QUESTION BANK (HOTS)-- XI (HOTS) MATHEMATICS

TRIGONOMETRIC FUNCTIONQuestion – 1 If 𝛂, 𝛃 are the acute angles and cos2𝛂 = 3cos2β−13−cos2 β , show that tan 𝛂 = √ 2 tan𝛃.

Question--2: If tan2A = 2tan2B + 1, prove that cos2A + sin2B =0.

Question--3 Solve: √2 sec∅ + tan∅ = 1

Question – 4 If cos(A+B) sin(C-D) = cos(A-B) sin(C+D) , then show that tanA tanB tanC + tanD = 0

Question-- 5 If sinθ = n sin(θ+2𝛂), prove that

tan(θ+𝛂) = 1+n1−n tan𝛂.Question – 6 Prove that

(a) sin3x + sin3(2π3 +x) + sin3(4 π3 +x) = - 34 sin3x.

Question-- 7: Solve the equation: sin3θ + cos2θ = 0

Question-- 8: Prove that : 1+sinx−cosx1+sinx+cosx = tan(x/2)

Question-9 If 𝛂 , 𝛃 are the distinct roots of acosθ + bsinθ = c, prove that sin(𝛂+𝛃) = 2aba ²+b ² .

Question-10 Solve the equation: tan2x + sec2x = 1

SETS , RELATION & FUNCTION (HOTS)

Question- 1 : Let A = {(x,y):y=ex ,x∈R} and B = {(x,y):y=e-x ,x∈R}. Is A∩B empty? If not find the ordered pair belonging to A∩B.

Question- 2: Two finite sets have m and n elements. The number of subsets of the first set is 112 more than that of the second set. Find the values of m and n resp.

Question-3: Prove that for non-empty sets (AUBUC)∩(A∩B’∩C’)’∩C’ = B∩C’.

Question:4 If n(A) = 3, n(B) = 6 and the number of elements in AUB and in A∩B.

Question-5: In an examination, 80% students passed in Mathematics,72% passed in science and 13% failed in both the subjects, if 312 students passed in both the subjects. Find the total number of students who appeared in the examination.

Question-6: If R is a relation from set A={11,12,13} to set B={8,10,12} defined by y=x-3,then write R-1.

Question-7: if f(x) = ,find f(-1) & f(3).

Question-7: Y=f(x) = , then find x=f(y).

Question-8: If f(x) = , find f[f(x)].

COMPLEX NUMBERS (HOTS)

Question-1: If iz3 +z2 – z+ i = 0, then show that |z| = 1.

Question-2: If a+ib = c+ic−i , then a2+b2 = 1 and b/a = 2 cc ²−1

Question-3: If x = - 5 +2√(-4) , find the value of x4+9x3+35x2 – x+4.

Question-4: Show that a real x will satisfy equation 1−ix1+ix = a – ib, if a2+b2 = 1 where a, b are real.

Question-5: A variable complex z is such that arg ( z−1z+1

) = π2 , show that x2+y2 – 1=0

Question-6: Find the values of x and y if x2 – 7x +9yi and y2i+20i – 12 are equal.

Question-7: Prove that arg(z) = 2 – arg(z) ,z ≠0π

Question-8: If z = 11−cosφ−isinφ , then find Re(z).

SEQUENCES & SERIES (HOTS)

Question:1 If the roots of (b-c)x2+(c-a)x+(a-b) = 0 are equal, then a,b,c are in A.P.

Question:2 There are n A.M.’s between 7 and 85 such that (n-3)th mean : nth mean is 11 : 24.Find n.

Question:3 If g1, g2 be two G.M.’s between a and b and A is the A.M. between a & b, then prove that

g1 ²g2 +

g2 ²g1 = 2A.

Question:4 If a is the A.M. of b, c and two geometric means between b , c and G1,G2, then prove that G1

3 = g23

Question:5 An A.P. consists of n(odd)terms and it’s middle term is m. Prove that Sn = mn.

