CBSE Class 12 Maths 2026 Question Paper 65/2/1 with Solutions
All 45 questions from the CBSE Class 12 Mathematics board paper, Set 65/2/1 (2026),
with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.
A relation R on set A = { 1 , 2 , 3 } defined as R = {( 1 , 1 ) , ( 2 , 2 ) , ( 1 , 2 )} is
(A) Reflexive only(B) Reflexive and Transitive(C) Symmetric and Transitive(D) Transitive only
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Answer: (D) Transitive only
( 3 , 3 ) ∈ / R , so R is not reflexive.( 1 , 2 ) ∈ R but ( 2 , 1 ) ∈ / R , so R is not symmetric.The only pairs that chain are ( 1 , 1 ) , ( 1 , 2 ) giving ( 1 , 2 ) ∈ R and ( 1 , 2 ) , ( 2 , 2 ) giving ( 1 , 2 ) ∈ R , so R is transitive. Hence R is transitive only.
If A and B are square matrices of same order, then which of the following statements is/are always true ? (i) ( A + B ) ( A − B ) = A 2 − B 2 (ii) A B = B A (iii) ( A + B ) 2 = A 2 + A B + B A + B 2 (iv) A B = 0 ⇒ A = 0 or B = 0
(A) Only (i) and (iii)(B) Only (ii) and (iii)(C) Only (iii)(D) Only (iii) and (iv)
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Answer: (C) Only (iii)
Matrix multiplication is not commutative, so (ii) fails in general and hence ( A + B ) ( A − B ) = A 2 − A B + B A − B 2 = A 2 − B 2 in general; (i) fails. ( A + B ) 2 = ( A + B ) ( A + B ) = A 2 + A B + B A + B 2 always; (iii) is true.Two non-zero matrices can have product zero, e.g. [ 0 0 1 0 ] [ 0 0 1 0 ] = 0 ; (iv) fails. Only (iii) is always true.
If A = 1 − 1 0 a 2 5 b c 3 is a symmetric matrix, then the value of 3 a + b + c is
(A) 2(B) 6(C) 4(D) 0
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Answer: (A) 2
For a symmetric matrix a ij = a j i . a = a 12 = a 21 = − 1 , b = a 13 = a 31 = 0 , c = a 23 = a 32 = 5 .3 a + b + c = − 3 + 0 + 5 = 2 .
If A = [ cos x sin x − sin x cos x ] and A + A ′ = I , then the value of x ∈ [ 0 , 2 π ] is
(A) 0 (B) 4 π (C) 3 π (D) 2 π
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Answer: (C) 3 π
A ′ = [ cos x − sin x sin x cos x ] , so A + A ′ = [ 2 cos x 0 0 2 cos x ] .A + A ′ = I gives 2 cos x = 1 , i.e. cos x = 2 1 .In [ 0 , 2 π ] , x = 3 π .
For a square matrix A, ( 3 A ) − 1 =
(A) 3 A − 1 (B) 9 A − 1 (C) 3 1 A − 1 (D) 9 1 A − 1
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Answer: (C) 3 1 A − 1
( 3 A ) ( 3 1 A − 1 ) = A A − 1 = I .So ( 3 A ) − 1 = 3 1 A − 1 .
If − 1 − 2 0 − 2 a 4 5 − 1 2 a = − 86 , then the sum of all possible values of a is
(A) 4(B) 5(C) − 4 (D) 9
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Answer: (C) − 4
Expanding along R 1 : − 1 ( 2 a 2 + 4 ) + 2 ( − 4 a − 0 ) + 5 ( − 8 − 0 ) = − 2 a 2 − 8 a − 44 . − 2 a 2 − 8 a − 44 = − 86 ⇒ a 2 + 4 a − 21 = 0 ⇒ ( a + 7 ) ( a − 3 ) = 0 .a = − 7 or a = 3 ; sum = − 4 .
