Category: Class 11 Mathematics Objective Questions

Mathematics Multiple Choice Questions on “Applications of Quadratic Equations”.

1. If, (a + 1)x² + 2(a+1)x + (a – 2) = 0, then, for what parameter of ‘a’ the given equation have real and distinct roots?
a) (-∞, ∞)
b) (-1, ∞)
c) [-1, ∞)
d) (-1, 1)
Answer: b
Clarification: For, real and distinct roots, D > 0
Where, D = b² – 4ac
In the equation, (a + 1)x² + 2(a+1)x + (a – 2) = 0
D = [2(a+1)]² – 4 (a + 1)(a – 2)
= 4a² + 4 + 8a – 4{a² – 2a + a – 2}
= 4a² + 4 + 8a – 4a² + 4 a + 8 > 0
=> 12a + 12 > 0
=> 12a > -12
=> a > -1
Therefore, a € (-1, ∞)

2. If, (a + 1)x² + 2(a+1)x + (a – 2) = 0, then, for what parameter of ‘a’ the given equation have equal roots?
a) (-∞, -1)
b) [-1, ∞)
c) (0, 1)
d) Not possible
Answer: d
Clarification: For, equal roots, D = 0
Where, D = b² – 4ac
In the equation, (a + 1)x² + 2(a+1)x + (a – 2) = 0
D = [2(a+1)]² – 4 (a + 1)(a – 2)
= 4a² + 4 + 8a – 4{a² – 2a + a – 2}
= 4a² + 4 + 8a – 4a² + 4 a + 8 > 0
=> 12a + 12 = 0
=> 12a = -12
=> a = -1
So, from here it is clear that a = -1 is not possible because the equation is becoming linear.

3. If, (a + 1)x² + 2(a+1)x + (a – 2) = 0, then, for what parameter of ‘a’ the given equation have imaginary roots?
a) (-∞, -1)
b) (-1, ∞)
c) (-1, 1)
d) (-∞, ∞)
Answer: a
Clarification: For, imaginary roots, D > 0
Where, D = b² – 4ac
In the equation, (a + 1)x² + 2(a+1)x + (a – 2) = 0
D = [2(a+1)]² – 4 (a + 1)(a – 2)
= 4a² + 4 + 8a – 4{a² – 2a + a – 2}
= 4a² + 4 + 8a – 4a² + 4 a + 8 < 0
=> 12a + 12 < 0
=> 12a < -12
=> a < -1
Therefore, a € (-∞, -1)

4. If x₁, x₂ are real roots of ax² – x + a = 0. Then, find the set of all values of parameter ‘a’ for which |x₁ – x₁| < 1?
a) (1 – 5a)/ a² < 0
b) (1 – 5a)/ a² = 0
c) (1 – 5a)/ a² > 0
d) (1 – 5a)/ a < 0
Answer: a
Clarification: |x₁ – x₂| < 1
= (x₁ – x₂)² < 1
= (x₁ + x₂)² – 4 x₁ x₂ -1 < 0
= (1/a)² – 4 – 1 < 0
= (1 – 5a)/a² < 0.

5. What is the value of x if (a + 2b – 3c)x² + (b + 2c – 3a)x + (c + 2a – 3b) = 0 where a, b, c are in A.P?
a) 1/2
b) 1/4
c) 2/3
d) 3/4
Answer: b
Clarification: If the coefficients of (x² + x + c) = 0, then
x will always be = 1
Therefore, here, (a + 2b – 3c) + (b + 2c – 3a) + (c + 2a – 3b) = 0
So, x = 1.
As, one of its root is 1 so, we will calculate the other one.
As, a, b, c are in A.P so,
b = (a + c)/2
Thus, product of the roots αβ = (c + 2a – 3b)/(a + 2b – 3c)
As, a root say α = 1, then,
β = (c + 2a – 3(a + c)/2) / (a + 2(a + c)/2 – 3c)
We get the value of β = 1/4

6. What will be the value of f(x) if, 2A, A + B, C are integers and f(x) = Ax² + Bx + C = 0?
a) Natural Number
b) Unpredictable
c) Integer
d) Complex Number
Answer: c
Clarification: f(x) = Ax² + Bx + C = 0
So, f(x) = Ax² + (A + B)x – Ax + C
= Ax² – Ax + (A + B)x + C
= 2Ax(x – 1)/2 + (A + B)x + C
Therefore, f(x) is an integer.

7. What will be the sum of the real roots of the equation x² + 5|x| + 6 = 0?
a) Equal to 5
b) Equal to 10
c) Equal to -5
d) Does not exist
Answer: d
Clarification: Since, x², 5|x| and 6 are positive,
So, x² + 5|x| + 6 = 0 does not have any real root
Therefore, sum does not exist.