Question:6 Sum of infinity the series

12 +

12+4 +

12+4+6 +………..

Question:7 Prove that 23rd term of sequence 17, 1615 , 15

25 , 14

35 ,………is the first negative term.

Question:8 Prove that 1+23 +

63 ² +

1033 +

1434 + ………..∞ = 3.

Question: 9 If there are distinct real numbers a,b,c are in G.P. and a+b+c = bx , show that x ≤ -1 or x ≥ 3.

Question:10 If first term of H.P. is 1/7 and 2nd term is 1/9, prove that 12th term is 1/29.

STRAIGHT LINES & CONIC SECTION (HOTS)

Question 1 Find the equation of the straight lines which pass through the origin and trisect the intercept of line 3x+4y=12 b/w the axes.

Question 2 Find the equation of the line through the intersection of the lines x -3y+1=0 and 2x+5y -9=0 and whose distant from the origin is √ 5 .

Question 3 The points (1,3) and (5,1) are the opposite vertices of a rectangle.The other two vertices lie on the line y = 2x+c. Find c and the remaining vertices.

Question 4 The consecutive sides of a parallelogram are 4x+5y=0 and 7x+2y=0. If the equation of one diagonal be 11x+7y=9, find the equation of other diagonal.

Question 5 One side of a rectangle lies along the line 4x+7y+5=0. Two of vertices are (-3,1) & (1,1). Find the equation of other three sides.

Question 6 The extremities of the base of an isosceles ∆ are the points (2a,0) & (0,a). The equation of the one

of the sides is x=2a. Find the equation of the other two

sides and the area of the ∆.

Question 7 One side of a square is inclined to x-axis at an angle and one of its extremities is at origin. If αthe sides of the square is 4, find the equations of the diagonals of the square.

Question 8 Prove that the diagonals of the //gm. Formed by the four lines.

x/a + y/b = 1 ……(i) , x/b + y/a = 1 …….(ii) ,

x/a + y/b = -1 ……(iii) , x/b + y/a = -1 ……..(iv)

are perp. to each other.

Question 9 On the portion of the line x+3y – 3 =0 which is intercepted b/w the co-ordinates axes, a square is constructed on the side of the line away from the origin. Find the co-ordinates of the intersection of its diagonals. Also find the equations of its sides.

Question 10 If one diagonal of a square is along the line 8x-15y=0 and one of its vertex is at (1,2), then find the equation of sides of the square passing through this vertex.

Question.11 Find the eqn. of parabola whose focus is at (-1,-2) & the directrix is x – 2y + 3=0.

Question.12 Find the eqn. of ellipse whose axes are parallel to the coordinate axes having it’s centre at the point (2,-3) and one vertex at (4,-3) & one focus at (3,-3).

Question.13 Find the eqn. of hyperbola, the length of whose latus-rectum is 8 and e = 3√ 5 .

Question. 14 Find the eqn. of circle whose radius is 5 and which touches the circle x2+y2 – 2x – 4y – 20 = 0 externally at the point (5,5).

Question.15 Find the eqn. of circle of radius 5 which lies within the circle x2+y2+14x+10y – 26 = 0 and which touches the given circle at the point (-1,3).

: L.H.S. 1−tan ² A1+ tan ²B + sin2B = −2 tan ²B2(1+tan2B) + sin2B , by putting above

result and simplify it.

Solution √2 + sin∅ = cos∅ ⇨ cos∅ - sin∅ = √2 dividing by √a2+b2 =√2 ∵ a=1,b=1 (cos∅ - sin∅ )/ √2 = 1 ⇨ cos( /4)п cos∅ - sin( /4) п sin∅ =1 ⇨ cos(∅+ /4) = cos0п 0 ∅+ /4 = 2nп п±0, n∈Z ⇨ ∅ = 2n – /4.п п

Hint: According to required result , we have to convert given part into tangent function By using cos2𝛂 = 1−tan ²α1+tan ²α ∴ we will get 1−tan ²α1+tan ²α = 3( 1− tan² β1+tan ² β