If e − x + e − y = 2 , then d x d y is
(A) e x − y (B) e y − x (C) − e x − y (D) − e y − x
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Answer: (D) − e y − x
Differentiating: − e − x − e − y d x d y = 0 . d x d y = − e − y e − x = − e y − x .
For f ( x ) = x + x 1 ( x = 0 )
(A) local maximum value is 2(B) local minimum value is − 2 (C) local maximum value is − 2 (D) local minimum value < local maximum value
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Answer: (C) local maximum value is − 2
f ′ ( x ) = 1 − x 2 1 = 0 ⇒ x = ± 1 ; f ′′ ( x ) = x 3 2 .f ′′ ( − 1 ) = − 2 < 0 : local maximum at x = − 1 , value f ( − 1 ) = − 2 .f ′′ ( 1 ) = 2 > 0 : local minimum at x = 1 , value f ( 1 ) = 2 .So the local maximum value is − 2 (and it is less than the local minimum value 2).
If ∫ 0 2 a 1 + 4 x 2 1 d x = 6 π , then the value of a is
(A) 4 3 (B) 2 3 (C) 3 (D) 2 3
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∫ 0 2 a 1 + 4 x 2 d x = 2 1 [ tan − 1 2 x ] 0 2 a = 2 1 tan − 1 4 a .2 1 tan − 1 4 a = 6 π ⇒ tan − 1 4 a = 3 π ⇒ 4 a = 3 .a = 4 3 .
Which of the following expressions will give the area of region bounded by the curve y = x 2 and line y = 16 ?
(A) ∫ 0 4 x 2 d x (B) 2 ∫ 0 4 x 2 d x (C) ∫ 0 16 y d y (D) 2 ∫ 0 16 y d y
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The region lies between y = 0 and y = 16 , with x from − y to y . It is symmetric about the y-axis, so area = 2 ∫ 0 16 x d y = 2 ∫ 0 16 y d y .
The general solution of the differential equation d x d y = x y is
(A) log y = log x + C (B) y + x = C (C) y − x = C (D) log y + log x = C
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Separating: y d y = x d x . Integrating: 2 y = 2 x + C 1 . y − x = C .
The integrating factor of the differential equation 2 x d x d y − y = 3 is
(A) x (B) x 1 (C) e x (D) e − x
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Standard form: d x d y − 2 x 1 y = 2 x 3 , so P = − 2 x 1 . I.F. = e ∫ − 2 x 1 d x = e − 2 1 l o g x = x 1 .
If ∣ a ∣ = 5 and − 2 ≤ λ ≤ 1 , then the sum of greatest and the smallest value of ∣ λ a ∣ is
(A) − 5 (B) 5(C) 10(D) 15
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Answer: (C) 10
∣ λ a ∣ = ∣ λ ∣ ⋅ 5 .For − 2 ≤ λ ≤ 1 , 0 ≤ ∣ λ ∣ ≤ 2 . Greatest value = 10 (at λ = − 2 ), smallest = 0 (at λ = 0 ); sum = 10 .
Vector of magnitude 3 making equal angles with x and y axes and perpendicular to z axis is
(A) i ^ + 2 2 j ^ (B) 3 k ^ (C) 2 3 2 i ^ + 2 3 2 j ^ (D) 3 i ^ + 3 j ^ + 3 k ^
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Answer: (C)
2 3 2 i ^ + 2 3 2 j ^
Perpendicular to the z-axis: n = 0 . Equal angles with x and y axes: l = m . l 2 + m 2 = 1 ⇒ l = m = 2 1 (taking positive values).Vector = 3 ( 2 1 i ^ + 2 1 j ^ ) = 2 3 2 i ^ + 2 3 2 j ^ .
Direction cosines of the line given by equations : 4 2 x − 1 = 3 1 − y = 6 − z are
(A) 2 , − 3 , − 6 (B) 7 2 , 7 − 3 , 7 − 6 (C) 7 2 , 7 − 3 , 7 6 (D) 61 4 , 61 − 3 , 61 − 6
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Answer: (B) 7 2 , 7 − 3 , 7 − 6
Write in standard form: 2 x − 2 1 = − 3 y − 1 = − 6 z − 0 . Direction ratios 2 , − 3 , − 6 ; 4 + 9 + 36 = 7 . Direction cosines 7 2 , 7 − 3 , 7 − 6 .
