Class 11 Sample Paper 2022 Solution Maths Term 2

Directorate of Education, GNCT of Delhi
Solution of Practice Paper
(Term-2)
2021-22
Class – XI
Mathematics (Code: 041)

SECTION – A

VALUE POINTS
Q. No.
1 1
√2
Use the formula 2cos A cos B= cos(A+B)+cos (A-B)

OR
2 2
1 1 sin θ+cos θ
LHS= 2 + 2 = 2 2
cos θ sin θ sin θ cos θ

4
= = 4 cosec 2 2 θ
( sin 2θ )2

∵cosec 2 ϕ≥1 so 4 cosec 2 2 θ≥4

2 1
Given |3 x − 2|≤
2
−1 1
⸫ => ≤(3x-2)≤ (∵|x|≤a => -a≤x≤a)
2 2
−1 1
=> +2≤3x-2+2≤ +2 (adding 2 on each term)
2 2
3 5 3 1 1 5 1
≤3x≤ => x ≤3x ≤ x (dividing each term by 3)
2 2 2 3 3 2 3
1 5 1 5
=> ≤x≤ i.e., xϵ [ , ]
2 6 2 6
3 INVOLUTE contains 4 vowels (I,O,U,E) and 4 consonants (N,V,L,T).
3 vowels can be selected out of 4 vowels by 4 C 3 ways
and 2 consonants can be selected out of 4 consonants by 4 C 2 ways
⸫ Number of words formed using 3 vowels and 3 consonants
= 4 C 3 x 4 C 2 x5!= 4x6x120=2880

Hence total number of words is 2880 which contains 3 vowels and 2 consonants

4 The given equation of parabola is y 2 =8 x => y 2 =4.2 . x
⸫ a=2, focus =(2,0)
Let P(x,y) be any point on the parabola then PS=4
|√ | |√ |
d= ( x 2 − x1 )2 + ( y 2 − y 1 )2 => ( x 2 − x1 )2 + ( y 2 − y 1 )2 =4
2
=> x − 4 x+4+8 x =16
=> x 2 +4 x − 12=0=>(x+6)(x-2)=0=> x=-6,x=2=> y 2 =8.2
2
y =8. ( − 6 ) not possible =>y=±4
⸫ The point P is (2,4) (2,-4)

5

Let f(x)=e x sinx +x n cosx
d d d
⸫f́ ( x )= {e x sinx +x n cosx } = ( e x sinx )+ ( x n cosx )
dx dx dx
d x x d d n n d
we know f́ ( u , v ) = v f́ ( u ) +u f́ ( v )=> f́ ( x )= sinx e +e sin x +cosx x +x cosx = sinx . e x +
dx dx dx dx
x n−1 n
e . cosx +cosx n x + x ( − sinx )
=e x ( sinx +cosx )+ x n −1 [ n cosx − x sinx ]

52
6 Number of ways of drawing 4 cards from 52 cards = C 4
In a deck of 52 cards , there are 13 diamonds and 13 spades.
⸫ Number of ways of drawing 3 diamond and one spade is = 13C 3 x 13C 1
13 13
C 3 × C 1 13 x 12 x 11 x 13 x 24 286
Thus the probability of obtaining 3 diamond and one spade = 52 = =
C4 6 x 52 x 51 x 50 x 49 20825

SECTION – B
7
|x − 1|
Given f(x)=
x−1
Domain: Clearly,f(x) is defined for all x ε R except x=1

⸫ Domain of f=R-{1}
x −1
Range :Now f(x)= =1, when x>1
x −1
( x −1 )
and f(x)=- =-1, when x<1
x −1
⸫ Range of f={-1,1}
8

cotx 5π
Given f(x)= and α + β = ,
1+cotx 4

⸫ f ( α )= ( 1+cot
cot α
α ) ( 1+cot β )
cot β

( )
cos α cos β
sin α sin β

( 1+
cos α
sin α ) 1+
cos β
sin β

cos α .cos β cos α . cos β
= =
( sin α +cos α )( sin β+cos β ) sin α sin β+sin α cos β +cos α sin β+cos α cos β

cos α . cos β cos α .cos β 2cos α . cos β
= = =
( cos α cos β+sin α sin β ) + ( sin α cos β+cos α sin β ) cos ( α − β ) +sin ( α + β ) 2 [ cos ( α − β ) +sin ( α + β ) ]

cos ( α + β )+cos ( α − β )
=
( )
cos
5π
4
+cos ( α − β )
=
−1
√2
+cos ( α − β )
=
1
2 [ cos ( α − β ) +sin ( α + β ) ]
[ ( )] [
2 cos ( α − β ) +sin
5π
4
2 cos ( α − β ) −
1
]
√2
2

OR

tan 70 =tan ( 50 +20 )
0 0 0

=>
tan 50 0 +tan 200 tan A+tan B
0
tan 70 = [ because tan ( A+B )=
0
1 − tan 50 tan 20 0
1 − tan A tan B
=> tan 70 ( 1− tan 50 tan 20 ) tan 50 +tan 20
0 0 0 0 0

=> tan 70 0− tan 700 tan 500 tan 20 0=tan 50 0 + tan 20 0
=> tan 70 0− tan ( 900 − 200 ) tan 50 0 tan 20 0=tan 50 0 +tan 200
=> tan 70 0− cot 200 tan 50 0 tan 200 =tan 500 +tan 200 [because tan ( 90 0 − A )=cotA
=> tan 70 0= tan 50 0 + tan 20 0 +tan 500
=> tan 70 0= tan 200 +2 tan 500 hence proved

9 A committee of 7 has to be formed from 9 boys and 4 girls.
9 4
(i) exactly 3 girls= C 4 × C 3
9! 4! 9 x8 x7 x6 x5!
= × = =72x7=504
4 ! 5 ! 3 ! 1! 3 x 2 x 5!

