Class 11 Math: CBSE Sample Question Papers- Term II (2021-22)- 1 | Sample Papers for Class 11 Medical and Non-Medical - JEE PDF Download

General Instructions:

• This question paper contains three sections A, B and C. Each part is compulsory.
• Section - A has 6 short answer type (SA1) questions of 2 marks each.
• Section - B has 4 short answer type (SA2) questions of 3 marks each.
• Section - C has 4 long answer type questions (LA) of 4 marks each.
• There is an internal choice in some of the questions.
• Q14 is a case-based problem having 2 sub parts of 2 marks each.

Q.1. A wheel makes 360 revolutions in 1 min. Through how many radians does it turn in 1 second.

No. of revolution in one minute = 360

No. of revolutions in one second =360/60 = 6
∴ 6 revolutions = 6 × 2p = 12p radians

Hence, no. of radians turned in one second

= 12p

Find the degree measure of the angle subtended at the centre of circle of radius 100 cm by an arc of length 22 cm (use π =22/7)

Given, l = length of arc = 22 cm

r = radius of circle = 100 cm

Let θ be the angle subtended at the centre,

then

Q.2. If parabola y² = px passes through point (2, −3), find the length of latus rectum.

Given equation of parabola is y² = px       ...(i)

Since (i) passes through (2, −3).
∴ (-3)² = p.2

comparing above by y² =4ax,
a = 9/8
∴ Length of latus rectum = 4a = 9/2 units

Q.3. Differentiate:

Q.4. If sin A = 3/5 and π/2 < A < π. Find cos A, tan 2A.

Since, sin A > 0 and π/2 < A < π, then A is in II Quadrant.

 

Q.5. Solve the inequality: 

The given inequality   Adding (−4) to each term,

Multiplying by (-2/7) to each term

Q.6. If the letters of the word ‘ALGORITHM’ are arranged at random in a row what is the probability the letter ‘GOR’ must remain together as a unit ?

Number of letters in the word ‘ALGORITHM’ is 9

If ‘GOR’ remain together, then considered it as 1 number

∴ Number of letters = 6 + 1 = 7

Number of words, if ‘GOR’ remain together = 7!

Total number of words from the letters of the word ‘ALGORITHM’ = 9!
∴ Required probability

A card is selected from a pack of 52 cards.
(i) What is the sample space?
(ii) Calculate the probability that card is an ace of spade?

(i) Sample space i.e. n(S) = 52

(ii) Number of ace of spade in a pack = 1 i.e. n(E) = 1
Probability that card is an ace = n(E)/n(S) = 1/52.

Q.7. Find the centre and radius of the circle 2x² + 2y² – x = 0.

Given equation of the circle is
2x² + 2y² – x = 0

Q.8. Find the co-ordinate of the vertices, foci eccentricity and length of latus rectum of the hyperbola.

Find the distance between the directrices of the hyperbola x² – y² = 8.

The given hyperbola

here a = 5, b = 2

Q.9. From a class of 40 students, in how many ways can five students be chosen for an excursion party.

From a class of 40 students, five students can be chosen for an excursion party in ⁴⁰C₅ ways

Q.10. From a group of 7 boys and 5 girls, a team consisting of 4 boys and 2 girls is to be made. In how many different ways it can be done?

From a group of 7 boys, 4 boys can be choose in ⁷C₄ ways. 

From a group of 5 girls, 2 girls can be choose in ⁵C₂ ways. 

∴ Required number of ways to choose a team consisting 4 boys and 2 girls are

Q.11. Using first principle find the derivative of x cos x function

Let f (x) = x cos x
∴ f(x+h) = (x+h) cos (x+h)

Q.12. Find the equation of a circle passing through the point (7, 3) having radius 3 units and whose centre lies on the line y = x – 1.