In a linear programming problem, the linear function which has to be maximized or minimized is called
(A) a feasible function(B) an objective function(C) an optimal function(D) a constraint
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Answer: (B) an objective function
The linear function to be optimised in an LPP is called the objective function.
For the feasible region shown below, the non-trivial constraints of the linear programming problem are
(A) x + y ≤ 5 , x + 3 y ≤ 9 (B) x + y ≤ 5 , x + 3 y ≥ 9 (C) x + y ≥ 5 , x + 3 y ≤ 9 (D) x + y ≥ 5 , 3 x + y ≤ 9
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Answer: (C) x + y ≥ 5 , x + 3 y ≤ 9
The line through ( 0 , 5 ) and ( 5 , 0 ) is x + y = 5 ; the line through ( 0 , 3 ) and ( 9 , 0 ) is x + 3 y = 9 . The shaded region (vertices ( 3 , 2 ) , ( 5 , 0 ) , ( 9 , 0 ) ) lies on the side of x + y = 5 away from the origin: x + y ≥ 5 . It lies on the origin side of x + 3 y = 9 : x + 3 y ≤ 9 .
For two events A and B such that P ( A ) = 0 and P ( B ) = 1 , P ( A ′ / B ′ ) =
(A) 1 − P ( A / B ) (B) 1 − P ( A ′ / B ) (C) P ( B ′ ) 1 − P ( A ∩ B ) (D) P ( B ′ ) 1 − P ( A ∪ B )
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Answer: (D) P ( B ′ ) 1 − P ( A ∪ B )
P ( A ′ / B ′ ) = P ( B ′ ) P ( A ′ ∩ B ′ ) .A ′ ∩ B ′ = ( A ∪ B ) ′ , so P ( A ′ ∩ B ′ ) = 1 − P ( A ∪ B ) .P ( A ′ / B ′ ) = P ( B ′ ) 1 − P ( A ∪ B ) .
For two vectors a and b Assertion (A) : ∣ a × b ∣ 2 + ( a ⋅ b ) 2 = ∣ a ∣ 2 ∣ b ∣ 2 Reason (R) : ∣ a × b ∣ = ( a ⋅ b ) tan θ , ( θ = 2 π )
(A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (B) Both A and R are true, but R is not the correct explanation of A.
∣ a × b ∣ 2 + ( a ⋅ b ) 2 = ∣ a ∣ 2 ∣ b ∣ 2 sin 2 θ + ∣ a ∣ 2 ∣ b ∣ 2 cos 2 θ = ∣ a ∣ 2 ∣ b ∣ 2 for all vectors, so A is true.( a ⋅ b ) tan θ = ∣ a ∣∣ b ∣ cos θ ⋅ c o s θ s i n θ = ∣ a ∣∣ b ∣ sin θ = ∣ a × b ∣ for θ = 2 π , so R is true.A holds for all angles (including θ = 2 π ) and follows directly from sin 2 θ + cos 2 θ = 1 , not from R; so R is not the correct explanation of A.
Assertion (A) : A line can have direction cosines < 1 , 1 , 1 > Reason (R) : cos θ = 1 is possible for θ = 0 .
(A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (D) Assertion (A) is false, but Reason (R) is true.
For direction cosines l 2 + m 2 + n 2 = 1 , but 1 2 + 1 2 + 1 2 = 3 = 1 ; A is false. cos 0 = 1 ; R is true.
Check whether f : R − { 3 } → R defined as f ( x ) = x − 3 x − 2 is onto or not.
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Answer: Not onto (1 has no pre-image).