8. What is the number of solution(s) of the equation |√x – 2| + √x(√x – 4) + 2 = 0?
a) 2
b) 4
c) No solution
d) Infinitely many solutions
Answer: a
Clarification: We have |√x – 2| + √x(√x – 4) + 2 = 0,
|√x – 2| + (√x)² – 4√x + 2 = 0
|√x – 2| + |√x -2|² – 2 = 0
|√x – 2| = -2, 1
Thus, √x – 2 = +1, -1 or x = 1, 9

9. Which is the largest negative integer which satisfies (x² – 1)/(x – 2)(x – 3)?
a) -4
b) -3
c) -1
d) -2
Answer: d
Clarification: By wavy curve method
(x² – 1)/(x – 2)(x – 3) > 0
So, x = -1, 1, 2, 3
Thus, x € (-∞, -1) ∪ (1, 2) ∪ (3, ∞)
Therefore, the largest negative integer is -2.

10. Which one is the complete set of values of x satisfying log _x2 (x + 1) > 0?
a) (1, ∞)
b) (-1, 0) – {0}
c) (-1, 1) – {0}
d) (1, 0) ∪ (1, ∞)
Answer: d
Clarification: If, x² > 1, then x + 1 > 0
So, x > 0
x € (1, ∞)
If, 0 < x < 1, the 0 < x + 1 < 1
x € (-1, 0)
Thus, x € (1, ∞) ∪ (1, ∞)

11. What is the set of values of p for which the roots of the equation 3x² + 2x + p(p – 1) = 0 are of opposite sign?
a) (-∞, 0)
b) (0, 1)
c) (1, ∞)
d) (0, ∞)
Answer: b
Clarification: Since the roots of the given equation are of opposite sing,
So, products f the roots < 0
p(p – 1) / 3 < 0
p(p – 1) < 0
0 < p < 1
For real and distinct roots ½ – √21/ 6 < p < ½ + √21/6

12. For what value of θ, 1 lies between the roots of the quadratic equation 3x² – 3sinθ x – 2cos²θ = 0?
a) 2nπ + π/6 < θ < 2nπ + 5π/6
b) 2nπ + π/3 < θ < 2nπ + 5π/3
c) 2nπ + π/6 ≤ θ ≤ 2nπ + 5π/6
d) 2nπ + π/3 ≤ θ ≤ 2nπ + 5π/3
Answer: a
Clarification: Let, f(x) = 3x² – 3sinθ x – 2cos²θ
The coefficient of x² > 0
f(1) < 0
So, 3 – 3sinθ – 2cos²θ < 0
=> 2sin²θ – 3sinθ + 1 < 0
=> (2sinθ – 1)(sinθ – 1) < 0
=> ½ 2nπ + π/6 < θ < 2nπ + 5π/6

13. A real number ‘a’ is called a good number if the inequality (2x² – 2x – 3) / (x² + x + 1) ≤ a is satisfied for all real x. What is the set of all real numbers?
a) (-∞, 10/3]
b) (10/3, ∞)
c) [10/3, ∞)
d) [-10/3, ∞)
Answer: c
Clarification: We have, (2x² – 2x – 3) / (x² + x + 1) ≤ a ᵾ x € R
=> 2x² – 2x – 3 ≤ a(x² + x + 1) ᵾ x € R
=> (2 – a)x² – (2 – a)x – (3 – a) ᵾ x € R
2 – a < 0 and (2 – a)x² – 4(2 – a)(3 – a) ≤ 0 ᵾ x € R
So, a > 2 and a ≤ 2 or a ≥ 10/3
=> a ≥ 10/3
Therefore, a € [10/3, ∞)

14. Let S denotes the set of all real values of the parameter ‘a’ for which every solution of the inequality log_1/2 x² ≥ log_1/2 (x + 2) is the solution of the inequality 49x² – 4a⁴ ≤ 0. What is the value of S?
a) (-∞, -√7) ∪ (√7, ∞)
b) (-∞, -√7] ∪ [√7, ∞)
c) (-√7, √7)
d) [-√7, √7]
Answer: b
Clarification: We have, log_1/2 x² ≥ log_1/2 (x + 2)
=> x² ≤ x + 2
=> -1 ≤ x ≤ 2
And, 49x² – 4a⁴ ≤ 0 i.e. x² ≤ 4a⁴ / 49
=> -2a²/7 ≤ a ≤ 2a²/7
From the above equations,
-2a²/7 ≤ -1 and 2 ≤ 2a²/7
i.e. a² € 7/2 and a² ≥ 7
=> a € (-∞, -√7] ∪ [√7, ∞)
So, S = (-∞, -√7] ∪ [√7, ∞)

