Week 3 Practice Assignment

Week 3 Practice Assignment


Mathematics I


1. Multiple Choice Questions (MCQ)

1. What will be the equation of the tangent to the curve f(x)=2x2+9x+20 f(x) = 2x^2 + 9x + 20 at point (−3,11)(-3, 11)?

Options:

  • y=3x y = 3x
  • y=−3x+2 y = -3x + 2
  • y=−3x+20 y = -3x + 20
  • y=−x3+2 y = -\frac{x}{3} + 2
  • y=x3+20 y = \frac{x}{3} + 20
  • y=−x3 y = -\frac{x}{3}

Solution: Slope at x=−3 x = -3 : m=2×2×(−3)+9=−3 m = 2 \times 2 \times (-3) + 9 = -3 Equation of tangent: y=−3x+c y = -3x + c Passes through (−3,11)(-3, 11): 11=−3×(−3)+c  ⟹  c=2 11 = -3 \times (-3) + c \implies c = 2 Answer: y=−3x+2 y = -3x + 2 123


2. Find the length of the line segment on the straight line y=2 y = 2 bounded by the curve y=4x2 y = 4x^2 .

Options:

  • 12 \frac{1}{\sqrt{2}}
  • 2 \sqrt{2}
  • 1+2 1 + \sqrt{2}
  • 1+12 1 + \frac{1}{\sqrt{2}}

Solution: Intersection points: 4x2=2  ⟹  x=±12 4x^2 = 2 \implies x = \pm \frac{1}{\sqrt{2}} Length: ∣12−(−12)∣=22=2 \left| \frac{1}{\sqrt{2}} - \left(-\frac{1}{\sqrt{2}}\right) \right| = \frac{2}{\sqrt{2}} = \sqrt{2} Answer: 2 \sqrt{2} 14


3. Mr. Mehta has two sons. Both sons send money to their father each month separately as M1(x)=(x−2)2 M_1(x) = (x-2)^2 and M2(x)=(x+2)2 M_2(x) = (x+2)^2 . If x x denotes the month, then choose the curve which best represents the total amount P(x) P(x) received by Mr. Mehta every month.

Solution: P(x)=(x−2)2+(x+2)2=2x2+8 P(x) = (x-2)^2 + (x+2)^2 = 2x^2 + 8 Answer: The curve passing through (0,8),(1,10),(4,40)(0,8), (1,10), (4,40)123


4. A civil engineer found that the durability d d of the road she is laying depends on two functions y1 y_1 and y2 y_2 as follows: d=ay1y2 d = a y_1 y_2 where a>0 a > 0 . Functions y1 y_1 and y2 y_2 depend on the amount of plastic (x x ) mixed in bitumen, and their variations are shown in the graph given below. Find the values of functions y1 y_1 and y2 y_2 such that the durability of the road is maximum.

Solution: From graph: y1=6−x y_1 = 6 - x , y2=x+1 y_2 = x + 1 d=a(6−x)(x+1)=−ax2+5ax+6a d = a(6-x)(x+1) = -a x^2 + 5a x + 6a Maximum at x=52 x = \frac{5}{2} : y1=6−52=72 y_1 = 6 - \frac{5}{2} = \frac{7}{2} , y2=52+1=72 y_2 = \frac{5}{2} + 1 = \frac{7}{2} 1


5. Let A A be the set of all points on the curve defined by the function f1(x)=x2−x−42 f_1(x) = x^2 - x - 42 and let B B be the set of all points on the curve f2 f_2 defined by the reflection of the curve f1 f_1 with respect to the X X -axis. If C C is the set of all points on the axes, then choose the correct option regarding the cardinality of set D D where D=(A∩B)∪(A∩C)∪(B∩C) D = (A \cap B) \cup (A \cap C) \cup (B \cap C) .

Options:

  • infinite
  • 8
  • 4
  • 6
  • 2
  • zero

Solution: After analysis, the cardinality of D D is 4. Answer: 41


6. Let f1(x)=x2−25 f_1(x) = x^2 - 25 . Let A A be the set of all points inside the region by the curves representing f1(x) f_1(x) and its reflection f2(x) f_2(x) with respect to the X X -axis (excluding the points on the curve). Choose the correct option.

Options:

  • The cardinality of A A is 2.
  • The cardinality of A A is 4.
  • Y Y -coordinates of the points in set A A belong to the interval (−25,25)(-25, 25).
  • Y Y -coordinates of the points in set A A belong to the interval [−25,25][-25, 25].
  • X X -coordinates of the points in set A A belong to the interval [−5,5][-5, 5].
  • X X -coordinates of the points in set A A will be all real numbers because f1 f_1 is a quadratic function.

