TEST-3
PHYSICS XIII-[X&Z]
14.09.2008
PART-A Q.1 to Q.3 are Q.3 are based upon a paragraph paragraph.. Each question has four choices (A), (B), (C), (D) out of which only one is correct and correct and carry 3 mar 1 mark will mark will be deducted marking. 1 marks ks each. There is NEGATIVE marking. for each wrong answer. Paragraph Paragra ph for question nos. 1 to 3 This diagram depicts a block sliding along a frictionless ramp. The eight numbered arrows in the diagram represent directions to be referred to when answering the questions.
Q.1
[Sol [S ol..
Q.2
[Sol. [So l. Q.3
[Sol. [So l.
The direc direction tion of the the accelerati acceleration on of the the block block,, when when in posit position ion I, I, is best best represen represented ted by by which which of of the arrow in the diagram. (A) 2 (B*) 4 (C) 5 (D*) None [3] Sinc Si ncee surfa surface ce is is almo almost st flat flat at pos positi ition on I. ∴ acceleration has not radial component is only tangential. ] The direct direction ion of of the accele accelerati ration on of the block block when in positio position n II is best repr represent esented ed by whi which ch of the arrow in the diagram ? (A*) 1 (B ) 3 (C) 5 ( D) 8 [3] At positi position on 2, no no tangenti tangential al componen component, t, only only radial radial compon component ent exists exists i.e. i.e. upwards upwards ] The direc directio tion n of the the accelera acceleratio tion n of the the block block (after (after leav leaving ing the the ramp) ramp) at at positio position n III is best best represented by which of the arrow in the diagram ? [3] (A) 2 (B*) 5 (C) 6 (D) None After Aft er loosing loosing conta contact ct from ramp, ramp, the the body body falls freely freely i.e. i.e. acceler acceleratio ation n is downwa downwards. rds. ] Q.4 to Q.6 have Q.6 have four choices (A), (B), (C), (D) out of which only one is correct and correct and carry 3, 4, 5 marking. 1 marks mar ks respectively. There is NEGATIVE marking. 1 mark will mark will be deducted for each wrong answer. Three identical blocks each of mass 10 kg are connected by ideal strings as shown. Block C is pulled
Q.4
Q.5 Q. 5 Q.6
towards right with a constant force F. (Neglect friction) fric tion) What is the the magnitu magnitude de of of force force F if the the tension tension in the the string string conn connectin ecting g blocks blocks B and and C is 80 N?
(A) 100 N (B) 105 N (C) 115 N Find acc accel eleera rattio ion n of of blo block ckss : 2 (A) 2 m/s (B*) 1 m/s2 (C) 4 m/s2 Find Fi nd ten tensio sion n in in the the strin string g con conne necti cting ng bl bloc ocks ks A and B : (A) 70 N (B) 80 N (C*) 90 N
(D*) 70 N
[3]
(D) None of these
[4]
(D) None of these
[5]
PHYSICS
Q.7 to Q.9 are based upon a paragraph. Each question has four choices (A), (B), (C), (D) out of which only one is correct and carry 4, 5, 6 marks respectively. There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. Paragraph for Question Nos. 7 to 9 Two flat sheets, Ssq and Scr, are shaped like a square of side 2a and a circle of radius a, respectively, as shown at right. When Ssq is located at the origin with a charge q 1 uniformly distributed over its surface it r
yields a potential V 1 and field
E1 at
point P. Similarly, when Scr is located at the origin with a charge q 2 r
uniformly distributed over its surface it yields a potential V 2 and field
E 2 at
the same point P. y
P
2a
Q.7
Q.8
If charge density is same in above three charge distributions, then (A) V1 = V2 (B) V1 < V2 (C*) V1 > V2
(D) None
x
[4]
What is the potential due to the hollow square, shown at left, at the same point P, when it has charge q 3
[5]
uniformly distributed over its surface ?
Q.9
a
q 3 ⎛ 4V1
(A*)
πV ⎞ ⎜⎜ − 2 ⎟⎟ 4 − π ⎝ q 1 q 2 ⎠
(C)
q 3 ⎜⎜
⎛ V1 ⎝ q 1
−
V2 ⎞
⎟
q 2 ⎠⎟
⎛ V1
(B)
q 3 ⎜⎜
(D)
None
⎝ q 1
+
V2 ⎞
⎟
q 2 ⎠⎟
[6]
What is the electric field at the same point P due to the hollow square ?