15. If, x⁴ + 4x³ + 6ax² + 6bx + c is divisible by x³ + 3x² + 9x + 3. Then, what is the value of a + b + c?
a) 4
b) 6
c) 7
d) 10
Answer: c
Clarification: Here, f(x) = x⁴ + 4x³ + 6ax² + 6bx + c so, let its roots be, α, β, γ, δ and
g(x) = x³ + 3x² + 9x + 3 so, let its roots be, α, β, γ
So, from here we can conclude,
α + β + γ + δ = -4 and α + β + γ = -3
Thus, δ = -1
This means (x + 1)(x³ + 3x² + 9x + 3)
On solving this equation in simpler form we get,
x⁴ + 4x³ + 12x² + 12x + 3
=> 6a = 12 => a = 2
=> 6b = 12 => b = 2
=> c = 3
=> a + b + c = 7

Mathematics Written Test Questions and Answers for Class 11 on “Sum to n Terms of Special Series”.

1. Find the sum of first n terms.
a) (frac{n(n+1)}{2})
b) ((frac{n(n+1)}{2})^3)
c) (frac{n(n+1)(2n+1)}{6})
d) ((frac{n(n+1)}{2})^2)
Answer: a
Clarification: Sum of first n terms = 1+2+3+4+……+n
=> (n/2) (a + a_n) = (n/2) (1+n) = (frac{n(n+1)}{2}).

2. Find the sum of squares of first n terms.
a) (frac{n(n+1)}{2})
b) ((frac{n(n+1)}{2})^3)
c) (frac{n(n+1)(2n+1)}{6})
d) ((frac{n(n+1)}{2})^2)
Answer: c
Clarification: Sum of squares of first n terms = 1²+2²+3²+……………+n²
k³–(k – 1)³=3k²–3k + 1
On substituting k = 1, 2, 3, ……, n and adding we get,
n³ = 3 (sum_{i=0}^n k^2 – 3 sum_{i=0}^n k + n)
n³ = 3 (sum_{i=0}^n k^2 – 3 frac{n(n+1)}{2}) + n
(sum_{i=0}^n k^2 = frac{n(n+1)(2n+1)}{6}).

3. Find the sum of cubes of first n terms.
a) (frac{n(n+1)}{2})
b) ((frac{n(n+1)}{2})^3)
c) (frac{n(n+1)(2n+1)}{6})
d) ((frac{n(n+1)}{2})^2)
Answer: c
Clarification: Sum of cubes of first n terms = 1³+2³+3³+……………+n³
(k + 1)⁴–k⁴ = 4k³ + 6k² + 4k + 1.
On substituting k = 1, 2, 3, ……, n and adding we get,
4n³+n⁴+6n²+4n = (4sum_{i=0}^n k^3 + 6 sum_{i=0}^n k^2 + 4 sum_{i=0}^n k + n)
4n³+n⁴+6n²+4n = (4sum_{i=0}^n k^3 + 6frac{(n(n+1)(2n+1))}{6} + 4frac{n(n+1)}{2} + n)
(sum_{i=0}^n k^3 = (frac{n(n+1)}{2})^2).

4. Find the sum 1²+2²+3²+……………+10².
a) 325
b) 365
c) 385
d) 435
Answer: c
Clarification: We know, sum of squares of first n terms is given by (frac{(n(n+1)(2n+1))}{6}).
Here, n=10. So, sum = 10*11*21/6 = 385.

5. Find the sum 1³+2³+3³+……………+8³.
a) 1225
b) 1184
c) 1475
d) 1296
Answer: d
Clarification: We know, sum of cubes of first n terms is given by ((frac{n(n+1)}{2})^2).
Here, n=8. So, sum = (8*9/2)² = 1296.

6. Find the sum to n terms of the series whose n^th term is n (n-2).
a) (frac{n(n-1)(2n+4)}{6})
b) (frac{n(n+1)(2n-5)}{6})
c) (frac{(n-2)(2n-5)}{3})
d) (frac{n(n+1)(2n-5)}{3})
Answer: b
Clarification: Given, n^th term is n(n-2)
So, a_k = k(k-2)
Taking summation from k=1 to k=n on both sides, we get
(sum_{i=0}^n a_k = sum_{i=0}^n k^2 – 2 sum_{i=0}^n k = frac{n(n+1)(2n+1)}{6} – 2frac{n(n+1)}{2} = frac{n(n+1)(2n-5)}{6}).

7. Find the sum of series up to 6^th term whose n^th term is given by n² + 3ⁿ.
a) 91
b) 1284
c) 1183
d) 1092
Answer: c
Clarification: Given, n^th term is n² + 3ⁿ
So, a_k = k² + 3^k
Taking summation from k=1 to k=n on both sides, we get
(sum_{i=0}^na_k = sum_{i=0}^nk^2 + sum_{i=0}^n3^k)
(sum_{i=0}^nk^2 = n(n+1) (2n+1)/6)
(sum_{i=0}^n3^k = 3*(3^n-1)/ (3-1) = (3/2) (3^n-1))
So, (sum_{i=0}^na_k = n(n+1) (2n+1)/6 + (3/2) (3^n-1))
Sum up to 6^th term = 6*7*13/6 + (3/2) (3⁶-1) = 91+1092 = 1183.