)−1

3−1−tan ² β1+ tan ² β

= 3−3 t an ² β−1−tan ²β3+3 tan ² β−1+ tan ² β

Hint: We can write above given result as cos (A+B)cos(A−B) =

sin(C+D)sin(C−D)

By C & D cos ( A+B )+cos (A−B)cos ( A+B )−cos(A−B) =

sin (C+D )+sin(C−D)sin (C+D )−sin(C−D)

(ii) sin(θ+2α)sinθ = 1n , by C & D sin (θ+2α )+sinθ

sin (θ+2α )−sinθ = 1+n1−n

[Hint: Use sin3A = 3sinA – 4sin3A ⇨ 4sin3A = 3sinA - sin3A ⇨ sin3A = ¼[3sinA - sin3A]]

[Answer number of students failed in both the subjects = n(M’∩S’)=13% of x=0.13x

n(U) – n{(MUS)’} = 1.52x – 312 ⇨x=480.] Answer: ex = e-x ⇨ e2x =1⇨ x=0, for x=0,y=1⇨ A and B meet on (0,1) and A∩B≠∅.

[ Answer: A ⊆ B ⇨n(AUB)=n(B), n(A∩B)=n(A).]Answer: 2m-2n =112⇨ 2n(2(m-n) – 1)= 24(23 – 1).

Answer: m = mid term = T(n+1)/2 = a+(n+12 - 1)d ⇨ 2m= 2a+(n-1)d , Sn = n/2[2a+(n-1)d]=mn.Answer: 2a = b+c , c = ar3 ⇨ r = ¿ , put in G1= br & G2= br2[Hint: 2A = a+b, b/g2 = g2/g1 = g1/a ⇨ a = g12 / g2 , b = g22 /g1.][Hint: 1

2+4+6+……….n terms = 12(1+2+3+………nterms) = 1

2 n(n+1)2

=1

n(n+1) ⇨ S∞=1 [ 11.2 + 1

2.3+......∞ =1 as given short-cut method on blog)].[Hint: a=17, d=-4/5 and let nth term be first negative term ⇨ 17+(n-1)(-4/5) < 0 ⇨ n > 89/4 ⇨ n=23]

Answer: we can write above series as 1 + 23 [1+33 +532+

…….],where a=1,d=2 and r= 1/3, then use formula of combined A.P.& G.P(Arithmetic –geometric series) [ a

1−r + dr(1−r ) ² ] or you can

do by another method S - 13 S = {1+23 + 63² +1033 + 1434 + ………..∞} – {13 + 232 +

633 +

1034 +

1435 + ………..∞}

⇨ 2S3 = 1+ (23 - 13) + 432 {1+ 13+ 132+…..∞} =2 ⇨ S=3 a/(1-r)[Hint: take D ≥ 0 , a+ar+ar2 = (ar)x ⇨ r2+(1-x)r+1=0]

[Hint: as 1/a,1/b,1/c are in H.P. ⇨ a,b,c are in A.P. therefore first and second terms are in A.P. will be 7,9 a=7, d= 2 , then find a12][Hint: a+arn-1 = 66 , (ar)( arn-2) = 128 and Sn = 126 ⇨ r=2 and n = 6]AQ : QB = 2 : 1 ⇨ Q=(4/3 ,2) then equation of line OP and OQ passing through (0,0) is 3x – 8y =0 and 3x – 2y =0 resp.]

[ Hint: Let the line AB be trisected at P and Q, then AP : PB = 1:2 , A(4,0) , B(0,3) BY using section formula we get P(8/3 , 1)

[ Hint: (x -3y+1) +k(2x+5y -9)=0 -------(1) , then find distant from (0,0) on the line (1) is √ 5 ⇨ k=7/8 ,put in (1)

Answer is 2x+y – 5=0]

[Hint: D(α ,β ¿ C(5,1)

y=2x+c M (3,2)

A(1,3) B(X,2X-4)

M(3,2) (by mid point formula) , it lies on BD ∴ c = -4 , use(AB)2 + (BC)2 = (AC)2 ⇨ x=4 or 2 ∴ B(4,4) then D (2,0), if B(2,0) then D(4,4)]

11x+7y=9

[ Hint: D C

7x+2y=0 P

O 4x+5y=0 B

B(5/3,-4/3) , D(-2/3,7/3) by solving equations of OB & BD and OD & BD resp.