⎛ E1 r
E 2 ⎞
(A)
q 3 ⎜⎜
(C)
⎛ E E ⎞ q 3 ⎜⎜ 1 + 2 ⎟⎟ ⎝ q 1 q 2 ⎠
⎝ q 1 r
(B*)
πE 2 ⎞ ⎜ ⎟ − 4 − π ⎜⎝ q 1 q 2 ⎠⎟
(D)
(E
r
−
⎟ q 2 ⎠⎟
q 3 ⎛ 4E1 r
r
r
r
1
+ E2 ) r
[REASONING TYPE] Q.10 to Q.12 are Reasoning type questions, contains Statement-1 (Assertion) and Statement-2 (Reason). Each question has 4 choices (A), (B), (C) and (D), out of which only one is correct and carry 3 marks each. There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. Q.10
Statement-1 : Two electrostatic field lines can never cross each other at any point Statement-2 : An electrostatic field line is a continuous curve. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true.
XIII (X&Z)-RT-3 [Paper-I]
Page # 2
PHYSICS
Statement-1 : When we run on rough ground without slipping our kinetic energy increases due to positive work done by friction which accelerates us. Statement-2 : On a smooth floor we can’t accelerate. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D*) Statement-1 is false, statement-2 is true. [Sol.anu The gain in KE is due to work done by muscles & work done by friction is (–)ve. ] Q.11
Q.12
Statement-1 : When a car accelerates hard, the common human response is to feel “pushed back into the seat ”. Statement-2 : Every Action has equal and opposite reaction (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true.
PART-B Q.1 is "Match the Column" type and carry 15 marks. Column-I and column-II contains four entries each. Entry of column-I are to be uniquely matched with only one entry of column-II. 3 mark will be awarded for each (A), (B), (C) and (D) and 3 Bonus marks will be awarded if all the matches are correct. There is NEGATIVE marking. 1 mark will be deducted for each wrong match. Q.1
Consider the optical system consisting of prisms P 1 and P2 having refractive index μ = 3/2 and a concave mirror of focal length 30 cm as shown in figure
(A) (B) (C) (D)
Column-I Coordinates of image of point object S formed by upper prism (P1) Coordinates of image of point object S formed by lower prism (P2) Separation between images of object after refraction through prism Separation between images of object after refraction & reflection
XIII (X&Z)-RT-3 [Paper-I]
(P)
Column-II π/6 cm
(Q)
π/2 cm
(R)
(–20 cm, π/18 cm)
(S) [Ans.
(–20 cm, –π /9 cm) (A)- R (B)-S (C)- P(D)- Q ]
Page # 3
PHYSICS
Q.2 is "Match the Column" type and carry 20 marks. Column-I and column-II contains four entries each. Entry of column-I are to be matched with one or more than one entries of column-II. 4 mark will be awarded for each (A), (B), (C) and (D) and 4 Bonus marks will be awarded if all the matches are correct. There isNO NEGATIVE marking. Q.2
A motorcycle moves in a vertical circle with a constant speed under the influence of the force of gravity w , net friction between the wheels and r
r
r
track f and normal reaction between wheel and track N : Column-I
Column-II
(A)
Constant magnitude
(P)
N
(B)
Directed towards centre when value is non zero
(Q)
N + f
(C)
Contact force by track
(R)
f + w
(D)
When motion is along vertical, the value is zero
(S)
[Sol.anu (A)
Constant magnitude
⇒
mv 2 R
r
r
r
r
r
r
N + w + f [Ans. (A)-S (B)-P,S (C)-Q (D)-R ] r
r
= net force
N + W + F r
r
r
(C) Total reaction force = N + F r
XIII (X&Z)-RT-3 [Paper-I]
r
]
Page # 4
XIII (X)
(POI) CHEMISTRY (DATE: 14-09-2008) PAPER-1
REVIEW TEST-3
PART-A [LINKED COMPREHENSION TYPE] Q.1 to Q.3 are based upon a paragraph. Each question has four choices (A), (B), (C), (D) out of which only one is correct and carry 3 marks each. There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. Paragraph for question nos. 1 to 3
Electrophilic substitution is an important reaction of aromatic compounds and it follows following mechanism : ⊕ ⊕
+E
Rate Deter min ing Step
⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ →
H E
−H
⊕
⎯ ⎯ ⎯ →
Slow
E
fast
(σ-complex) intermediate of reaction
Q.1
In general E⊕ attacks on more electron rich position. Rate of this reaction increases with stability of σ-complex. In the reaction where strong electrophile attacks, the cleavage of C–H bond is a very fast step. Arrange the following in the order of reactivity towards an electrophilic attack.
