - p) = 0 -5(p+2) + 1(-3) + 4( 1 - p) = 0 -5p - 10 - 3 + 4 - 4p = 0 -9p - 9 = 0 -9p = 9 p = 9/-9 p = -1

42.

Find the value of p for which the points (-5, 1), (1, p) and (4, -2) arecollinear.

Answer»

x₁ = -5 x₂ = 1x₃ = 4

y₁ = 1y₂ = py₃ = -2

For the points to be collinear :-

x₁(y₂ -y₃) + x₂(y₃ - y₂) + x₃(y₁ - y₂) = 0

-5{ p -(-2)} + 1 ( - 2 - 1) + 4( 1 - p) = 0

-5(p+2) + 1(-3) + 4( 1 - p) = 0

-5p - 10 - 3 + 4 - 4p = 0

-9p - 9 = 0

-9p = 9

p = 9/-9

p = -1

Discussion

43.	372UNDERSTANDING ISC MATHEMATICS-Xu5. If p, q are roots of 4x2-(5a + 1)x + 5 a = 0 and q = 1 + p. Find the values of a, p and a
Answer» thank you

Discussion

44.	ot meant for exanatoations:-2x-1 y +12
Answer» x+1/2 + y-1/3 = 83(x+1) + 2(y-1) = 483x + 3 + 2y - 2 = 483x + 2y = 47........(1) x-1/3 + y+1/2 = 92(x-1) + 3(y+1) = 962x - 2 + 3y + 3 = 542x + 3y = 53...... (2) Multiply eq(1) by 2 and eq(2) by 36x + 4y = 94.....(3)6x + 9y = 159...(4) eq(4) - eq(3)5y = 65y = 65/5 = 13 Put value of y in eq(1)3x + 213 = 473x = 47 - 263x = 21x = 21/3 = 7 Therefore, value of x = 7 and y = 13 tq paina 48 Ela vachindi

44.

ot meant for exanatoations:-2x-1 y +12

Answer»

x+1/2 + y-1/3 = 83(x+1) + 2(y-1) = 483x + 3 + 2y - 2 = 483x + 2y = 47........(1)

x-1/3 + y+1/2 = 92(x-1) + 3(y+1) = 9*62x - 2 + 3y + 3 = 542x + 3y = 53...... (2)

Multiply eq(1) by 2 and eq(2) by 36x + 4y = 94.....(3)6x + 9y = 159...(4)

eq(4) - eq(3)5y = 65y = 65/5 = 13

Put value of y in eq(1)3x + 2*13 = 473x = 47 - 263x = 21x = 21/3 = 7

Therefore, value of x = 7 and y = 13

tq

paina 48 Ela vachindi

Discussion

45.	4x +5x-32 y
Answer» =4x+5x-32yAdd 4x and 5x=9x-32y

Discussion

46.	Solve(i)4x+ 5x = 0
Answer» 4x^2+5x=0x(4x+5)=0x=0,-5/4

Discussion

47.	5x - (4x-7)(3x - 5) = 6 -3(4x - 9)(x - 1)
Answer» answer of the question is 8. 8 is the good answer x is equal to 8 because you have to use the distributive property 8 is the best and correct answer ha 8is the right answer of the following questions x = 8 is the right answer . answer is 8 of this question 8 is the correct answer

47.

5x - (4x-7)(3x - 5) = 6 -3(4x - 9)(x - 1)

Answer»

answer of the question is 8.

8 is the good answer

x is equal to 8 because you have to use the distributive property

8 is the best and correct answer ha

8is the right answer of the following questions

x = 8 is the right answer .

answer is 8 of this question

8 is the correct answer

Discussion

48.	x-5/3+4x-5x-1
Answer» x-5/3+4x-5x-1= x-x - 5/3-1= 0-8/3= - 8/3

Discussion

49.	8 ( 4 x %2B 5 y ) ^ 2 - 12 ( 4 x %2B 5 y )
Answer» 8(4x + 5y)^2 - 12(4x+ 5y). = 8(16x^2 + 25 y^2 + 40xy ) - (48x + 60y) = 128x^2 + 200y^2 + 320xy - 48x - 60y

49.

8 ( 4 x %2B 5 y ) ^ 2 - 12 ( 4 x %2B 5 y )

Answer»

8(4x + 5y)^2 - 12(4x+ 5y).

= 8(16x^2 + 25 y^2 + 40xy ) - (48x + 60y)

= 128x^2 + 200y^2 + 320xy - 48x - 60y

Discussion

50.	Write the coefficient of x^2 in the following polynomial:2-x^{2}+x^{3}
Answer»

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

\left. \begin{array} { l l } { 2 } &amp; { f ( x ) = \sqrt { x - 1 } + \sqrt { 6 - x } } \\ { 4 } &amp; { f ( x ) = \frac { 1 } { \sqrt { x - [ x ] } } } \end{array} \right.

inorde,2), (9, 4) and (p, 3) are the vertices of parallelogram,find the value of p.

[ ( \frac { 2 } { 5 } ) ^ { 0 } \times ( \frac { 4 } { 5 } ) ^ { 0 } ] \div ( \frac { 1 } { 7 } ) ^ { 0 }

(3^0 %2B 2^0)/5^0

5^0*(3^0 %2B 2^0)

raph for each linear equatioDraw theiven below:(ii) x t 30(iv) 2x -7 0(vi) y + 6 0(viii) 3y +5(x) y 0ii) x-5=0ii) y-2-0

Draw the graph of the linear equation x +2 y=8. At which point(s) does the graph cut thex-axis and y-axis.