8. Find the sum up to 7^th term of series 2+3+5+8+12+………………….
a) 70
b) 490
c) 340
d) 420
Answer: a
Clarification: S_n = 2+3+5+8+12+……………………………+ a_n
S_n = 2+3+5+8+12+ ……. + a_n-1 + a_n
Subtracting we get, 0 = 2+1+2+3+4+………………………….. – a_n
=>a_n = 2+1+2+3+4+…………….+(n-1) = 2+(n-1)n/2 = (1/2) (n²-n+4)
n^th term is (1/2) (n²-n+4)
So, a_k = (1/2) (k²-k+4)
Taking summation from k=1 to k=n on both sides, we get
(sum_{i=0}^na_k = (1/2)sum_{i=0}^nk^2 – (1/2)sum_{i=0}^nk + 2n) = n(n+1) (2n+1)/(2*6) – n(n+1)/4 + 2n
Here, n=7. So, (sum_{i=0}^na_k) = (7*8*15)/12 – (7*8)/4 + 2*7 = 70.

9. Find the sum to 6 terms of each of the series 2*3+4*6+6*11+8*18+………………..
a) 784
b) 882
c) 928
d) 966
Answer: d
Clarification: General term of above series is a_k = 2k*(k²+2) = 2k³+4k
Taking summation from k=1 to k=n on both sides, we get
(sum_{i=0}^na_k = 2sum_{i=0}^nk^3 + 4sum_{i=0}^nk = 2(frac{n(n+1)}{2})^2 + 4frac{n(n+1)}{2})
= n²(n+1)²/2+2n(n+1)
= 36*49/2 + 2*6*7
= 966.

10. Find the sum of series 6²+7²+…………………..+15².
a) 55
b) 1185
c) 1240
d) 1385
Answer: b
Clarification: 6²+7²+………………..…..+15²
= (1²+2²+3²+……..+15²) – (1²+2²+3²+4²+5²)
= 15*16*31/6 – 5*6*11/6
= 1240-55 = 1185.

11. Find the sum of series 6³+7³+………………..…..+20³.
a) 43875
b) 83775
c) 43775
d) 43975
Answer: a
Clarification: 6³+7³+………………..…..+20³
= (1³+2³+3³+……..+20³) – (1³+2³+3³+4³+5³)
= (20*21/2)² – (5*6/2)²
= (210)² – (15)²
= 225*195
= 43875.

12. Find the sum of series 1²+3²+5²+…………………………..+11².
a) 279
b) 286
c) 309
d) 409
Answer: b
Clarification: 1²+3²+5²+…………………………..+11²
= (1²+2²+3²+……+11²) – (2²+4²+6²+8²+10²)
= (1²+2²+3²+……11²) – 2²(1²+2²+3²+4²+5²)
= 11*12*23/6 – 4*5*6*11/6
= 506 – 220
= 286.

13. Find the sum of series 1³+3³+5³+…………………………..+11³.
a) 2556
b) 5248
c) 6589
d) 9874
Answer: a
Clarification: 1³+3³+5³+…………………………..+11³
= (1³+2³+3³+……+11³) – (2³+4³+6³+8³+10³)
= (1³+2³+3³+……11³) – 2³(1³+23+3³+4³+5³)
= (11*12/2)² – 8(5*6/2)²
= 66²-8*15²
= 4356 – 1800
= 2556.

Mathematics Written Test Questions and Answers for Class 11,

Mathematics Question Papers for Class 11 on “First Order Derivative – 2”.

1. If A (x₁, y₁) and B (x₂, y₂) be two points on the curve y = ax² + bx + c, then as perLagrange’s mean value theorem whichof the following is correct?
a) At least one point C(x₃, y₃) where the tangent will be intersecting the chord AB
b) At least one point C(x₃, y₃) where the tangent will be overlapping to the chord AB
c) At least two points where the tangent will be parallel to the chord AB
d) At least one point C(x₃, y₃) where the tangent will be parallel to the chord AB
Answer: d
Clarification: Here, y = f(x) = ax² + bx + c
As f(x) is a polynomial function, it is continuous and differentiable for all x.
So, according to geometrical interpretation of mean value theorem there will be at least one point C (x₃, y₃) between A (x₁, y₁) and B (x₂, y₂) where tangent will be parallel chord AB.