Solution: A A is infinite. Y Y -coordinates: (−25,25)(-25, 25) X X -coordinates: (−5,5)(-5, 5) Correct option: Y Y -coordinates of the points in set A A belong to the interval (−25,25)(-25, 25)14


2. Multiple Select Questions (MSQ)

7. Choose the correct set of options regarding the function f(x)=x2+6x+8 f(x) = x^2 + 6x + 8 .

Options:

  • y=−3 y = -3 is the axis of symmetry.
  • −2-2 and −4-4 are the zeroes of the above function.
  • The maximum value of the above function is −1-1.
  • Slope of the function at (−3,−1)(-3, -1) is zero.
  • 2x+6 2x + 6 is the slope of this curve at any given x x .
  • The function is symmetric around x=3 x = 3 .

Solution:

  • Correct: −2-2 and −4-4 are the zeros; slope at (−3,−1)(-3, -1) is zero; slope at any x x is 2x+6 2x + 6 .
  • Incorrect: y=−3 y = -3 is not the axis of symmetry; maximum value is not −1-1 (it’s a minimum); not symmetric around x=3 x = 3 123

8. A quadratic function f f is such that its value decreases over the interval (−∞,−2)(-\infty, -2) and increases over the interval (−2,∞)(-2, \infty), and f(0)=f(−4)=23 f(0) = f(-4) = 23 . Then, f f can be:

Options:

  • −3x2−12x+23-3x^2 - 12x + 23
  • 3x2+12x+233x^2 + 12x + 23
  • 5(x−2)2+35(x-2)^2 + 3
  • 5(x+2)2+35(x+2)^2 + 3
  • ax2+4ax+23,a>0a x^2 + 4a x + 23, a > 0
  • ax2+4ax+23,a<0a x^2 + 4a x + 23, a < 0

Solution:

  • Correct: 3x2+12x+23 3x^2 + 12x + 23 , 5(x+2)2+3 5(x+2)^2 + 3 , ax2+4ax+23,a>0 a x^2 + 4a x + 23, a > 0 1

9. Suppose one root of a quadratic equation of the form ax2+bx+c=0 a x^2 + b x + c = 0 , with a,b,c∈R a, b, c \in \mathbb{R} , is 2+3 2 + \sqrt{3} . Then choose the correct set of options.

Options:

  • There can be infinitely many such quadratic equations.
  • There is no such quadratic equation.
  • There is a unique quadratic equation satisfying the properties.
  • x2−4x+1=0 x^2 - 4x + 1 = 0 is one such quadratic equation.
  • x2−2x−3=0 x^2 - 2x - 3 = 0 is one such quadratic equation.

Solution:

  • Correct: There can be infinitely many such quadratic equations; x2−4x+1=0 x^2 - 4x + 1 = 0 is one such equation14

10. A company’s profits are known to be dependent on the months of a year. The profit pattern (in lakhs of Rupees) from January to December is P(x)=−2x2+25x P(x) = -2x^2 + 25x . Here, x x represents the month number, starting from 1 (for January) and ending at 12 (for December). On this basis, choose the correct option.

Options:

  • The maximum profit in a month is Rs. 78 lakhs.
  • The maximum profit in a month is Rs. 78.125 lakhs.
  • The maximum profit in a month is Rs. 77 lakhs.
  • The maximum profit is recorded in June.
  • The profit in December is 144 lakhs.
  • None of the above.

Solution:

  • Correct: The maximum profit in a month is Rs. 78 lakhs (in June)1

11. Raghav sells 2000 packets of bread for Rs. 20,000 each day, and makes a profit of Rs. 4,000 per day. He finds that if the cost price increases by Rs. x x per packet, he can increase the selling price by Rs. 2x 2x per packet. However, when this price increase happens, he loses 200x 200x of his customers. Choose the correct options.

Options:

  • For the maximum profit per day, cost price is Rs. 12 per packet.
  • For the maximum profit per day, cost price is Rs. 4 per packet.
  • For the maximum profit per day, the sale price increases by Rs. 4 per packet.
  • For the maximum profit per day, Raghav will lose 400 customers.
  • The maximum difference in profit per day could be Rs. 3200.
  • The maximum difference in profit per day could be Rs. 7200.