Let y = x − 3 x − 2 . Then x y − 3 y = x − 2 ⇒ x ( y − 1 ) = 3 y − 2 ⇒ x = y − 1 3 y − 2 . This is not defined for y = 1 . Indeed x − 3 x − 2 = 1 would need x − 2 = x − 3 , which is impossible. So 1 ∈ R has no pre-image; f is not onto.
OR
Check whether f : Z × Z → Z × Z (where Z is the set of integers) defined as f ( x , y ) = ( 2 y , 3 x ) is injective or not.
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Answer: Injective.
Let f ( x 1 , y 1 ) = f ( x 2 , y 2 ) . Then ( 2 y 1 , 3 x 1 ) = ( 2 y 2 , 3 x 2 ) , so 2 y 1 = 2 y 2 and 3 x 1 = 3 x 2 . Hence y 1 = y 2 , x 1 = x 2 , i.e. ( x 1 , y 1 ) = ( x 2 , y 2 ) . f is injective (one-one).
If x = a sin 3 t , y = b cos 3 t , then find d x d y at t = 4 π .
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Answer: − a b
d t d x = 3 a sin 2 t cos t , d t d y = − 3 b cos 2 t sin t .d x d y = 3 a s i n 2 t c o s t − 3 b c o s 2 t s i n t = − a b cot t .At t = 4 π : d x d y = − a b .
Find the absolute maximum value of f ( x ) = cos x + sin 2 x , x ∈ [ 0 , π ]
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Answer: 4 5 (at x = 3 π )
f ′ ( x ) = − sin x + 2 sin x cos x = sin x ( 2 cos x − 1 ) .f ′ ( x ) = 0 ⇒ x = 0 , π or cos x = 2 1 ⇒ x = 3 π .f ( 0 ) = 1 , f ( 3 π ) = 2 1 + 4 3 = 4 5 , f ( π ) = − 1 .Absolute maximum value = 4 5 .
OR
If the volume of a solid hemisphere increases at a uniform rate, prove that its surface area varies inversely as its radius.
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Answer: Proved.
Let r be the radius. V = 3 2 π r 3 and d t d V = k (constant). d t d V = 2 π r 2 d t d r = k ⇒ d t d r = 2 π r 2 k .Total surface area of a solid hemisphere S = 2 π r 2 + π r 2 = 3 π r 2 . d t d S = 6 π r d t d r = 6 π r ⋅ 2 π r 2 k = r 3 k .So the rate of change of the surface area varies inversely as the radius.
If A B = j ^ + k ^ and A C = 3 i ^ − j ^ + 4 k ^ represent the two vectors along the sides AB and AC of △ A B C , prove that the median A D = 2 A B + A C , where D is midpoint of BC. Hence, find the length of median AD.
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A D = A B + B D = A B + 2 1 B C = A B + 2 1 ( A C − A B ) = 2 A B + A C .A B + A C = 3 i ^ + 0 j ^ + 5 k ^ , so A D = 2 3 i ^ + 2 5 k ^ .∣ A D ∣ = 2 1 9 + 25 = 2 34 .
Find the co-ordinates of the point on the line r = − j ^ + 3 k ^ + λ ( 2 i ^ − 2 j ^ + k ^ ) such that the sum of co-ordinates is 3.
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Answer: ( 2 , − 3 , 4 )
A general point on the line is ( 2 λ , − 1 − 2 λ , 3 + λ ) . Sum of co-ordinates: 2 λ − 1 − 2 λ + 3 + λ = 2 + λ = 3 ⇒ λ = 1 . Point: ( 2 , − 3 , 4 ) .
Find : ∫ 9 x − x 2 x + 2 d x
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Answer: − 9 x − x 2 + 2 13 sin − 1 ( 9 2 x − 9 ) + C
Write x + 2 = − 2 1 ( 9 − 2 x ) + 2 13 , where 9 − 2 x = d x d ( 9 x − x 2 ) . ∫ 9 x − x 2 − 2 1 ( 9 − 2 x ) d x = − 9 x − x 2 .9 x − x 2 = ( 2 9 ) 2 − ( x − 2 9 ) 2 , so ∫ 9 x − x 2 d x = sin − 1 2 9 x − 2 9 = sin − 1 9 2 x − 9 .Answer: − 9 x − x 2 + 2 13 sin − 1 ( 9 2 x − 9 ) + C .