2. If (limlimits_{x rightarrow a}frac{(a^x-x^a)}{x^x-a^a}) = -1 then, what is the value of a?
a) 1
b) 2
c) 3
d) 4
Answer: a
Clarification: Let, y = x^x
Thus, log y = x log x
Differentiating both sides with respect to x, we get,
1/y dy/dx = (x*1/x) + log x
=>dy/dx = y(1 + log x)
Or, dx^x/dx = x^x(1 + log x)
Using, L’Hospital’s rule,
(limlimits_{x rightarrow a}frac{(a^x-x^a)}{x^x-a^a}) = (limlimits_{x rightarrow a}frac{(a^x*log⁡a-x^{a-1})}{x^x(1+logx)})
= (limlimits_{x rightarrow a}frac{(a^a*log⁡a-a^{a})}{a^a(1+loga)})
= (log a – 1)/(log a + 1)
As per the question,
(log a – 1)/(log a + 1) = -1
Or, (log a – 1) = -log a – 1
Or, 2 log a = 0
Or, log a = 0
So, a = 1

3. If functions f(x) and g(x) are continuous in [a, b] and differentiable in (a, b) then which of the following is correct if there exists at least one point c, a < c < b, such that (begin{vmatrix}f(a) & f(b) \g(a) & g(b) end {vmatrix})?
a) (b + a)(begin{vmatrix}f(a) & f”(c) \g(a) & g”(c) end {vmatrix})
b) (b – a)(begin{vmatrix}f(a) & f”(c) \g(a) & g”(c) end {vmatrix})
c) (b + a)(begin{vmatrix}f(a) & f'(c) \g(a) & g'(c) end {vmatrix})
d) (b – a)(begin{vmatrix}f(a) & f'(c) \g(a) & g'(c) end {vmatrix})
Answer: d
Clarification: Let, F(x) = (begin{vmatrix}f(a) & f(b) \g(a) & g(b) end {vmatrix}) = f(a)g(x) – f(x)g(a) …..(1)
=> F’(x) = f’(a)g’(x) – f’(x)g(a)
Since, f(x) and g(x) are continuous in [a, b] and differentiable in (a, b),
So, F(x) is continuous in [a, b] and differentiable in (a, b)
Also from (1), F(a) = f(a)g(a) – f(a)g(a) = 0
And F(b) = f(a)g(b) – f(b)g(a)
Now, by the mean value theorem, there exists at least one point c, a < c < b, such that,
F’(c) = (F’(b) – F’(a)) / (b – a)
=> f(a) g’(c) – g(a) f’(c) = (f(a)g(b) – f(b)g(a) – 0)/b – a
Or, f(a)g(b) – f(b)g(a) = (b – a)( f(a) g’(c) – g(a) f’(c))
=>(begin{vmatrix}f(a) & f(b) \g(a) & g(b) end {vmatrix}) = (b – a)(begin{vmatrix}f(a) & f'(c) \g(a) & g'(c) end {vmatrix})

4. What is the number of critical points of f(x) = |x² – 1| / x²?
a) 0
b) 1
c) 2
d) 3
Answer: c
Clarification:Clearly f (x) is not differentiable at x = 1 and x = -1
And x = 0 is not a critical point not in the domain.
Therefore 1 and -1 are critical points.
Thus, there are 2 critical points.

5. What will be the value of dy/dx if x = asec²θ and y = atan³θ at θ = π/4?
a) 1/2
b) 3/4
c) 3/2
d) 1/4
Answer: c
Clarification: Since, x = asec²θ,
Therefore, dx/dθ = a*d/dx(sec²θ)
= 2asecθ*secθ tanθ
Again, dy/dθ = a*d/dx(tan³θ)
= a * 3 tan²θ * d/dθ(tanθ)
= 3a tan²θ sec²θ
Therefore, dy/dx = (dy/dθ)/(dx/dθ)
= 3a tan²θ sec²θ/2asecθ*secθ tanθ
Thus, at θ = π/4 we have,
dy/dx = 3/2(tan π/4)
= 3/2

6. What is the number of critical points for f(x) = max(sinx, cosx) for x belonging to (0, 2π)?
a) 2
b) 5
c) 3
d) 4
Answer: c
Clarification: We know that in the range of (0, 2π) the graph of sinx and cosx intersects each other in three points.
And we know that these points of intersection are only the critical points
Thus, there are 3 critical points.

