TBJEE 2021 Question Paper - Maths

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2021
Question Paper Series

P

MATHEMATICS JM

Time : 45 Minutes Maximum Marks : 120

Total Marks : 120 (4 × 30)

Answer all questions

This Question Paper consists of 16 pages. Each Multiple Choice Question (MCQ)
is provided with four options (A), (B), (C) and (D). Identify the correct option and
darken/fill the corresponding circle (A)/(B)/(C)/(D) with Blue/Black Ballpoint Pen
on the OMR Answer Sheet.

For each question, 4 marks will be awarded for correct answer and for each
wrong answer 1 mark will be deducted.

Τ šøìÅ¥¹ l¡üv¡¹ ƒà*

&Òü šøÅ¥šy[i¡ìt¡ 16 [i¡ ³å[‰t¡ šõË¡à "àìá¡ú šø[t¡[i¡ MCQ-&¹ Îàì= W¡à¹[i¡ δ±à¤¸ l¡üv¡¹ (A), (B), (C) &¤} (D)
ëƒ*Úà "àìá¡ú Î[k¡A¡ l¡üv¡¹[i¡ [>¤¢àW¡> A¡¹ &¤} OMR Answer Sheet-&¹ [>‹¢à[¹t¡ \àÚKàÚ l¡üv¡¹[i¡
(A)/(B)/(C)/(D) >㺠¤à A¡àìºà Ballpoint Pen [ƒìÚ ®¡[t¢¡ A¡¹¡ú

šøìt¡¸A¡ šøìÅ¥¹ Î[k¡A¡ l¡üv¡ì¹¹ \>¸ 4 >´¬¹ ëƒ*Úà Òì¤
&¤} šøìt¡¸A¡ ®å¡º l¡üv¡ì¹¹ \>¸ 1 >´¬¹ A¡ài¡à ™à줡ú

™t¡Û¡o š™¢”z >à ¤ºà Òì¤, t¡t¡Û¡o š™¢”z ë³àÒ¹ Jåºì¤ >à¡

1. Let, ρ be the relation on ℝ (set of all real numbers) defined by
ρ = { (a , b ) : a , b ∈ ℝ, a 2 + b 2 = 1 } , then ρ is

(A) symmetric and transitive

(B) symmetric but neither reflexive nor transitive

(C) transitive but neither reflexive nor symmetric

(D) None of the above

1ú ³ì> A¡¹, ρ = { (a , b ) : a , b ∈ ℝ, a 2 + b 2 = 1 } &A¡[i¡ δšA¢¡ ë™Jàì> ℝ γÑz ¤àÑz¤ Î}J¸à¹ ëÎi¡, ρ
δšA¢¡[i¡ Òì¤

(A) [Îì³[i¡öA¡ &¤} i¡öà[X[i¡®¡

(B) [Îì³[i¡öA¡ [A¡”ñ [¹ìóá[G®¡ ¤à i¡öà[X[i¡®¡ >Ú

(C) i¡öà[X[i¡®¡ [A¡”ñ [¹ìóá[G®¡ ¤à [Îì³[i¡öA¡ >Ú

(D) l¡üšì¹¹ ëA¡àì>à[i¡Òü >Ú

2. If [x] denotes the greatest integer less than or equal to x, then the range of the function
f (x ) = [x ] − x is

(A) [0,1) (B) (–1,0]

(C) (–∞,∞) (D) (–1,1)

2ú ™[ƒ [x], &A¡[i¡ Î줢àZW¡ šèo¢Î}J¸à ë¤àc¡àÚ ™à x - &¹ ë=ìA¡ ëáài¡ ¤à γà> ( ≤ x ) ÒÚ, t¡ì¤ f (x ) = [x ] − x
"ìšÛ¡A¡[i¡¹ [¤Ñz๠"e¡º[i¡ Òì¤

(A) [0,1) (B) (–1,0]

(C) (–∞,∞) (D) (–1,1)

JM—Series P 2

3. z is a complex number such that z − 1 + z + 1 ≤ 4 . Then z lies in Argand plane