Evaluate : ∫ 12 π 12 5 π 1 + c o t x d x
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Answer: 6 π
I = ∫ π /12 5 π /12 s i n x + c o s x s i n x d x .Using ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x with a + b = 2 π : I = ∫ π /12 5 π /12 c o s x + s i n x c o s x d x . Adding: 2 I = ∫ π /12 5 π /12 1 d x = 12 5 π − 12 π = 3 π . I = 6 π .
OR
Evaluate : ∫ 6 − π 2 π ( sin ∣ x ∣ + cos ∣ x ∣ ) d x
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For x < 0 : sin ∣ x ∣ + cos ∣ x ∣ = − sin x + cos x ; for x ≥ 0 : sin x + cos x . ∫ − π /6 0 ( cos x − sin x ) d x = [ sin x + cos x ] − π /6 0 = 1 − ( − 2 1 + 2 3 ) = 2 3 − 3 .∫ 0 π /2 ( sin x + cos x ) d x = [ − cos x + sin x ] 0 π /2 = 1 − ( − 1 ) = 2 .Total = 2 3 − 3 + 2 = 2 7 − 3 .
If d x d ( F ( x )) = e x + 1 1 , then find F ( x ) given that F ( 0 ) = log 2 1 .
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Answer: F ( x ) = x − log ( e x + 1 ) , i.e. log e x + 1 e x
F ( x ) = ∫ e x + 1 d x = ∫ 1 + e − x e − x d x .Put 1 + e − x = t , − e − x d x = d t : F ( x ) = − log ( 1 + e − x ) + C . F ( 0 ) = − log 2 + C = log 2 1 ⇒ C = 0 .F ( x ) = − log ( 1 + e − x ) = log e x + 1 e x = x − log ( e x + 1 ) .
Solve the following differential equation :x d x d y = y − x sin 2 ( x y ) , given that y ( 1 ) = 6 π
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Answer: cot ( x y ) = log ∣ x ∣ + 3
d x d y = x y − sin 2 x y (homogeneous). Put y = v x , d x d y = v + x d x d v .v + x d x d v = v − sin 2 v ⇒ s i n 2 v d v = − x d x .Integrating: − cot v = − log ∣ x ∣ + C 1 , i.e. cot x y = log ∣ x ∣ + C . y ( 1 ) = 6 π : cot 6 π = 3 = 0 + C ⇒ C = 3 .Solution: cot ( x y ) = log ∣ x ∣ + 3 .
OR
Find the general solution of the differential equation : y log y d y d x + x = y 2 .
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Answer: x log y = − y 2 + C
d y d x + y l o g y 1 x = y 2 l o g y 2 , linear in x.I.F. = e ∫ y l o g y d y = e l o g ( l o g y ) = log y . x log y = ∫ y 2 l o g y 2 log y d y = ∫ y 2 2 d y = − y 2 + C .General solution: x log y = − y 2 + C .
Solve the following linear programming problem graphically : Maximize Z = 10500 x + 9000 y Subject to constraintsx + y ≤ 50 2 x + y ≤ 80 x , y ≥ 0
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Answer: Maximum Z = 495000 at x = 30 , y = 20
Lines: x + y = 50 through ( 50 , 0 ) , ( 0 , 50 ) ; 2 x + y = 80 through ( 40 , 0 ) , ( 0 , 80 ) ; they meet at ( 30 , 20 ) . The feasible region (origin side of both lines, first quadrant) has corners O ( 0 , 0 ) , A ( 40 , 0 ) , B ( 30 , 20 ) , C ( 0 , 50 ) . Z ( O ) = 0 , Z ( A ) = 420000 , Z ( B ) = 315000 + 180000 = 495000 , Z ( C ) = 450000 .Maximum Z = 495000 at ( 30 , 20 ) .