7. If y = (3x – 4)/(x+2) then what s the value of dy/dx?
a) dy/dx
b) y
c) 1/ (dy/dx)
d) A constant
Answer: c
Clarification: It is given that y = (3x – 4)/(x + 2) ……….(1)
Now differentiating both the sides, we get that,
dy/dx = (x + 2)*3 – (3x – 4)/(x + 2)²
= 10/(x + 2)²
Again from (1) we get,
xy + 2y = 3x – 4
or, x = – 2(y + 2)
Thus dx/dy = -2* ((y – 3) – (y + 2))/ (y – 3)²
Or, y – 3 = (3x – 4)/(x + 2) – 3
= -10/(x + 2)
Thus, dx/dy = 10/(-10/(x + 2))²
= (x + 2)2/10, where,x ≠ 0 i.e. dx/dy ≠ 0
Therefore, dy/dx*dx/dy = 10/(x + 2)2 * -10/(x + 2)
= 1
=> dy/dx = 1/(dy/dx)

8. What is the value of (dy/dx)² + 1 if x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)?
a) Tan²θ
b) Cosec²θ
c) Cot²θ
d) Sec²θ
Answer: d
Clarification: Here, x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)
=> x = 2a cos²θ*sin2θ and y = 2a sin²θ*cos2θ
Now differentiating x and y with respect to θ we get,
dx/dθ = 2a[cos²θ*2cos2θ + sin2θ*2cos2θ cos2θ]
= 4a cosθ (cosθ cos2θ – sinθ sin2θ )
= 4a cosθ cos(θ + 2θ)
= 4a cosθ cos3θ
dy/dθ = 2a[cos2θ*2cosθ sinθ + sin²θ (-2sin2θ)]
= 4a sinθ (cosθ cos2θ – sinθ sin2θ )
= 4a sinθ cos(θ + 2θ)
= 4a sinθ cos3θ
Thus, dy/dx = (dy/dθ)/(dx/dθ)
= (= 4a cosθ cos3θ)/( 4a sinθ cos3θ)
= tanθ
So, (dy/dx)² + 1 = 1 + tan²θ = sec²θ

9. If, y = 1/(1 + x + x² + x³), then what is the value of y’ at x = 0?
a) 0
b) 1
c) -1
d) ½
Answer: c
Clarification: Given, y = 1/(1 + x + x² + x³)
Assuming x ≠ 1
Or, y = (x – 1)/(x – 1)( x³ + x² + x + 1)
Differentiating both the sides with respect to x, we get,
dy/dx = [(x⁴ – 1)*1 – (x – 1)*4x³]/ (x⁴ – 1)²
Thus, putting x = 0 in the above equation, we get,
(dy/dx) = -1/(-1)²
= -1

Mathematics Multiple Choice Questions on “Relations”.

1. A relation is a subset of cartesian products.
a) True
b) False
Answer: a
Clarification: A relation from a non-empty set A to a non-empty set B is a subset of cartesian product A X B. First element is called the preimage of second and second element is called image of first.

2. Let A={1,2,3,4,5} and R be a relation from A to A, R = {(x, y): y = x + 1}. Find the domain.
a) {1,2,3,4,5}
b) {2,3,4,5}
c) {1,2,3,4}
d) {1,2,3,4,5,6}
Answer: a
Clarification: We know, domain of a relation is the set from which relation is defined i.e. set A.
So, domain = {1,2,3,4,5}.

3. Let A={1,2,3,4,5} and R be a relation from A to A, R = {(x, y): y = x + 1}. Find the codomain.
a) {1,2,3,4,5}
b) {2,3,4,5}
c) {1,2,3,4}
d) {1,2,3,4,5,6}
Answer: a
Clarification: We know, codomain of a relation is the set to which relation is defined i.e. set A.
So, codomain = {1,2,3,4,5}.

4. Let A={1,2,3,4,5} and R be a relation from A to A, R = {(x, y): y = x + 1}. Find the range.
a) {1,2,3,4,5}
b) {2,3,4,5}
c) {1,2,3,4}
d) {1,2,3,4,5,6}
Answer: b
Clarification: Range is the set of elements of codomain which have their preimage in domain.
Relation R = {(1,2), (2,3), (3,4), (4,5)}.
Range = {2,3,4,5}.

5. If set A has 2 elements and set B has 4 elements then how many relations are possible?
a) 32
b) 128
c) 256
d) 64
Answer: d
Clarification: We know, A X B has 2*4 i.e. 8 elements. Number of subsets of A X B is 2⁸ i.e. 256.
A relation is a subset of cartesian product so, number of possible relations are 256.

6. Is relation from set A to set B is always equal to relation from set B to set A.
a) True
b) False
Answer: b
Clarification: A relation from a non-empty set A to a non-empty set B is a subset of cartesian product A X B. A relation from a non-empty set B to a non-empty set A is a subset of cartesian product B X A.
Since A X B ≠ B X A so, both relations are not equal.

7. If A={1,4,8,9} and B={1, 2, -1, -2, -3, 3,5} and R is a relation from set A to set B {(x, y): x=y²}. Find domain of the relation.
a) {1,4,9}
b) {-1,1, -2,2, -3,3}
c) {1,4,8,9}
d) {-1,1, -2,2, -3,3,5}
Answer: c
Clarification: We know, domain of a relation is the set from which relation is defined i.e. set A.
So, domain = {1,4,8,9}.