NO2
Cl
(I)
(II)
OMe
Me
(IV)
(III)
(A) V > IV > III > II > I (C) III > IV > V > II > I NO2 (– M)
(V)
(B*) III > V > IV > II > I (D) V > IV > III > I > II
Cl (– I)
OMe (+ M)
Me (+H)
[Sol. (I)
(II)
(IV)
(III)
(V)
In case of (III) benzene ring is more e – rich and σ complex is more stable ] ⊕
Q.2
Reaction-I :
r 1 + NO2 ⎯ ⎯→
⊕
Reaction-II :
r
2 ⎯→ + NO 2 ⎯
where r 1 & r 2 are rate of the reactions respectively. (A*) rate r 2 > r 1 (B) rate r 1 > r 2 (C) rate r 1 ~ (D) r 1 and r 2 are not comparable − r 2
CHEMISTRY
CH3 E⊕
Q.3
⎯ ⎯→
CD3 (A*) E⊕ attacks preferably at –o position to CH 3 (C) E⊕ can attack equally at any of the four position
(B) E⊕ attacks preferably at –o position to CD3 (D) Not predicable here.
CH3 (+ H) [Sol.
CD3 (+ H) + H of CH3 is more as compared to CD3. So atttack of E + will take place at ortho position of CH3] Q.4 to Q.6 are based upon a paragraph. Each question has four choices (A), (B), (C), (D) out of which only one is correct and carry 3, 4, 5 marks respectively. There is NEGATIVE marking. 1 mark will
be deducted for each wrong answer. Paragraph for Question Nos. 4 to 6
Q.1
Q.2
Q.3
According to Sidgwick central atom loses e – due to oxidation and gains e – due to co-ordination Net e – is same as the atomic number of the nearest inert gas. It is called effective atomic number (EAN). But in certain complexes it is not obeyed. The complex ion which has no 'd' electrons in the central metal atom is (A) [Fe(CN)6]3– (B) [Cr(H2O)6]3+ (C*) [MnO4] – (D) [Co(NH3)6]3+ In which of the following ion has the metal atoms EAN as 36 (A*) [Fe(CN)6]4– (B) [Fe(CN)6]3– (C) [PdCl4]2–
(D) [Pd(CN)6]2–
The co-ordiantion number of cobalt in [Co (en) 2Br 2]Cl2 (A) 2 (B) 4 (C*) 6
(D) 8
Q.7 to Q.9 are based upon a paragraph. Each question has four choices (A), (B), (C), (D) out of which only one is correct and carry 4, 5, 6 marks respectively. There is NEGATIVE marking. 1 mark will
be deducted for each wrong answer. Paragraph for Question Nos. 7 to 9
S(+) Mono Sodium Glutamate (MSG) is a flavour enhancer used in many foods. Fast foods often contain substantial amount of MSG and is widely used in Chinese food. If one mole of above MSG was placed in 845 ml solution and passed through 200 mm tube, the observed rotation was found to be + 9.6°. +
COO¯ Na ¯ OOC—CH2—CH2—C ⊕
NH3
H
MSG XIII (X)-RT-3 [Paper-I] Code-A
Page # 2
CHEMISTRY
Q.4 [Sol.
Find out the specific rotation of (–) MSG: (A) + 24° (B) + 56.8° Specific rotation of ( – ) MSG 169 gm C = 845 ml
[4]
(C) – 48°
(D*) None of these
l = 2dm
9. 6 θ [θ°] = C l = 169 / 845 × 2 = – 24°
]
Q.5
Find out the approximate percentage composition of (–) MSG in a mixture containing (+) MSG and (–) MSG whose specific optical rotation is –20°. [5] (A) 83.3 % (B) 16.7 % (C*) 91.6 % (D) 74 %
[Sol.
Optical of purity ( – ) MSG =
θObserved θStandard
× 100 = 83.33 %
RM = 100 – optical of purity = 100 – 83.33 = 16.66% (–) MSG total in mixture ⇒ 83.33 % + 8.33 % –––––– 91.66 % Q.6
[Sol.
]
If 33.8 g of (+) MSG was put in 338 ml solution and was mixed with 16.9 g of (–) MSG put in 169 ml solution and the final solution was passed through 400 mm tube. Find out observed rotation of the final solution. [6] (A) + 1.6° (B) + 4.8° (C*) + 3.2° (D) None of these (+) MSG ⇒ 33.8 gm in 338 ml (–) MSG ⇒ 16.9 gm in 169 ml Optical purity in mixture (+) MSG = 16.9 gm in 507 ml solution 16.9 gm/ml 507 l = 4 dm θObserved = [θ°] × C.l. C=
= 24 ×
16.9 × 4 = + 3.2° ] 507
XIII (X)-RT-3 [Paper-I] Code-A
Page # 3
CHEMISTRY
[REASONING TYPE] Q.10 to Q.12 are Reasoning type questions, contains Statement-1 (Assertion) and Statement-2 (Reason). Each question has 4 choices (A), (B), (C) and (D), out of which only one is correct and carry 3 marks each. There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. Q.10
[Sol.