(ii) at most 3 girls
(a)No girl and 7 boys
(b) 1 girl and 6 boys
(c)2 girls and 5 boys
(d) 3 girls and 4boys

⸫ Committee consisting of atmost 3 girls
= 4 C 0 × 9C 7 + 4 C 1 ×9 C 6 + 4 C 2× 9C 5 + 4 C 3 × 9C 4
=1x36+4x84+6x126+126x4=36+336+756+504
=1632

10 1:3 externally

OR
To show ABCD is a parallelogram we need to show opposite sides are equal
Note that
AB=
√ ( −1− 1)2+( − 2− 2)2+( −1− 3 )2 =√ 4+16+16 =6
BC=√ ( 2+1 )2 + ( 3+2 )2 + ( 2+1 )2=√ 9+25+9 =√ 43
CD=√ ( 4 − 2 )2 + ( 7 −3 )2 + ( 6 − 2 )2 =√ 4+16+16 =6
DA=√ ( 1− 4 )2 + ( 2− 7 )2 + ( 3− 6 )2 =√ 9+25+9 =√ 43
since AB=CD and BC=AD, ABCD is a parallelogram
Now it is required to prove that ABCD is not a rectangle . For this we need to shoe that diagonals
are unequal, we have
AC= √ ( 2− 1 )2 + ( 3 −2 )2 + ( 2− 3 )2 =√ 1+1+1 =√ 3
BD=√ ( 4+1 )2 + ( 7+2 )2 + ( 6+1 )2 =√ 25+81+49 =√ 155
since AC≠BD ,ABCD is not a rectangle

SECTION – C
11 We have the following inequalities
3x+4y ⩽ 60 ----(i)
x+3y ⩽ 30 ---------(ii)
x⩾ 0 ---------------(iii)
y⩾ 0 ----------------(iv)

Take inequality (i)
3x+4y ⩽ 60
In equation form it can be written as 3x+4y=60 make table and plot graph. Similarly for line
x+3y ⩽ 30. x⩾ 0 ⩾ 0 plot graph.

Here the common shaded region represent the solution region for system of inequalities

OR

Let x litres of 3% solution be added to 460 litres of 9% solution of acid

⸫ Total quantity of mixture=(460+x)litres
9 3
Total acid content in the (460+x)litres of mixture = (460x +x )
100 100
It is given that acid content in the resulting mixture must be more than 5% but less than 7% acid
Therefore
9 3
5% of (460+x)<460x +x )<7% of (460+x)
100 100
5 9 3 7
=> x (460+x)<460x +x < x (460+x)
100 100 100 100
=>5 X (460+x)<460X 9+3x <7 x (460+x) [ multiplying by 100]
=>2300 +5x<4140+3x<3220+7x
Taking first two inequalities ,

2300 +5x<4140+3x
=>5x-3x<4140-2300=>x<920-----------(i)
Taking last two inequalities
4140+3x<3220+7x after solving we get x>230--------(ii)
Hence the number of litres of 3% solution of acid must be more tham 230 L and less than 920L

12 The given equation of circle is 4 x 2 +4 y 2 − 12 x −16 y −21=0
2 2 21
=> x + y −3 x − 4 y − =0
4

(
=> x − 3 x+
2
)4 +( y − 4 y+4 )− 94 − 4 − 214 =0
9 2

=>( x − ) + ( y −2 ) −
2
3 15 2
− 4=0
2 2

( 32 ) +( y −2 ) =(√ 232 )
2 2
2
=> x −

3
therefore Centre=( ,2) and radius r 1=
2
Let ‘r 2’ be the radius of the concentric circle
√ 23
2

therefore its equation will be

( )
x−
3 2
2
2 2
+ ( y −2 ) =r2

1 2 2
also π r1 =π r 2
2
2 2
=> π r 1 =2 π r 2

23
=> =2 π r 22
2

2 23
=>r 2 =
4

=>r 2 =
√ 23
2

( )
2
3 2 23
therefore x − + ( y −2 ) =
2 4

2 9 2 7 23
=> x −3 x+ + y − 4 y+4 − =
4 2 4

2 2 7
=> x + y −3 x − 4 y+4 − =0
2

=>2 x 2 +2 y 2 − 6 x −8 y+4+1=0 is the required equation of the circle.

13 let f(x)=tan √ x
sin √ (x+h) sin √ x
−
cos √( x+h) cos √ x
f’(x)= lim
h→0 h
sin √( x+h)cos √ x−sin √ x cos √( x+h)
= lim
h→0 h cos √ (x +h)cos √ x

sin( √( x +h)− √ x) √ x+h− √ x
lim ×
h→0 hcos √ x cos √( x+h) √ x +h− √ √ x
1 sin( √( x+h)− √ x) ( √( x+h)− √ x ) 2 √x
= lim ×lim ×lim = sec
h→0 cos √ x cos √( x+h) h→0 √ x+h− √ x h→0 h 2√ x

i) P(complex or very complex)
=P( E 1 or E 2 )
14 =P P(E 1∪ E2 )
= P(E 1 ) + P(E 2 ) - P(E 1∩ E2 )
=0.15+0.20-0=0.35

(ii) P(neither very complex nor very simple)
= P(E ' 1 ∩' E5 )
= P(E 1∪ E5 )'
=1- P(E 1∪ E5 )
= 1-[ P(E 1 ) + P(E 5 ) ]
=1-(0.15+0.08)
=1-0.23
=0.77