8. If A={1,4,8,9} and B={1, 2, -1, -2, -3, 3,5} and R is a relation from set A to set B {(x, y): x=y²}. Find codomain of the relation.
a) {1,4,9}
b) {-1,1, -2,2, -3,3}
c) {1,4,8,9}
d) {-1,1, -2,2, -3,3,5}
Answer: d
Clarification: We know, codomain of a relation is the set to which relation is defined i.e. set B.
So, codomain = {-1,1, -2,2, -3,3,5}.

9. If A={1,4,8,9} and B={1, 2, -1, -2, -3, 3,5} and R is a relation from set A to set B {(x, y): x=y²}. Find range of the relation.
a) {1,4,9}
b) {-1,1, -2,2, -3,3}
c) {1,4,8,9}
d) {-1,1, -2,2, -3,3,5}
Answer: b
Clarification: Range is the set of elements of codomain which have their preimage in domain.
Relation R = {(1,1), (1, -1), (4,2), (4, -2), (9,3), (9, -3)}.
Range = {-1,1, -2,2, -3,3}.

10. Let A={1,2} and B={3,4}. Which of the following cannot be relation from set A to set B?
a) {(1,1), (1,2), (1,3), (1,4)}
b) {(1,3), (1,4)}
c) {(2,3), (2,4)}
d) {(1,3), (1,4), (2,3), (2,4)}
Answer: a
Clarification: A relation from set A to set B is a subset of cartesian product of A X B. In ordered pair, first element should belong to set A and second element should belongs to set B.
In {(1,1), (1,2), (1,3), (1,4)}, 1 and 2 should also be in the set B which is not so as given in question.
Hence, {(1,1), (1,2), (1,3), (1,4)} is not a relation from set A to set B.

Mathematics Multiple Choice Questions on “Inequalities”.

1. 7>5 is ______________________
a) linear inequality
b) quadratic inequality
c) numerical inequality
d) literal inequality
Answer: c
Clarification: Since here numbers are compared with inequality sign so, it is called numerical inequality.

2. x>5 is _____________________
a) double inequality
b) quadratic inequality
c) numerical inequality
d) literal inequality
Answer: d
Clarification: Since a variable ‘x’ is compared with number ‘5’ with inequality sign so it is called literal inequality.

3. ax + b > 0 is _____________________
a) double inequality
b) quadratic inequality
c) numerical inequality
d) linear inequality
Answer: d
Clarification: Since it has highest power of x ‘1’ and has inequality sign so, it is called linear inequality.
It is not numerical inequality as it does not have numbers on both sides of inequality.
It does not have two inequality signs so it is not double inequality.

4. ax²+bx+c > 0 is _____________________
a) double inequality
b) quadratic inequality
c) numerical inequality
d) linear inequality
Answer: b
Clarification: Since it has highest power of x ‘2’ and has inequality sign so, it is called quadratic inequality.
It is not numerical inequality as it does not have numbers on both sides of inequality.
It does not have two inequality signs so it is not double inequality.

5. ax²+bx+c ≥ 0 is a strict inequality.
a) True
b) False
Answer: a
Clarification: Since it has equality sign along with inequality sign so it is a slack inequality not strict inequality.

6. If Ram has x rupees and he pay 40 rupees to shopkeeper then find range of x if amount of money left with Ram is at least 10 rupees is given by inequation __________________
a) x ≥ 10
b) x ≤ 10
c) x ≤ 50
d) x ≥ 50
Answer: d
Clarification: Amount left is at least 10 rupees i.e. amount left ≥ 10.
x-40 ≥ 10 => x ≥ 50.

7. If Ram has x rupees and he pay 40 rupees to shopkeeper then find range of x if amount of money left with Ram is at most 10 rupees is given by inequation __________________
a) x ≥ 10
b) x ≤ 10
c) x ≤ 50
d) x ≥ 50
Answer: c
Clarification: Amount left is at most 10 rupees i.e. amount left ≤ 10.
x-40 ≤ 10 => x ≤ 50.

Mathematics Multiple Choice Questions on “Slope of a Line”.

1. What is the distance between (1, 3) and (5, 6)?
a) 3 units
b) 4 units
c) 5 units
d) 25 units
Answer: c
Clarification: We know, distance between two points (x₁, y₁) and (x₂, y₂) is (sqrt{(x_1-x_2)^2+(y_1-y_2)^2}).
So, distance between (1, 3) and (5, 6) is (sqrt{(1-5)^2+(3-6)^2}=sqrt{(4)^2+(3)^2}) = 5 units.

2. What is the distance of (5, 12) from origin?
a) 6 units
b) 8 units
c) 10 units
d) 13 units
Answer: d
Clarification: We know, distance between two points (x₁, y₁) and (x₂, y₂) is (sqrt{(x_1-x_2)^2+(y_1-y_2)^2}).
So, distance between (5, 12) from origin (0, 0) is (sqrt{(5-0)^2+(12-0)^2} = sqrt{(5)^2+(12)^2}) = 13 units.