Find he CE LCMI for the numben 300and 190.

8Express the following numbers in the form 2, where p and q are both integers9 0:() 03(i) 52(iii) 0-404040...

(1) JUL09)9. Find the value of the following:() - 12 + (-7) + 36 + (-32) + (- 49)D. The sum of two integers is -23. If one of the integers is 145, find the other.. If one of the integers is -43, find the other

15. Draw the graph of the linear equation x +2 y-8. At which point(s) does the graph cut thex-axis and y-axis.

15. Draw the graph of the linear equation x +2 y -8. At which point(s) does the graph cut thex-axis and y-axis.

f*(l*(i*m))

ptg 드 2them. tool the Valuep^{3}+q^{3}+6 p^{2}+5 q^{2}+6 p q

0 2f \ L -

1.Draw the graph of a linear equation + y = 7. At what point does the graphCut the x--axis and y--axis.

Draw the graph of the linear equation x +2 y 8. At which point(s) does thexis and y-axis.graph cut the

Write a polynomial function with the real zeros-1.5,-0.3, and 1.75.

6. Prove that. points A (3, 4), B (0,1) andC(2,3)are counle cs (a, al, (p,6) and D (3,5) are the vertices of a paraitelogram AdCo,taken in orde Ftuq), C (p.6) and D (3,5) are the vertices of a parallelogram ABCD, taken in order. Find ouFHnvalue of p and q.

Find the distance of the point P(6,-6) from the origin.

A numbeu Cenkikts of tuue digit whose omnumben the diguintee Change theieFihd the humber

humber? Can you write it in the form P, where p and gare integers9and q#0?

ations for 0° &lt; 0 &lt; 90१ ;24/3¢0s20 = 3W 9/\ ८056 -- 5४00 _ 1-5cosf+sin® 1.3

9. In Fig. 12.27, AB and CD are two diameters of acircle (with centre O) perpendicular to each otherand OD is the diameter of the smaller circle. IfOA 7 cm, find the area of the shaded region.

Find the Distance between the pair of points(-5, 7) and (-1, 3).

-varnABC isc

ii) 331 % of 372

If (x, 3) and (3, 5) are the extremities of a diameter ofa circle with centre at (2, y). Then the value of x and yare-(A) x 1, y 4 (B) x 4, y 1(C) x 8, y 2 (D) of these

In the given figure, BE 1 AC. AD is any line from A toBC intersecting BE at H. P, Q and R are mid-points ofAH,AB and BC respectively, then prove that

I6. AB is a diameter of a circle. AH and BKAB.are perpendiculars from A and B respectively to the tangent at P. Provethat AH + BK

ABCD is a square. E, FL ABCDisas0G = DH and BE = CF = DG = AH. Prove that EFGH is a square.INCERT]R G and H are points on AB, BC, CD and DA respectively such thatH anE CF14.

Ifa graph y-pix), cut x axis on two points, then find the number of zeros

If p(x) = ax - 3x +76 and 1, -2 are zeroes of pix),then find the value of a and b.

1 y=pix)5. The number of real zeros of y=p (x) is............. in the given figure.A. 1B.2C.3D. OXX'

Level-IIrf the point Pix y) be equidistant from the points A(e+ b, b- a) and B(a -b, a+ b), thenprove that bx -ay.1.

is a diameter of a circle. AH and BK are perpendiculars from A and B respectively to the tangent at P6. AB is athat AH + BKAB.Prove

6. AB is diam16.a diameter of a circle. AH and BK are perpendiculars from A and B respectively to the tangent at P: Provethat AH + BKAB.tnnother triangle whose

The tuo numbease in thahumber Js 89 hih themum

(2) 1 () 0 (0) 1 (d) 14. Let P(O) and () be extremities of two perpendiculardiameters of the ellipse r? + 4 y2 = 4. Then distance ofcentre O from the chord PQ is :V1+3 cose(a) Acos cos o(b)To(c) V5(d) of these

32+48The value ot + 12s equal to

Find the value of p for which the points (-5, 1), (1, p) and (4, -2) arecollinear.

372UNDERSTANDING ISC MATHEMATICS-Xu5. If p, q are roots of 4x2-(5a + 1)x + 5 a = 0 and q = 1 + p. Find the values of a, p and a

ot meant for exanatoations:-2x-1 y +12

4x +5x-32 y

Solve(i)4x+ 5x = 0

5x - (4x-7)(3x - 5) = 6 -3(4x - 9)(x - 1)

x-5/3+4x-5x-1

8 ( 4 x %2B 5 y ) ^ 2 - 12 ( 4 x %2B 5 y )

Write the coefficient of x^2 in the following polynomial:2-x^{2}+x^{3}

\left. \begin{array} { l l } { 2 } & { f ( x ) = \sqrt { x - 1 } + \sqrt { 6 - x } } \\ { 4 } & { f ( x ) = \frac { 1 } { \sqrt { x - [ x ] } } } \end{array} \right.

f(l(i*m))

ations for 0° < 0 < 90१ ;24/3¢0s20 = 3W 9/\ ८056 -- 5४00 _ 1-5cosf+sin® 1.3