Then find point P (1/2,1/2) & equation of OC i.e, OP is y=x.]

[ Hint:

D slope=-4/7 C(1,1)

Slope=7/4 slope=7/4

A(-3,1) 4x+7y+5=0 B

Equation of BC is 7x – 4y -3=0 , equation of AD & CD are 7x – 4y +25=0 & 4x+7y=11=0 resp.]

[ Hint: y

C

B(0,a)

x+2y-2a=0

o A(2a,0) X

by solving CA2 = CB2 ⇨ Y=(5a)/2 i.e, C is (2a,5a/2), equation of BC is 3x – 4y+4a=0 & area of ∆ACB is 5a2/2 sq.units.]

[ Hint Y

B(h,k)

(-4sinα, 4cosα)

C 4 4 A(4cosα, 4sinα)

M O L

Angle COM=900-α , angle AOL=α ]

Take ∆OLA , find A as OL/4=cosα & AL/4=sinα and in ∆OMC ,find point C, then find equation of OB & AC

( by using mid pointof OB & AC) equations of OB & AC are x(cosα+sinα) – y(cosα – sinα)=0 , x(cosα - sinα) + y(cosα + sinα)=4 resp.][Hint: Find all co-ordinates & for perpendicularity of diagonals show

( −aba+b , −aba+b ) D line (iii) C ( aba−b , −aba−b )

Slope=1

Line(iv)

Slope=-1 line(ii) (−aba−b ,

aba−b ) A line(i) B ( ab

a+b , aba+b )

product of slopes of AC & BD = -1×1=−1]

[Hint:

Y C

D(4,3)

(0,1)B 45 P(2,2)

X+3y-3=0

A(3,0)

P is the mid point of BD

angle ABD=450 , use formula tan 450 =| 3m+13−m |

⇨ m =1/2 or -2 , equations of BD ,AC, CD AD & BC are x-2y+2=0, 2x+y-6=0, x+3y-13=0, 3x-y-9=0 & 3x-y+1=0 resp.]

[ Hint: use formula tan 450 = m1−

815

1+m1815

⇨ m1 = 23/7 ,

(1,2) A B

m1 8x-15y=0

450 m2 =8/15

D C

Equations of AD & AB ( Perp. to each other) are 23x-7y-9=0 , 7x+23y-53=0]

Answer.2 Let (x1,y1) be the pt. of intersection of axis and directrix. By mid point formula x1=4, y1=-11, A be the vertex & F is the focus , slope of AF is -4 , then slope of directrix is ¼

Eqn. of directrix is x-4y-48=0 , . FP2=PM2 ⇨ 16x2+y2+8xy+96x-554y-1879=0.

Answer.3 same as Q.2 , eqn. of line per. To x-y+1=0 is x+y+k=0

Required eqn. is x2+y2-14x+2y+2xy+17=0.

Answer.1 Let P(x,y) be any point on the parabola whose focus is F(-1,-2) & the directrix x-2y+3 =0. Draw PM is per. From P on directrix , by defn. FP2=PM2 ⇨ (X+1)2 + (Y+2)2 =( X−2Y +3

√1+4 )2

Answer.5 Let 2a , 2b be the major & minor axes , it’s eqn. is

(x-2)2/a2 + (y+3)2/b2 = 1, C is the centre , F1, A are the one focus & vertex resp. , CF1 = ae =1 , CA=a =2 ⇨ e=1/2 , we know that b2 = a2(1-e2) ⇨ b2 = 3, find eqn. of ellipse.

Answer.8 use b2 = a2(1-e2), eqn. is x2/25 – y2/20 = 1