Solution:

  • Correct: For the maximum profit per day, cost price is Rs. 12 per packet; maximum difference in profit per day could be Rs. 32001

3. Numerical Answer Type (NAT)

12. A farmer has a wire of length 576 metres. He uses it to fence his rectangular field to protect it from animals. If he fences his field with four rounds of wire, and the field has the maximum area possible to accommodate such a fencing, what is the area (in square metres) of the field?

Solution: Perimeter per round: 2(l+m) 2(l + m) Total perimeter: 4×2(l+m)=576  ⟹  l+m=72 4 \times 2(l + m) = 576 \implies l + m = 72 Area: A=l(72−l) A = l(72 - l) Maximum area: Amax=1296 A_{max} = 1296 Answer: 1296123


13. Consider the quadratic function f(x)=x2−2x−8 f(x) = x^2 - 2x - 8 . Two points P P and Q Q are chosen on this curve such that they are 2 units away from the axis of symmetry. R R is the point of intersection of axis of symmetry and the X X -axis. And S S is the vertex of the curve. Based on this information, answer the following:

a) What is the height of △PQR \triangle PQR taking PQ PQ as the base? Solution: Axis of symmetry: x=1 x = 1 Points: x=−1,3  ⟹  y=−5 x = -1, 3 \implies y = -5 Height: 0−(−5)=5 0 - (-5) = 5 Answer: 5

b) What is the height of △PQS \triangle PQS taking PQ PQ as the base? Solution: Vertex: y=−9 y = -9 Height: −5−(−9)=4 -5 - (-9) = 4 Answer: 41


14. What will be the value of parameter k k , if the discriminant of equation 4x2+9x+10k=0 4x^2 + 9x + 10k = 0 is 1?

Options:

  • 8280 \frac{82}{80}
  • 4180 \frac{41}{80}
  • 12 \frac{1}{2}
  • 41160 \frac{41}{160}
  • 1
  • None of the above.

Solution: Discriminant: 81−160k=1  ⟹  k=12 81 - 160k = 1 \implies k = \frac{1}{2} Answer: 12 \frac{1}{2} 1


15. A boat has a speed of 30 km/hr in still water. In flowing water, it covers a distance of 50 km in the direction of flow and comes back in the opposite direction. If it covers this total of 100 km in 10 hours, then what is the speed of flow of the water (in km/hr)?

Options:

  • 5−537 5 - 5\sqrt{37}
  • −106 -10\sqrt{6}
  • 106 10\sqrt{6}
  • 203 20\sqrt{3}
  • −203 -20\sqrt{3}
  • 2

Solution: Let speed of flow = x x 5030+x+5030−x=10 \frac{50}{30 + x} + \frac{50}{30 - x} = 10 Solving, x=106 x = 10\sqrt{6} Answer: 106 10\sqrt{6} 1


16. A stunt man performs a bike stunt between two houses of the same height as shown in Figure 1. His bike (lowest part of the bike) makes an angle of θ \theta at house A A with the horizontal at the beginning of the stunt, follows a parabolic path and lands at house B B with an angle of (180−θ) (180 - \theta) with the horizontal. If the maximum height achieved by the bike is 12.5 ft from the ground and tan⁡θ=1 \tan \theta = 1 , then find the distance between the two houses.

Options:

  • 1 ft
  • 2.5 ft
  • 5 ft
  • 10 ft
  • 15 ft
  • 20 ft

Solution: Distance between houses: 10 ft Answer: 10 ft1


4. Multiple Select Questions (MSQ) Continued

17. Given that f1(x)=−x2−6x f_1(x) = -x^2 - 6x and f2(x)=x2+6x+10 f_2(x) = x^2 + 6x + 10 . Let f(x) f(x) be a function such that the domain of f(x) f(x) is [α,β][\alpha, \beta], where f1(α)=f2(α) f_1(\alpha) = f_2(\alpha) and f1(β)=f2(β) f_1(\beta) = f_2(\beta) , then choose the set of correct options.

Options:

  • Range of f(x) f(x) is [−1,3][-1, 3].
  • Range of f(x) f(x) is $5$.
  • Domain of f(x) f(x) is [−5,5][-5, 5].
  • Domain of f(x) f(x) is [−5,−1][-5, -1].
  • Inadequate information provided for finding the range of f(x) f(x) .
  • Inadequate information provided for finding the domain of f(x) f(x) .

Solution:

  • Correct: Domain of f(x) f(x) is [−5,−1][-5, -1]; inadequate information for range1

18. If f(x)=2x2+(5+k)x+7 f(x) = 2x^2 + (5 + k)x + 7 , g(x)=5x2+(3+k)x+1 g(x) = 5x^2 + (3 + k)x + 1 , h1(x)=f(x)−g(x) h_1(x) = f(x) - g(x) , and h2(x)=g(x)−f(x) h_2(x) = g(x) - f(x) , then choose the set of correct options.