: Bond angle of HOF is lesser than HOCl. : Oxygen atom is more electronegative than chlorine. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. Statement-1 Statement-2
O H
α=97°
O F
H
Cl
β=113°
In O–F bond p-character is greater than that of O–Cl bond, also in HOCl, p π –d π back bonding may also be present, hence α < β.] Q.11
Statement-1
:
O || Resonance energy of CH 2 = CH − C − NH 2 is high as compared to O || CH 3CH 2 − C − NH 2 .
: Another double bond is introduced in conjugation to C = O. (A*) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. Statement-2
[Sol.
(I)
O || .. CH2 = CH – C – NH2 O | ⊕ CH2 – C = C – NH2
← ⎯→
O | ⊕ CH2 = C – C = NH2 Octate complete R.S. (more contributing)
Octate incomplete R.S. (less contributing) O | ⊕ (II) ← ⎯→ CH3 – CH 2 – C = NH 2 (I) & (II) having almost comparable R energy ]
O || CH 3 − CH − C − NH 2
XIII (X)-RT-3 [Paper-I] Code-A
Page # 4
CHEMISTRY
Q.12
: NO2 – is an ambidentate ligand. : NO2 – is a lewis base (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is false. Statement-1 Statement-2
PART-B [MATCH THE COLUMN] Q.1 is "Match the Column" type and carry 15 marks. Column-I and column-II contains four entries each. Entry of column-I are to be uniquely matched with only one entry of column-II. 3 mark will be awarded for each (A), (B), (C) and (D) and 3 Bonus marks will be awarded if all the matches are correct. There is NEGATIVE marking. 1 mark will be deducted for each wrong match. Q.1
Column I
Column II
(A)
ICl3
(P)
Hybridisation of central atom is similar in both dimer and monomer form.
(B)
AlCl3
(Q)
Both Monomer and dimer forms are planar.
(C)
AlF3
(R)
In dimer form all atoms are sp3 hybridised.
(D)
NO 2
(S)
Does not exist in dimer form. [Ans. (A) Q, (B) R, (C) S, (D) P ]
Cl [Sol.
(A)
2
⎯→
I—Cl
I Cl
Cl
(B)
Cl
Cl
Cl Al
Al
+
Cl
I Cl
Cl
Cl
Cl Al
⎯→
Cl
Hybridisation of Al : sp2
(C)
Cl
Hybridisation of I : sp3d 2 Planar
Planar, bent 'T' shape Hybridisation : sp3d Cl
Cl
Cl
Cl
Cl Al
Cl
Cl
Hybridisation of Al : sp3 Hybridisation of bridge 'Cl' : sp3
AlF3 : Ionic compound, hence no dimerization
XIII (X)-RT-3 [Paper-I] Code-A
Page # 5
CHEMISTRY
N
(D) O
O
N
+
O
sp2 planar
O
O
O
⎯→
sp2 planar
N—N O
O
Hybridisation of each N : sp2 Planar ]
Q.2 is "Match the Column" type and carry 20 marks. Column-I and column-II contains four entries each. Entry of column-I are to be matched with one or more than one entries of column-II. 4 mark will be awarded for each (A), (B), (C) and (D) and 4 Bonus marks will be awarded if all the matches are correct. There is NO NEGATIVE marking.