(A) on the boundary and in the interior of an ellipse

(B) on the boundary and in the interior of a circle

(C) in the interior of a hyperbola

(D) None of the above

3ú \[i¡º ¹à[Å z ™[ƒ z −1 + z +1 ≤ 4 δšA¢¡[i¡ìA¡ [·ý¡ A¡ì¹, t¡àÒìº "à¹Ksi¡ t¡ìº z -&¹ "¤Ñ‚à>
Òì¤

(A) &A¡[i¡ l¡üš¤õìv¡¹ l¡üšì¹ &¤} [®¡t¡ì¹

(B) &A¡[i¡ ¤õìv¡¹ l¡üšì¹ &¤} [®¡t¡ì¹

(C) &A¡[i¡ š¹à¤õìv¡¹ [®¡t¡ì¹

(D) l¡üšì¹¹ ëA¡à>[i¡Òü >Ú

4. Solution of the differential equation x 2 (x dx + y dy ) + 2y (x dy − y dx ) = 0 , subject to the
condition y(1) = 0, is

(A) (x 2 + y 2 )(x − 2)2 = 4x 2 (B) (x 2 + y 2 )(x + 2)2 = 9x 2

(C) (x 2 − y 2 )(x + 2)2 = 4x 2 (D) (x 2 − y 2 )(x − 2)2 = 9x 2

4ú x 2 (x dx + y dy ) + 2y (x dy − y dx ) = 0 &Òü "¤A¡º γãA¡¹o[i¡¹ y (1) = 0 Åt¢¡ ë³ì> ë™ Î³à‹à> Òì¤,
t¡à Òº

(A) (x 2 + y 2 )(x − 2)2 = 4x 2 (B) (x 2 + y 2 )(x + 2)2 = 9x 2

(C) (x 2 − y 2 )(x + 2)2 = 4x 2 (D) (x 2 − y 2 )(x − 2)2 = 9x 2

JM—Series P 3 [ P.T.O.

5. In the quadratic equation ax 2 + bx + c = 0, if ∆ = b 2 − 4ac and α + β , α2 + β2 , α3 + β3
are in GP, where α, β are the roots of ax 2 + bx + c = 0, then

(A) ∆ ≠ 0 (B) b∆ = 0

(C) c ∆ = 0 (D) ∆ = 0

5ú ax 2 + bx + c = 0 [‡Qàt¡ γãA¡¹ìo¹ ¤ã\‡Ú Òº α * β, ™[ƒ α + β , α2 + β2 , α3 + β3 GP ët¡ =àìA¡
&¤} ∆ = b 2 − 4ac ÒÚ, t¡àÒìº

(A) ∆ ≠ 0 (B) b∆ = 0

(C) c ∆ = 0 (D) ∆ = 0

 sin2 − 1 
6. The value of tan−1   is
 cos 2 

π π
(A) −1 (B) 1 −
2 4

π π
(C) 2 − (D) −1
2 4

 sin2 − 1 
6ú tan−1   - &¹ ³à> Òº
 cos 2 

π π
(A) −1 (B) 1 −
2 4

π π
(C) 2 − (D) −1

2 4

JM—Series P 4

7. The mean and standard deviation of 100 observations were found to be 40 and 10
respectively. If at the time of calculation two observations were wrongly taken as 30 and
70 in place of 3 and 27 respectively, then the correct standard deviation is

(A) 8·24 (B) 9·24

(C) 10·24 (D) 7·24

7ú 100 [i¡ š™¢ì¤Û¡ìo¹ KØl¡ (mean) &¤} γ¸A¡ [¤W¡å¸[t¡ (standard deviation) Òº ™=àyû¡ì³ 40 &¤} 10 .
™[ƒ [ÒìÎ줹 Î³Ú ƒå[i¡ š™¢ì¤Û¡o 30 &¤} 70 ëA¡ ®å¡º A¡ì¹ ™=àyû¡ì³ 3 &¤} 27 &¹ š[¹¤ìt¢¡ ë>*Úà ÒìÚ
=àìA¡, t¡àÒìº Î[k¡A¡ γ¸A¡ [¤W塸[t¡ Òì¤

(A) 8·24 (B) 9·24

(C) 10·24 (D) 7·24

8. The statement ( p → r ) ∨ (q → r ) is logically equivalent to

(A) ( p ∧ q ) ∨ r

(B) ( p ∨ q ) → r

(C) ( p ∧ q ) → r

(D) ( p → q ) → r

8ú ( p → r ) ∨ (q → r ) l¡ü[v¡û¡[i¡¹ γt塺¸ l¡ü[v¡û¡[i¡ Òì¤

(A) ( p ∧ q ) ∨ r

(B) ( p ∨ q ) → r

(C) ( p ∧ q ) → r

(D) ( p → q ) → r

JM—Series P 5 [ P.T.O.