The probability of hitting the target by a trained sniper is three times the probability of not hitting the target on a stormy day due to high wind speed. The sniper fired two shots on the target on a stormy day when wind speed was very high. Find the probability that (i) target is hit (ii) atleast one shot misses the target.
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Answer: (i) 16 15 (ii) 16 7
Let p = P(hit). p = 3 ( 1 − p ) ⇒ p = 4 3 , P(miss) = 4 1 . Shots are independent. (i) Target is hit (at least once) = 1 − P ( both miss ) = 1 − 16 1 = 16 15 . (ii) At least one shot misses = 1 − P ( both hit ) = 1 − 16 9 = 16 7 .
OR
Mother, Father and Son line up at random for a family picture. Let events E : Son on one end and F : Father in the middle. Find P(E/F).
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Answer: P ( E / F ) = 1
Sample space: 3 ! = 6 equally likely arrangements. F = {MFS, SFM}, so P ( F ) = 6 2 = 3 1 . In both arrangements of F the son is at an end, so E ∩ F = F and P ( E ∩ F ) = 3 1 . P ( E / F ) = P ( F ) P ( E ∩ F ) = 1 .
If P = 1 2 0 − 1 3 1 0 4 2 and Q = 2 − 4 2 2 2 − 1 − 4 − 4 5 , find (QP) and hence solve the following system of equations using matrices :x − y = 3 , 2 x + 3 y + 4 z = 17 , y + 2 z = 7
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Answer: QP = 6 I ; x = 2 , y = − 1 , z = 4
QP = 6 0 0 0 6 0 0 0 6 = 6 I , so P − 1 = 6 1 Q .The system is P X = B with X = x y z , B = 3 17 7 . X = P − 1 B = 6 1 6 + 34 − 28 − 12 + 34 − 28 6 − 17 + 35 = 6 1 12 − 6 24 = 2 − 1 4 .x = 2 , y = − 1 , z = 4 .
OR
Obtain the value of Δ = 1 + x 1 1 1 1 + y 1 1 1 1 + z in terms of x , y and z. Further, if Δ = 0 and x , y, z are non-zero real numbers, prove that x − 1 + y − 1 + z − 1 = − 1 .
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Answer: Δ = x y z + x y + y z + z x ; proved.
Expanding along R 1 : Δ = ( 1 + x ) [( 1 + y ) ( 1 + z ) − 1 ] − 1 [( 1 + z ) − 1 ] + 1 [ 1 − ( 1 + y )] . = ( 1 + x ) ( y + z + y z ) − z − y = x y z + x y + y z + z x .If Δ = 0 : x y + y z + z x = − x y z . Dividing by x y z = 0 : z 1 + x 1 + y 1 = − 1 , i.e. x − 1 + y − 1 + z − 1 = − 1 .
Find the sub intervals in which f ( x ) = cot − 1 ( sin x + cos x ) , x ∈ ( 0 , π ) is increasing and decreasing.
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Answer: Increasing on ( 4 π , π ) , decreasing on ( 0 , 4 π )
f ′ ( x ) = − 1 + ( s i n x + c o s x ) 2 c o s x − s i n x = 1 + ( s i n x + c o s x ) 2 s i n x − c o s x .The denominator is positive, so the sign of f ′ is that of sin x − cos x . f ′ ( x ) = 0 ⇒ tan x = 1 ⇒ x = 4 π in ( 0 , π ) .For 0 < x < 4 π , sin x < cos x : f ′ < 0 , f is decreasing. For 4 π < x < π , sin x > cos x : f ′ > 0 , f is increasing.
OR
A rectangle of perimeter 36 cm is revolved around one of its sides to sweep out a cylinder of maximum volume. Find the dimensions of the rectangle.