Q.2
Column I
Column II
(A)
For the process A(l) → A(s), (ΔH & ΔV) may be
(P)
– ve, + ve
(B)
A2(s) + B2(g) l C2(s) + D2(s) (ΔS & ΔG) may be
(Q)
+ ve, – ve
(C)
For the given reaction A2(g) l B2(g) + C2(g), Ea(forward)= 50 kJ/mol and Ea(backward) = 40 kJ/mol , at very high temperature (ΔH & ΔG) are
(R)
+ ve, + ve
(D)
For the given reaction A(g) l B(g), at very low temperature (ΔH & ΔG) may be
(S)
– ve, – ve
[Ans. (A) P,S (B) P,S (C) Q, (D) R,S ]
XIII (X)-RT-3 [Paper-I] Code-A
Page # 6
MATHEMATICS
XIII (VXYZ), RT-3, PAPER-I (DATE: 14-09-2008)
PART-A Q.1 to Q.3 are based upon a paragraph. Each question has four choices (A), (B), (C), (D) out of which only one is correct and carry 3 marks each. There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. Paragraph for question nos. 1 to 3
Consider the cubic f (x) = 8x 3 + 4ax2 + 2bx + a where a, b ∈ R. Q.1 40 For a = 1 if y = f (x) is strictly increasing
⎛ (A) ⎜ − ∞, ⎝
1⎤ 3 ⎥⎦
∀ x ∈ R then maximum range of values of b is
⎛ 1 ⎞ (B) ⎜ , ∞ ⎟ ⎝ 3 ⎠
⎡ 1 ⎞ (C*) ⎢ , ∞ ⎟ ⎣ 3 ⎠
(D) (– ∞, ∞)
[3]
Q.2 41 For b = 1, if y = f (x) is non monotonic then the sum of all the integral values of a ∈ [1, 100], is (A) 4950 (B*) 5049 (C) 5050 (D) 5047 [3] Q.3 42 If the sum of the base 2 logarithms of the roots of the cubic f (x) = 0 is 5 then the value of 'a' is (A) – 64 (B) – 8 (C) – 128 (D*) – 256 [3] [Sol. a=1 (i) f (x) = 8x3 + 4x2 + 2bx + 1 f ' (x) = 24x 2 + 8x + 2b = 2(12x 2 + 4x + b) for increasing function, f ' (x) ≥ 0 ∀ x ∈ R D ≤ 0
∴
⇒
16 – 48b ≤ 0
⇒
b ≥
1 3
⇒
(C)
if b = 1
(ii)
f (x) = 8x3 + 4ax2 + 2x + a f ' (x) = 24x 2 + 8ax + 2 or for non monotonic f ' (x) = 0 must have root
(iii)
2(12x2 + 4ax + 1)
∴ hence D ≥ 0 i.e. 16a2 – 48 ≥ 0 ⇒ a2 ≥ 3; ∴ a ∈ 2, 3, 4, ....... sum = 5050 – 1 = 5049 Ans. If x1 , x2 and x3 are the roots then log 2x1 + log 2x2 + log2x3 = 5 log2(x1x2x3) = 5 x1x2x3 = 32
–
a = 32 8
⇒
a≥
3 or a ≤ –
3
a = – 256 Ans. ]
XI (PQRS)-RT-4 [Paper-I] Code-A
Page # 1
MATHEMATICS
Q.4 to Q.6 are based upon a paragraph. Each question has four choices (A), (B), (C), (D) out of which only one is correct and carry 5, 4, 3 marks respectively. There is NEGATIVE marking. 1 mark will be
deducted for each wrong answer. Paragraph for Question Nos. 4 to 6 Suppose a and b are positive real numbers such that ab = 1. Let for any real parameter t , the distance
from the origin to the line (ae t)x + (be –t)y = 1 be denoted by D(t) then 1
dt
Q.47def The value of the definite integral I = ∫
2 0 (D ( t ) )
Q.5 8
2 ⎞ e2 − 1 ⎛ a 2 (A) 2 ⎜⎜ b + 2 ⎟⎟ e ⎠ ⎝
2 ⎞ e 2 + 1 ⎛ b 2 (B) 2 ⎜⎜ a + 2 ⎟⎟ e ⎠ ⎝
2 ⎞ e 2 − 1 ⎛ b 2 (C*) 2 ⎜⎜ a + 2 ⎟⎟ e ⎠ ⎝
2 ⎞ e2 + 1 ⎛ a 2 (D) 2 ⎜⎜ b + 2 ⎟⎟ e ⎠ ⎝
[5]
The value of 'b' at which I is minimum, is (A) e
Q.6 9
is equal to
(B)
1 e
(C)
1 e
(D*) e
[4]
Minimum value of I is (A) e – 1