9. The normals to the curve y = x 2 − x + 1 , drawn at the points with the abscissa x1 = 0 ,
5
x 2 = −1 and x3 = are
2
(A) parallel to each other (B) pairwise perpendicular

(C) concurrent (D) not concurrent

5
9ú y = x 2 − x + 1 ¤yû¡ì¹Jà¹ ë™ Î¤ [¤@ƒåìt¡ ®å \ x1 = 0 , x 2 = −1 , x3 = &¤} ëÎÒü [¤@ƒåP¡[ºìt¡ "[S¡t¡
2
"[®¡º´¬P¡[º
(A) š¹Ñš¹ γà”z¹àº (B) šø[t¡ ™åKº &ìA¡ "šì¹¹ l¡üš¹ º´¬

(C) γ[¤@ƒå (D) γ[¤@ƒå >Ú

ab
10. In a triangle ABC if sin A sin B = , then the triangle is
c2
(A) equilateral (B) isosceles

(C) right angled (D) None of the above

10ú ABC [y®å¡ì\ sin A sin B = ab2 Òìº, ABC [y®å¡\[i¡ Òì¤
c
(A) γ¤à× (B) γ[‡¤à×

(C) γìA¡àoã (D) l¡üšì¹¹ ëA¡à>[i¡Òü >Ú

11. The area of the region bounded by y =|x| and y = −|x|+2 is

(A) 4 sq.units (B) 3 sq.units

(C) 2 sq.units (D) 1 sq.unit

11ú y =|x| &¤} y = −|x|+2 ‡à¹à Îã³à¤‡ý¡ "e¡ìº¹ ëÛ¡yó¡º Òº
(A) 4 ¤K¢ &A¡A¡ (B) 3 ¤K¢ &A¡A¡

(C) 2 ¤K¢ &A¡A¡ (D) 1 ¤K¢ &A¡A¡

JM—Series P 6

x+p q r
12. If q x +r p = 0 , then the values of x are
r p x +q

(A) −( p + q + r ), ± p 2 + q 2 + r 2 + pq + qr + rp

(B) ± ( p 2 + q 2 + r 2 ), ± p 2 + q 2 + r 2 + pq + qr + rp

(C) 0, ± ( p + q + r )

(D) −( p + q + r ), ± p 2 + q 2 + r 2 − pq − qr − rp

x+p q r
12ú ™[ƒ q x +r p = 0 ÒÚ, t¡ì¤ x-&¹ ³à>P¡[º Òì¤
r p x +q

(A) −( p + q + r ), ± p 2 + q 2 + r 2 + pq + qr + rp

(B) ± ( p 2 + q 2 + r 2 ), ± p 2 + q 2 + r 2 + pq + qr + rp

(C) 0, ± ( p + q + r )

(D) −( p + q + r ), ± p 2 + q 2 + r 2 − pq − qr − rp

2
d 2y
{ ( 2
13. If y = loge x + x + a
2
)} , then ( x2 + a2 ) dx 2
+x
dy
dx
=?

(A) 2 (B) a2y

(C) –a 2y (D) –2

2 2
{ (
13ú ™[ƒ y = loge x + x2 + a2 )} ÒÚ, t¡àÒìº ( x + a ) ddxy + x dy
2 2
dx
=? 2

(A) 2 (B) a2y

(C) –a 2y (D) –2

JM—Series P 7 [ P.T.O.