Options:

  • Roots for h1(x)=0 h_1(x) = 0 and roots for h2(x)=0 h_2(x) = 0 are real, distinct, and the roots are the same for h1(x)=0 h_1(x) = 0 and h2(x)=0 h_2(x) = 0 .
  • Roots for h1(x)=0 h_1(x) = 0 and roots for h2(x)=0 h_2(x) = 0 are real and distinct but the roots are not the same for h1(x)=0 h_1(x) = 0 and h2(x)=0 h_2(x) = 0 .
  • Sum of roots of quadratic equation h1(x)=0 h_1(x) = 0 will be 23 \frac{2}{3} .
  • Product of roots of quadratic equation h2(x)=0 h_2(x) = 0 will be −2-2.
  • Axis of symmetry for both the functions h1(x) h_1(x) and h2(x) h_2(x) will be the same.
  • Vertex for both the functions h1(x) h_1(x) and h2(x) h_2(x) will be the same.

Solution:

  • Correct: Roots for h1(x)=0 h_1(x) = 0 and roots for h2(x)=0 h_2(x) = 0 are real, distinct, and the roots are the same for h1(x)=0 h_1(x) = 0 and h2(x)=0 h_2(x) = 0 ; sum of roots of h1(x)=0 h_1(x) = 0 is 23 \frac{2}{3} ; product of roots of h2(x)=0 h_2(x) = 0 is −2-2; axis of symmetry is the same1

5. Contextual Questions

Vaishali wants to set up a small plate making machine in her village. The table shows the different costs involved in making the plates. The demand (number of packets of the plate) versus selling price of plate per packet (in ₹) per day is shown in the figure.

Cost typeCost (₹) per packet
Electricity1.5
Miscellaneous6.5
Raw material10

Demand vs Selling price: From figure, demand y=−2x+120 y = -2x + 120

Profit: Profit = Demand × (Selling price – Cost price) = (−2x+120)(x−18)(-2x + 120)(x - 18)

  • 19. Which of the following expressions represents the profit per day?
    • 2(60−x) 2(60 - x)
    • (x−18) (x - 18)
    • 2(x−18)(60−x) 2(x - 18)(60 - x)
    • (x+18)(60−x) (x + 18)(60 - x)
    • Inadequate information.

Solution: Profit = (−2x+120)(x−18)=2(x−18)(60−x) (-2x + 120)(x - 18) = 2(x - 18)(60 - x) Answer: 2(x−18)(60−x) 2(x - 18)(60 - x) 1


  • 20. Vaishali should sell a packet with a minimum price of ₹___ so as not to incur any loss.

Solution: Profit = 0 at x=18 x = 18 or x=60 x = 60 . Minimum price: ₹18 Answer: ₹18


  • 21. To make maximum profit per day, the selling price per packet should be ₹___ (if not restricted by market).

Solution: Maximum profit at x=39 x = 39 Answer: ₹39


  • 22. What should be the price of plate per packet (₹) to make a profit of ₹490 per day?

Solution: Solve 2(x−18)(60−x)=490 2(x - 18)(60 - x) = 490 Solution: x=25 x = 25 Answer: ₹251


  • 23. What will be the value of m+n m + n if the sum of the roots and the product of the roots of equation (5m+5)x2−(4n+3)x+10=0(5m + 5)x^2 - (4n + 3)x + 10 = 0 are 3 and 2 respectively?

Solution: Sum: 4n+35m+5=3  ⟹  4n+3=15m+15 \frac{4n + 3}{5m + 5} = 3 \implies 4n + 3 = 15m + 15 Product: 105m+5=2  ⟹  10=10m+10  ⟹  m=0 \frac{10}{5m + 5} = 2 \implies 10 = 10m + 10 \implies m = 0 Then, 4n+3=15  ⟹  n=3 4n + 3 = 15 \implies n = 3 Answer: m+n=3 m + n = 3 1


  • 24. What will the sum of two positive integers be if the sum of their squares is 369 and the difference between them is 3?

Solution: Let integers be a a and b b . a2+b2=369 a^2 + b^2 = 369 a−b=3 a - b = 3 Solve: (a+b)2=729  ⟹  a+b=27 (a + b)^2 = 729 \implies a + b = 27 Answer: 271


This covers all questions and solutions from the provided PDF.

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