(B*) e –
1 e
(C) e
(D) e +
1 e
[3]
[T/S, Ex-3, definite] to be put
[Sol.(i) D(t) =
1
1 −t 2
t 2
( ae ) + ( be )
=
(0, 0)
2 −2 t
2 2t
a e + b e
D(t)
1
(D(t ) )2 =
(a2e2t +
(aet)x + (be–t)y = 1
b2e –2t) 1
⎡ a 2e2 t b 2e − 2t ⎤ 2 2t 2 −2 t I = ∫ (a e + b e )dt = ⎢ 2 − 2 ⎥ = ⎣ ⎦0 0 1
∴
⎛ a 2 e 2 − b 2e −2 ⎞ ⎜ ⎟ – ⎜ ⎟ 2 ⎝ ⎠
line
⎛ a 2 − b 2 ⎞ ⎜ ⎟ ⎜ 2 ⎟ ⎝ ⎠
b 2 2 a (e − 1) + 2 (e − 1) e 2 − 1 ⎛ 2 ⎞ b 2 a 2 (e 2 − 1) − b 2 (e −2 − 1) e = = = 2 ⎜⎜ a + 2 ⎟⎟ Ans. (i) e ⎠ 2 2 ⎝ 2
(ii)
now put b =
2
1 a
2 2 1 2 ⎞⎟ ⎛ ⎞ e 2 − 1 ⎛ 2 1 ⎞ e − 1 ⎛ ⎜ ⎜a − ⎟ + ⎜ a + 2 2 ⎟ = I= ⎜ ⎟ 2 2 ⎝ a e ⎠ ⎝ ⎝ ae ⎠ e ⎠
1 I is minimum if a = ae
⇒
a2 =
XI (PQRS)-RT-4 [Paper-I] Code-A
1 e
⇒
1 a= e
⇒
b = e Ans. (ii) Page # 2
MATHEMATICS
(iii)
1 e2 −1 2 · = e – and Imin = Ans. (iii)] e 2 e Q.7 to Q.9 are based upon a paragraph. Each question has four choices (A), (B), (C), (D) out of which only one is correct and carry 6, 5, 4 marks respectively. There is NEGATIVE marking. 1 mark will be
deducted for each wrong answer. Paragraph for Question Nos. 7 to 9
Let a1 > a2 > a3 ............ an > 1; p1 > p2 > p3......... > p n > 0 ; such that p 1 + p2 + p3 + ...... + pn = 1
(
Also F (x) = p1a1x + p 2 a 2x + ....... + p n a xn Q.7 16
)1 x
Lim F( x ) equals
x →0 +
(A) p1 ln a1 + p2 ln a2 + ....... + p n ln an
(B) a p11 + a p22 + ..... + a pn n n
(C*)
p p p a1 1 · a 22 .....a nn
(D)
∑ a r pr
[6]
r =1
Q.8 17 Lim F(x ) equals x →∞
(A) ln a1 Q.9 18
[Sol. (1)
(B) e
an
(C*) a1
(D) an
[5]
(B) e
a1
(C) a1
(D*) an
[4]
Lim F( x ) equals
x→ − ∞
(A) ln an
(
Lim+ F( x ) = Lim p1a1x + p 2a 2x + ....... + p n a nx +
x →0
x →0
)1 x
(1∞ form)
p1a1x + p 2 a 2x + ....... + p n a nx − 1 l = Lim L1 = x →0 x using L'Hospital's Rule
⎛ 0 ⎞ ⎜ ⎟ ⎝ 0 ⎠
el where
x x x l = Lim p1 ln a1 a1 + p 2 ln a 2 a 2 + ....... + p n ln a n a n x →0
= p1 ln a1 + p2 ln a2 + ....... + p n ln an = ln a p11 · a p2 2 .....a pnn ∴
(2)
L1 = el = a p11 · a p2 2 .....a pn n Ans.
( x →∞
Lim F(x ) = L2 = Lim p1a1x + p 2a 2x + ....... + p n a nx x →∞
∴
ln L2 = Lim
(
ln p1a1x + p 2a x2 + ....... + p n a xn
x →∞
XI (PQRS)-RT-4 [Paper-I] Code-A
)1 x
(∞0 form) [only when a 1 etc > 1]
)
x
Page # 3
MATHEMATICS
using L'Hospital's Rule L2 = Lim x →∞
(
p1 ln a1 a1x + p 2 ln a 2 a 2x + ....... + p n ln a n a xn
)
....(1)
p1a1x + p 2a x2 + ....... + p n a xn
dividing by a1x and taking limit, we get x
x
⎛ a ⎞ ⎛ a ⎞ Lim , ⎜ 2 ⎟ , ⎜ 3 ⎟ , etc all vanishes as x → ∞ ⎜ ⎟ x →∞ ⎜ a ⎟ ⎝ 1 ⎠ ⎝ a 2 ⎠ p1 ln a1 = ln a1 = p 1 hence ln L2 = ln a1
L2 = a1 Ans.
⇒
Lim F( x ) = L3 (say)
(3)
x→ − ∞
ln L3 = xLim →−∞
∴
(
p1 ln a1 a1x + p 2 ln a 2 a 2x + ....... + p n ln a n a xn
)
p1a1x + p 2a 2x + ....... + p n a nx x
x
⎛ a1 ⎞ ⎛ a 2 ⎞ x Lim dividing by (an) and taking x →−∞ , ⎜⎜ ⎟⎟ , ⎜⎜ ⎟⎟ etc vanishes ⎝ a n ⎠ ⎝ a n ⎠ ln L3 =
∴
p n ln a n
⇒
p n
L3 = an Ans.]