π
x tan x
14. ∫ sec x + tan x dx = ?
0

π π 
(A) −1 (B) π  + 1
2 2 

π π 
(C) +1 (D) π  − 1
2 2 

π
x tan x
14ú ∫ sec x + tan x dx = ?
0

π π 
(A) −1 (B) π  + 1
2  2 

π π 
(C) +1 (D) π  − 1
2  2 

e x (x 2 + 1)
15. ∫ (x + 1)2 dx = ?

x x −1 1
(A) e +C (B) e x +C
x +1 (x + 1)2

x +1 x 1
(C) e x +C (D) e 2
+C
x −1 x +1

e x (x 2 + 1)
15ú ∫ dx = ?
(x + 1)2

x x −1 1
(A) e +C (B) e x +C
x +1 (x + 1)2

x +1 x 1
(C) e x +C (D) e 2
+C

x −1 x +1

JM—Series P 8

16. If c 0 , c1, c 2 ,……, cn denote the coefficients in the expansion of (1 + x )n , then the value
of c1 + 2c 2 + 3c 3 + … + nc n is

(A) (n + 1)2n −1 (B) n 2n −1

n n −1
(C) (n + 1)2 (D) (n + 2)2

16ú (1 + x )n -&¹ [¤Ñzõ[t¡¹ ÎÒKP¡[º ™[ƒ c 0 , c1, c 2 ,……, cn ÒÚ, t¡àÒìº c1 + 2c 2 + 3c 3 + … + nc n -&¹ ³à>
Òì¤

(A) (n + 1)2n −1 (B) n 2n −1

n n −1
(C) (n + 1)2 (D) (n + 2)2

17. If sec ax + sec bx = 0 , then the values of x form

(A) two arithmetic progressions

(B) two geometric progressions

(C) one arithmetic progression and one geometric progression

(D) None of the above

17ú ™[ƒ sec ax + sec bx = 0 ÒÚ, t¡ì¤ x-&¹ ³à>γèÒ

(A) ƒå[i¡ γà”z¹ šøK[t¡ Kk¡> A¡ì¹

(B) ƒå[i¡ P¡ì>àv¡¹ šøK[t¡ Kk¡> A¡ì¹

(C) &A¡[i¡ γà”z¹ šøK[t¡ &¤} &A¡[i¡ P¡ì>àv¡¹ šøK[t¡ Kk¡> A¡ì¹

(D) l¡üšì¹¹ ëA¡à>[i¡Òü >Ú

JM—Series P 9 [ P.T.O.

1
18. If the function f (x ) = x 3 + bx 2 + ax + 5 satisfies Rolle’s theorem on [1,3] with c = 2 + ,
3
then

(A) a = 11, b = −6 (B) a = 11, b = 6

(C) a = −11, b = 6 (D) a = −11, b = −6

18ú ™[ƒ c = 2 + 1 - &¹ \>¸ [1,3] [¤Ñzàì¹ f (x ) = x 3 + bx 2 + ax + 5 "ìšÛ¡A¡[i¡ Rolle’s theorem ëA¡
3
[·ý¡ A¡ì¹, t¡àÒìº

(A) a = 11, b = −6 (B) a = 11, b = 6

(C) a = −11, b = 6 (D) a = −11, b = −6

19. A function whose graph is symmetrical about y-axis is given by

2
(A) f (x ) = loge (x + x + 1) (B) f ( x + y ) = f ( x ) + f (y ), ∀ x , y ∈ R

(C) f ( x ) = cos x + sin x (D) None of the above

19ú [>ì´•¹ ë™ "ìšÛ¡A¡[i¡¹ Køàó¡ y-"ìÛ¡¹ ÎàìšìÛ¡ šø[t¡Î³, ëÎ[i¡ Òº

2
(A) f (x ) = loge (x + x + 1) (B) f (x + y ) = f (x ) + f (y ), ∀ x , y ∈ R

(C) f (x ) = cos x + sin x (D) l¡üšì¹¹ ëA¡à>[i¡Òü >Ú

JM—Series P 10

20. Let, the variables x1 and x2 satisfy the following conditions :

3x1 + x 2 ≤ 15
3x1 + 4x 2 ≤ 24
x1, x 2 ≥ 0

Then the maximum value of the function Z = 4x1 + 3x 2 is

(A) 20 (B) 25

(C) 15 (D) 30

20ú ³ì> A¡¹, x1 &¤} x2 W¡º¹à[Å ƒå[i¡ >ãìW¡¹ Åt¢¡P¡[º šè¹o A¡ì¹ –

3x1 + x 2 ≤ 15
3x1 + 4x 2 ≤ 24
x1, x 2 ≥ 0

t¡àÒìº Z = 4x1 + 3x 2 "ìšÛ¡A¡[i¡¹ W¡¹³ ³à> Òì¤

(A) 20 (B) 25ø

(C) 15 (D) 30

21. How many 5-digit numbers divisible by 3 can be formed by using the digits 0,1,2,3,4
and 5, without repetition of digits?