Let equation of circle be

(x-h)²+(y-k)² = r²
⇒ (x-h)² + (y-k)² = 9   ...(i)
Now, the circle passes through the point (7, 3)

(7-h)² + (3-k)³ = 9  ... (ii)

⇒ 49 - 14h + h² + 9 - 6k + k²=9

⇒ h²+k²-14h-6k+49 = 0 ....(iii)

Now, y=x-1, k=h-1

on putting k=h-1 in Eq. (iii), we get

h²+(h-1)² - 14h - 6(h-1)+49 = 0

⇒ h²+h²-2h+1 - 14h - 6h + 6+49 = 0

⇒ 2h²-22h+56=0

⇒ h²-11h+28 =0

⇒ h²-7h-4h+28=0

⇒ h(h-7)-4(h-7)=0

⇒ (h-7)(h-4)=0

∴ h=4,7

When h = 7 then k = 7 – 1 = 6

∴ Centre (7, 6)

When h = 4 then k = 3

∴ Centre = (4, 3) 

So, the equation of circle when centre (7, 6), is
(x-7)² + (y-6)² = 9

⇒ x²-14x+49+y²-12y+36 =9

⇒ x²+y²-14x-12y+76 = 0

when centre (4, 3), then the equation of circle is,

(x-4)² + (y-3)² = 9

⇒ x²-8x+16+y²-6y+9 =9

⇒ x²+y-8x-6y+16 =0

Find the equation of a circle whose centre is (3, –1) and which cuts off a chord length 6 units on the line 2x – 5y + 18 = 0.

Given, centre of the circle is (3, –1),

⇒ OB² = 29+9
⇒ OB²= 38 units
So, the radius of circle is √38, units
∴ Equation of the circle with radius
r = √38 units and centre (3, –1) is

⇒ (x-3)² + (y+1)² = 38

⇒ x²-6x+9+y²+2y+1=38

x²+y²-6x+2y=28

x²+y²-6x+2y-28=0

Q.13.  Solve the following system of inequalities graphically
2x + y ≤ 24,
x + y ≥ 11,
2x + 5y ≤ 40,  x, y ≥ 0.

Inequality (x ≥ 0) represents the region on the right of Y-axis and Y-axis itself. Inequality (y ≥ 0) represents the region above X-axis and X-axis itself.

2x + y ≤ 24    ...(i)

x + y ≥ 11    ...(ii)

2x + 5y ≤ 40, x, y ≥ 0    ...(iii)

We first draw the graph of lines

2x + y = 24, x + y = 11 and 2x + 5y = 40
Now, 2x + y = 24, passes through A(12, 0) and B(0, 24)

Again, x + y = 11, passes through C(11, 0) and D(0, 11)

Further 2x + 5y = 40, passes through E(20, 0) and F(0, 8)
Shaded area PQAC is the solution area.

Q.14.  A child's game has 8 triangles of which 3 are blue and rest are red, and 10 squares of which 6 are blue and rest are red. One piece is lost at random.

(i) Find the probability that lost piece is triangle. 

(i) Since, total no. of triangles = 8
Triangles with blue colour = 3
Triangles with red colour = 8 – 3 = 5
and total no. of squares = 10
Squares with blue colour = 6
Squares with red colour = 10 – 6 = 4
Number of favourable outcomes for the event that lost figure is triangle, i.e., F (E) = 8
Total figures (square and triangle)

= 8 + 10 = 18

i.e., T(E) = 18
Probability (getting a triangle),
P(E) = F(E)/T(E)
= 8/18 = 4/9

(ii) Find the probabilities that lost piece is square of blue colour & triangle of red colour.

(ii) Number of favourable outcomes for the events that lost figure is square of blue colour, i.e., F(E) = 6 and T(E) = 18
∴ P(getting a blue square),
P(E) = F(E)/T(E)
= 6/18 = 1/3

Number of favourable outcomes for the event that lost figure is triangle of red colour = 5,
i.e., F(E) = 5 and
T(E) = 18

∴ P(lost figure is red triangle),
P(E) = 5/18