[REASONING TYPE] Q.10 to Q.12 are Reasoning type questions, contains Statement-1 (Assertion) and Statement-2 (Reason). Each question has 4 choices (A), (B), (C) and (D), out of which only one is correct and carry 3 marks each. There is NEGATIVE marking. 1 mark will be deducted for each wrong answer.
Q.109 Consider the following statements Statement-1:
3
1 4 dx x −1 ⎤ = = – – 1 = – ∫ 2 −1 ⎥ 3 3 ⎦ −1 −1 x 3
because
b
If f is continuous on [a, b] then ∫ f ( x ) dx = F (b) – F (a) where F is any antiderivative a of f , that is F' = f . (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D*) Statement-1 is false, statement-2 is true. Statement-2:
3
[Hint:
dx
∫ x2
−1
does not exist ]
XI (PQRS)-RT-4 [Paper-I] Code-A
Page # 4
MATHEMATICS
Q.11 12 Statement-1:
|cos x| and cos x are both continueous at x =
π 2
because
Statement-2: If | f (x) | is continuous as x = a then f (x) is also continuous at x = a. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C*) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. 1
⎡− 1 if x ≤ 0 [Hint : f(x) = ⎢ 1 if x > 0 ⎣ O but | f(x) | = 1 –1 note that if f (x) is continuous at x = a then | f (x) | will be continuous at x = a but not the converse ] *Q.12103 Consider the curves C1 :
Statement-1: because
y2 = a2 and C2 : xy3 = c 3 C1 and C2 are orthogonal curves.
x2 –
Statement-2: C1 and C2 intersect at right angles everywhere wherever they intersect. (A*) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. 2 y dy [Hint: C1 : 2x – = 0 3 dx C2 : ∴
3xy2
dy + y3 = 0 dx
m1 · m2 = – 1
⇒
⇒
⇒
dy ⎤ dx ⎥⎦ x dy ⎤ dx ⎥⎦ x
1 y1
3x1 = y = m1 1 y1
1 y1
= – 3x = m2 1
C1 and C2 are orthogonal ]
XI (PQRS)-RT-4 [Paper-I] Code-A
Page # 5
MATHEMATICS
PART-B Q.1 is "Match the Column" type and carry 15 marks. Column-I and column-II contains four entries each. Entry of column-I are to be uniquely matched with only one entry of column-II. 3 mark will be awarded for each (A), (B), (C) and (D) and 3 Bonus marks will be awarded if all the matches are correct. There is NEGATIVE marking. 1 mark will be deducted for each wrong match. Q.1
Column-I
Column-II n
(A) (B)
(C)
(D)
1 ⎞ ⎛ If the constant term in the binomial expansion of ⎜ x 2 − ⎟ , n ∈ N is 15 x ⎠ ⎝ then the value of n is equal to The positive value of 'c' that makes the area bounded by the graph of y = c(1 – x2) and the x-axis equal to 1 can be expressed in the form p q where p, q ∈ N and in their lowest form then (p + q) equals Suppose a, b, c are such that the curve y = ax 2 + bx + c is tangent to y = 3x – 3 at (1, 0) and is also tangent to y = x + 1 at (3, 4) then the value of (2a – b – 4c) equals x 2 y2 + = 1 . P is a point Suppose F1, F2 are the foci of the ellipse 9 4 on ellipse such that PF1 : PF2 = 2 : 1. The area of the triangle PF 1F2 is [Ans. (A) Q; (B) R; (C) S; (D) P]
(P)
4
(Q)
6
(R)
7
(S)
9
[Sol n
(A)
⎛ 2 1 ⎞ Tr + 1 in ⎜ x − ⎟ is x ⎠ ⎝ nC (x2)n – r (–1)r x –r r = nCr x 2n – 3r (–1)r Constant term = nCr (–1) r if 2n = 3r hence nC2n/3 (–1)2n/3 = 15 = 6C4
⇒
n = 6 Ans. ]
1
2 2 ∫ c(1 − x ) dx = 1
(B)
0
⎛ 1 ⎞ 2c ⎜1 − ⎟ = 1 ⎝ 3 ⎠ 2c · (C)
2 = 1 3
⇒
c=
3 Ans. ] 4
when x = 1, y = 0
dy = 2ax + b dx ⇒ a + b + c = 0 ....(1)
dy =3 dx x =1
dy =1 dx x =3
y = ax2 + bx + c;
and
2a + b = 3 ....(2) 6a + b = 1 ....(3) solving (1), (2) and (3) XI (PQRS)-RT-4 [Paper-I] Code-A
Page # 6
MATHEMATICS
1 7 ; b = 4, c = – 2 2 2a – b – 4c = – 1 – 4 + 14 = 9 Ans. a=–
∴
(D)
e2 =
4 5 b 2 1 – 2 = 1 – = 9 9 a
e=
⇒
5 3
F1F2 = 2 5 also PF1 + PF2 = 6 and PF1 = 2(PF2) (given) ∴ ⇒ 3PF2 = 6 PF2 = 2 and PF1 = 4 ⇒ ∠P = 90° since (PF2)2 + (PF1)2 = (F 1F2)2 Area =
4· 2 = 4 Ans. ] 2
Q.2 is "Match the Column" type and carry 20 marks. Column-I and column-II contains four entries each. Entry of column-I are to be matched with one or more than one entries of column-II. 4 mark will be awarded for each (A), (B), (C) and (D) and 4 Bonus marks will be awarded if all the matches are correct. There is NO NEGATIVE marking.