(A) 148 (B) 224

(C) 336 (D) 216

21ú 0,1,2,3,4 * 5 Î}J¸àP¡[º &A¡¤àì¹¹ ë¤[Å ¤¸¤Ò๠>à A¡ì¹ 3 ‡à¹à [¤®¡à\¸ 5 "ìS¡¹ A¡t¡P¡[º Î}J¸à Kk¡>
A¡¹à ë™ìt¡ šàì¹?

(A) 148¡ (B) 224

(C) 336 (D) 216

JM—Series P 11 [ P.T.O.

x [x ]
22. If [x] denotes the greatest integer less than or equal to x, then Lt =?
x →0 sin| x |

(A) 0 (B) 1

(C) Does not exist (D) –1

x [x ]
22ú ™[ƒ [x] &A¡[i¡ Î줢àZW¡ šèo¢Î}J¸à ë¤àc¡àÚ ™à x -&¹ ë=ìA¡ ëáài¡ ¤à γà> ( ≤ x ) ÒÚ, t¡ì¤ Lt =?
x →0 sin| x |
(A) 0 (B) 1

(C) "[Ñzâ« ë>Òü (D) –1

→ ^ ^ ^ → ^ → ^ ^ ^ → → →
23. If a = i + j + k , b = i and c = c1 i + 2 j + c 3 k , then a , b , c will be coplanar for

(A) c1 = 1 and c 3 = any real number

(B) c1 = 2 and c 3 = 1

(C) c1 = any real number and c 3 = 1

(D) c1 = any real number and c 3 = 2

→ ^ ^ → ^ → → → →
23ú ™[ƒ a = i + j^ + k , b = i &¤} c = c1 ^i + 2 ^j + c 3 k^ ÒÚ, t¡àÒìº a , b , c &A¡Òü γt¡ìº "¤[Ñ‚t¡ Òì¤
™J>

(A) c1 = 1 &¤} c 3 = ë™ìA¡à> ¤àÑz¤ Î}J¸à

(B) c1 = 2 &¤} c 3 = 1

(C) c1 = ë™ìA¡à> ¤àÑz¤ Î}J¸à &¤} c 3 = 1

(D) c1 = ë™ìA¡à> ¤àÑz¤ Î}J¸à &¤} c 3 = 2

JM—Series P 12

24. The ratio in which the yz plane divides the line joining the points (1,2,3) and (4,5,6)
is

(A) –1:2 (B) –2:5

(C) 2:3 (D) –1:4

24ú (1,2,3) &¤} (4,5,6) [¤@ƒåKà³ã ιºì¹JàìA¡ yz t¡º ë™ ">åšàìt¡ [¤®¡v¡û¡ A¡ì¹, t¡à Òº
(A) ¡–1:2 (B) –2:5

(C) 2:3 (D) –1:4

25. The equation of the plane passing through the line of intersection of the planes
2x + 3y − 5z + 7 = 0 , 7 x − 4y + 3z − 11 = 0 and parallel to the line joining the points
(3,1,–2) and (1,–2,4) is

(A) 333x – 124y + 49z – 361 = 0 (B) 124x – 333y + 49z + 361 = 0

(C) 49x + 124y + 331z + 61 = 0 (D) 330x + 120y + 40z + 361 = 0

25ú 2x + 3y − 5z + 7 = 0 &¤} 7x − 4y + 3z − 11 = 0 γt¡º‡ìÚ¹ ëრιºì¹JàKà³ã &¤} (3,1–2) *
(1,–2,4) [¤@ƒå‡ìÚ¹ Î}ì™àKA¡à¹ã ιºì¹J๠ÎìU γà”z¹àº γt¡ìº¹ γãA¡¹o Òº