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Answer: 12 cm by 6 cm (revolved about the 6 cm side)
Let the side revolved about be h (height) and the other side r (radius). 2 ( r + h ) = 36 ⇒ h = 18 − r . V = π r 2 h = π ( 18 r 2 − r 3 ) .d r d V = π ( 36 r − 3 r 2 ) = 0 ⇒ r = 12 (r = 0 ).d r 2 d 2 V = π ( 36 − 6 r ) = − 36 π < 0 at r = 12 , so V is maximum.Dimensions: 12 cm × 6 cm, the rectangle being revolved about its 6 cm side.
Find the domain of g ( x ) = cos − 1 ( x 2 − 1 ) . Hence, find the value of x for which g ( x ) = 3 π . Also, write the range of cos − 1 x other than its principal branch.
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Answer: Domain
[ − 2 , 2 ] ;
x = ± 2 3 ; e.g. range
[ π , 2 π ]
cos − 1 needs − 1 ≤ x 2 − 1 ≤ 1 ⇒ 0 ≤ x 2 ≤ 2 ⇒ − 2 ≤ x ≤ 2 .Domain = [ − 2 , 2 ] . g ( x ) = 3 π ⇒ x 2 − 1 = cos 3 π = 2 1 ⇒ x 2 = 2 3 ⇒ x = ± 2 3 = ± 2 6 (both in the domain).A branch other than the principal one: range [ π , 2 π ] (or [ − π , 0 ] , etc.).
A line passing through the points A(1, 2, 3) and B(5, 8, 11) intersects the line r = 4 i ^ + j ^ + λ ( 5 i ^ + 2 j ^ + k ^ ) . Find the co-ordinates of the point of intersection. Hence, write the equation of a line passing through the point of intersection and perpendicular to both the lines.
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Answer: ( − 1 , − 1 , − 1 ) ;
r = − i ^ − j ^ − k ^ + μ ( − 5 i ^ + 18 j ^ − 11 k ^ )
Line AB: r = i ^ + 2 j ^ + 3 k ^ + t ( 4 i ^ + 6 j ^ + 8 k ^ ) ; general point ( 1 + 4 t , 2 + 6 t , 3 + 8 t ) . Other line: general point ( 4 + 5 λ , 1 + 2 λ , λ ) . Equating: 3 + 8 t = λ and 1 + 4 t = 4 + 5 λ ⇒ 1 + 4 t = 19 + 40 t ⇒ t = − 2 1 , λ = − 1 . Check: 2 + 6 t = − 1 = 1 + 2 λ . Point of intersection ( − 1 , − 1 , − 1 ) . Direction perpendicular to both: ( 2 i ^ + 3 j ^ + 4 k ^ ) × ( 5 i ^ + 2 j ^ + k ^ ) = − 5 i ^ + 18 j ^ − 11 k ^ . Required line: r = − i ^ − j ^ − k ^ + μ ( − 5 i ^ + 18 j ^ − 11 k ^ ) , i.e. − 5 x + 1 = 18 y + 1 = − 11 z + 1 .
Smoking increases the risk of lung problems. A study revealed that 170 in 1000 males who smoke develop lung complications, while 120 out of 1000 females who smoke develop lung related problems. In a colony, 50 people were found to be smokers of which 30 are males. A person is selected at random from these 50 people and tested for lung related problems. Based on the given information, answer the following questions : (i) What is the probability that selected person is a female ? (1) (ii) If a male person is selected, what is the probability that he will not be suffering from lung problems ? (1) (iii) (a) A person selected at random is detected with lung complications. Find the probability that selected person is a female. (2) OR (iii) (b) A person selected at random is not having lung problems, find the probability that the person is a male. (2)
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Answer: (i) 5 2 (ii) 100 83 (iii)(a) 25 8 OR (iii)(b) 425 249
Let M, F: male, female; L: lung problems. P ( M ) = 50 30 = 5 3 , P ( F ) = 5 2 , P ( L ∣ M ) = 0.17 , P ( L ∣ F ) = 0.12 . (i) P ( F ) = 50 20 = 5 2 . (ii) P ( L ′ ∣ M ) = 1 − 0.17 = 0.83 = 100 83 . (iii)(a) P ( L ) = 5 3 ( 0.17 ) + 5 2 ( 0.12 ) = 0.102 + 0.048 = 0.15 ; P ( F ∣ L ) = 0.15 0.048 = 25 8 . (iii)(b) P ( L ′ ) = 5 3 ( 0.83 ) + 5 2 ( 0.88 ) = 0.498 + 0.352 = 0.85 ; P ( M ∣ L ′ ) = 0.85 0.498 = 425 249 .