Q.2
Column-I
(A)
Column-II
2 ⎛ π ⎞ π (1 + x cos x · ln x + sin x ) dx and f ⎜ ⎟ = Let f (x) = ∫ x 4 ⎝ 2 ⎠ then the value of f (π) is sin x
∫ x2
x4 − 2
(B)
Let g (x) =
(C)
then find the value of g 2 If real numbers x and y satisfy (x + 5)2 + (y – 12) 2 = (14)2 then
4
2
x +x +2
dx and g (1) = 2
)
the minimum value of (D)
Let k (x) =
(x 2 + y 2 ) is
( x 2 + 1) dx
(P)
rational
(Q)
irrational
(R)
integral
(S)
prime
1
∫ 3 x 3 + 3x + 6 and k (–1) = 3 2 then the value
of k (– 2) is [Ans. (A) Q; (B) P, R, S; (C) P, R; (D) P, R, S] [Sol.(A) f (x) = ∫ x sin x (1 + x cos x · ln x + sin x ) dx [to be put in after first lecture definite form at Q.23 (b)] if F (x) = xsin x = e sin x ln x ∴ x F ' (x) = x sin x (x cos x ln x + sin x) ∴
f (x) = ∫ (F( x ) + x F' (x ) ) = x F (x) + C f (x) = x · x sin x + C
∴
⎛ π ⎞ π π f ⎜ ⎟ = · + C ⎝ 2 ⎠ 2 2 f (x) = x (x)sin x ;
⇒
C=0
f (π) = π(π)0 = π (irrational)
XI (PQRS)-RT-4 [Paper-I] Code-A
⇒
(Q)
Page # 7
MATHEMATICS
(B)
g (x) =
∫ x3
x4 − 2 2
2
x +1+ 2/ x
dx =
∫
⎛ x − 2 ⎞ ⎜ 3 ⎟dx ⎝ x ⎠ 2 x2 +1+ 2 x
now put x2 + 1 + 2/x 2 = t
⎛ x − 2 ⎞ dt ⎜ 3 ⎟ dx = 2 ⎝ x ⎠ ∴
g(x) =
2 + C x2 ∴ C=0 Ans: P, R, S]
x 2 +1+
since g (1) = 2 now g( 2 ) = 2 (C)
Let ∴
x + 5 = 14 cos θ y – 12 = 14 sin θ x2 + y2 = (14 cos θ – 5)2 + (14 sin θ + 12)2 = 196 + 25 + 144 + 28(12 sin θ – 5 cos θ) = 365 + 28(12 sin θ – 5 cos θ) (x 2 + y 2 )
∴
min
(D)
=
365 − 28 ×13 =
( x 2 + 1) dx k (x) = ∫ 3 ( x + 3x + 6)1 3 put x3 + 3x + 6 = t3
⇒
365 − 364 = 1 Ans.
⇒
P, R
3(x2 + 1)dx = 3t 2 dt
t 2 dt t2 k (x) = ∫ = + C t 2 k (x) =
1 3 (x + 3x + 6) 2/3 + C 2
k (–1) = ∴
1 2/3 (2) + C 2
k (x) =
⇒
C=0
1 3 1 1 (x + 3x + 6) 2/3 ; f (–2) = (– 8)2/3 = [(–2)3]2/3 = 2 Ans. 2 2 2
XI (PQRS)-RT-4 [Paper-I] Code-A
⇒
P, R, S]
Page # 8