(A) 333x – 124y + 49z – 361 = 0 (B) 124x – 333y + 49z + 361 = 0

(C) 49x + 124y + 331z + 61 = 0 (D) 330x + 120y + 40z + 361 = 0

26. Two positive numbers x and y are such that x 2 + y 2 = a 2 , (a > 0) . Then the sum x + y
will be maximum when

a a
(A) x = ,y = (B) x = a , y = a
2 2

a a
(C) x = a , y = (D) x = ,y =a
2 2

26ú x &¤} y ƒå[i¡ ‹>àâ«A¡ Î}J¸à &¤} x 2 + y 2 = a 2 , (a > 0) ú x &¤} y &¹ ë™ ³àì>¹ \>¸ x + y ¹à[Å[i¡¹
³à> W¡¹³ Òì¤, t¡à Òº
a a
(A) x = ,y = (B) x = a , y = a
2 2

a a
(C) x = a , y = (D) x = ,y =a
2 2

JM—Series P 13 [ P.T.O.

27. If the straight lines ax + y + 1 = 0 , x + by + 1 = 0 , x + y + c = 0 (a, b, c are unequal and ≠ 1 )
1 1 1
are concurrent, then + + =?
1−a 1−b 1−c

(A) 0 (B) 2

(C) 1 (D) None of the above

27ú ™[ƒ ax + y + 1 = 0 , x + by + 1 = 0 , x + y + c = 0 (a, b, c "γà> &¤} ≠ 1 ) ιºì¹Jà [t¡>[i¡ γ[¤@ƒå
1 1 1
ÒÚ, t¡ì¤ + + =?
1−a 1−b 1−c

(A) 0 (B) 2

(C) 1 (D) l¡üšì¹¹ ëA¡à>[i¡Òü >Ú

28. The equations of tangents to the hyperbola 3x 2 − y 2 = 3 parallel to the straight line
2x − y + 4 = 0 are

(A) y = 2x ± 3 (B) y = 2x ± 1

(C) y = 2x ± 2 (D) y = 2x ± 5

28ú 3x 2 − y 2 = 3 š¹à¤õìv¡¹, 2x − y + 4 = 0 ιºì¹J๠Î[Òt¡ γà”z¹àº њŢA¡P¡[º¹ γãA¡¹o Òì¤

(A) y = 2x ± 3 (B) y = 2x ± 1

(C) y = 2x ± 2 (D) y = 2x ± 5

JM—Series P 14

29. Two variable straight lines ax cos α + by sin α = a and ax sin α − by cos α = b , (b > a > 0)
intersect at the point P, α being a parameter. Then the locus of P is an ellipse with
eccentricity

b2 − a 2 b2 − a 2
(A) (B)
a b

b2 − a 2
(C) (D) None of the above
b2 + a 2

29ú ax cos α + by sin α = a &¤} ax sin α − by cos α = b , (b > a > 0) ƒå[i¡ š[¹¤t¢¡>Å㺠ιºì¹J๠ëáƒ
[¤@ƒå P, ë™Jàì> α &A¡[i¡ š¸à¹à[³i¡à¹¡ú P-&¹ Îe¡à¹š=[i¡ Òº &A¡[i¡ l¡üš¤õv¡ ™à¹ l¡ü;ìA¡@ƒøt¡à

b2 − a 2 b2 − a 2
(A) (B)
a b

b2 − a2
(C) (D) l¡üšì¹¹ ëA¡à>[i¡Òü >Ú
b2 + a 2

30. Let, A is a 3×3 matrix and B is its adjoint matrix. If | B | = 144 , then | A | = ?

(A) ±2 (B) ±12

(C) ±8 (D) ±48

30ú ³ì> A¡¹, A &A¡[i¡ 3×3 ³¸à[i¡öG &¤} B Òº A - &¹ Î}ºN¥ (adjoint) ³¸à[i¡öG ™[ƒ | B | = 144 ÒÚ, t¡ì¤
| A |= ?

(A) ±2 (B) ±12

(C) ±8 (D) ±48

JM—Series P 15 [ P.T.O.

SPACE FOR ROUGH WORK

★ ★ ★

JM—Series P 16 FF21—700×3