A racing track is build around an elliptical ground whose equation is given by 9 x 2 + 16 y 2 = 144 . The width of the track is 3 m as shown below : Based on given information, answer the following questions : (i) Express y as a function of x from the given equation of ellipse. (1) (ii) Integrate the function obtained in (i) with respect to x . (1) (iii) (a) Find the area of the region enclosed within the elliptical ground excluding the track using integration. (2) OR (iii) (b) Write the co-ordinates of the points P and Q where the outer edge of the track cuts x axis and y axis in first quadrant and find the area of the triangle formed by points P, O, Q using integration. (2)
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Answer: (i)
y = ± 4 3 16 − x 2 (ii)
8 3 x 16 − x 2 + 6 sin − 1 4 x + C (iii)(a)
12 π sq m OR (iii)(b) P(7, 0), Q(0, 6); area 21 sq m
(i) 16 y 2 = 144 − 9 x 2 ⇒ y = ± 4 3 16 − x 2 (upper half: y = 4 3 16 − x 2 ). (ii) ∫ 4 3 16 − x 2 d x = 4 3 [ 2 x 16 − x 2 + 8 sin − 1 4 x ] + C = 8 3 x 16 − x 2 + 6 sin − 1 4 x + C . (iii)(a) Area = 4 ∫ 0 4 4 3 16 − x 2 d x = 3 [ 2 x 16 − x 2 + 8 sin − 1 4 x ] 0 4 = 3 ( 8 ⋅ 2 π ) = 12 π sq m. (iii)(b) Outer edge has semi-axes 4 + 3 = 7 and 3 + 3 = 6 : P(7, 0), Q(0, 6). Line PQ: 7 x + 6 y = 1 ⇒ y = 6 − 7 6 x . Area of △ P O Q = ∫ 0 7 ( 6 − 7 6 x ) d x = 42 − 21 = 21 sq m.
Sports car racing is a form of motorsport which uses sports car prototypes. The competition is held on special tracks designed in various shapes. The equation of one such track is given as follows :f ( x ) = { x 4 − 4 x 2 + 4 , x 2 + 40 , 0 ≤ x < 3 x ≥ 3 Based on given information, answer the following questions : (i) Find f ′ ( x ) for 0 < x < 3 . (1) (ii) Find f ′ ( 4 ) (1) (iii) (a) Test for continuity of f ( x ) at x = 3 . (2) OR (iii) (b) Test for differentiability of f ( x ) at x = 3 . (2)
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Answer: (i) 4 x 3 − 8 x (ii) 8 (iii)(a) continuous at x = 3 OR (iii)(b) not differentiable at x = 3
(i) f ′ ( x ) = 4 x 3 − 8 x for 0 < x < 3 . (ii) For x > 3 , f ′ ( x ) = 2 x , so f ′ ( 4 ) = 8 . (iii)(a) lim x → 3 − f ( x ) = 81 − 36 + 4 = 49 , lim x → 3 + f ( x ) = 9 + 40 = 49 , f ( 3 ) = 49 ; so f is continuous at x = 3 . (iii)(b) LHD = lim h → 0 + − h f ( 3 − h ) − f ( 3 ) = 4 ( 27 ) − 8 ( 3 ) = 84 ; RHD = lim h → 0 + h f ( 3 + h ) − f ( 3 ) = lim h → 0 + h 6 h + h 2 = 6 . LHD = RHD, so f is not differentiable at x = 3 .
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