3. The coordinates of a point dividing the line segment joining (1, 2) and (4, 5) internally in the ratio 2:1 is ____________________
a) (3, 4)
b) (4, 3)
c) (5, 4)
d) (5, 3)
Answer: a
Clarification: The coordinates of a point dividing the line segment joining (x₁, y₁) internally in the ratio m: n is ((frac{mx_2+nx_1}{m+n},frac{my_2+ny_1}{m+n})).
So, the coordinates of a point dividing the line segment joining (1, 2) and (4, 5) internally in the ratio 2:1 is ((frac{2*4+1*1}{2+1}, frac{2*5+1*2}{2+1})) = (3, 4).

4. In which ratio (3, 4) divides the line segment joining (1, 2) and (4, 5) internally?
a) 1:2
b) 2:1
c) 3:4
d) 4:3
Answer: b
Clarification: The coordinates of a point dividing the line segment joining (x₁, y₁) and (x₂, y₂) internally in the ratio m: n is ((frac{mx_2+nx_1}{m+n},frac{my_2+ny_1}{m+n})).
Let the ratio be k: 1.So, the coordinates of a point dividing the line segment joining (1, 2) and (4, 5) internally in the ratio k: 1 is ((frac{k*4+1*1}{k+1},frac{k*5+1*2}{k+1}))
=> ((frac{k*4+1*1}{k+1},frac{k*5+1*2}{k+1})) is same as (3, 4).
=> (4k+1)/(k+1) = 3
=> 4k+1 = 3k+3
=> k = 2
So, ratio is 2:1.

5. The coordinates of a point dividing the line segment joining (1, 2) and (4, 5) externally in the ratio 2:1 is ____________________
a) (4, 5)
b) (6, 8)
c) (7, 8)
d) (8, 6)
Answer: c
Clarification: The coordinates of a point dividing the line segment joining (x₁, y₁) and (x₂, y₂) externally in the ratio m: n is ((frac{mx_2-nx_1}{m-n},frac{my_2-ny_1}{m-n})).
So, the coordinates of a point dividing the line segment joining (1, 2) and (4, 5) externally in the ratio 2:1 is ((frac{2*4-1*1}{2-1},frac{2*5-1*2}{2-1})) = (7, 8).

6. _____________ is the midpoint of (1, 2) and (5, 8).
a) (2, 5)
b) (3, 5)
c) (5, 2)
d) (5, 3)
Answer: b
Clarification: We know, midpoint of (x₁, y₁) and (x₂, y₂) is ((frac{x1+x2}{ 2}, frac{y1+y2}{2})).
So, midpoint of (1, 2) and (5, 8) is ((1+5)/2, (2+8)/2) is (3, 5).

7. What is the area of triangle whose vertices are (-4, -4), (-3, 2), (3, -16)?
a) 24 sq. units
b) 27 sq. units
c) 32 sq. units
d) 37 sq. units
Answer: b
Clarification: We know, area of triangle joining vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is (1/2)* determinant (begin{pmatrix}-4 & -4 &1 \-3& 2 & 1\ 3& -16 & 1end{pmatrix}) is (frac{1}{2}){(-4)(2+16) – (-4)(-3-3) + (1)(48-6)} = (frac{1}{2})|(-72)+(-24)+42| = 27 square units.

8. If area of triangle formed by three points is zero then the three points must be collinear.
a) True
b) False
Answer: a
Clarification: Area of triangle formed by three points is zero then the three points must be collinear i.e. they must lie on the same line.

9. Angle made by line with ____________ measured anticlockwise is called inclination of the line.
a) positive x-axis
b) negative x-axis
c) positive y-axis
d) negative y-axis
Answer: a
Clarification: We know, inclination of line is always measured with positive x-axis in anticlockwise direction.

10. Slope of a line is given by _________ if inclination of line is α.
a) sinα
b) cosα
c) tanα
d) cotα
Answer: c
Clarification: Slope of a line is given by tanα if inclination of line is α. Slope is denoted by tangent of the inclination angle.

11. Find slope of line if inclination made by the line is 60°.
a) 1/2
b) 1/√3
c) √3
d) 1
Answer: c
Clarification: Slope of a line is given by tanα if inclination of line is α. If inclination is 60° the slope is tan 60° = √3.

12. What is the inclination of a line which is parallel to x-axis?
a) 0°
b) 180°
c) 45°
d) 90°
Answer: a
Clarification: If a line is parallel to x-axis then angle formed by it with x-axis is zero. So, its inclination is zero.

13. What is the inclination of a line which is parallel to y-axis?
a) 0°
b) 180°
c) 45°
d) 90°
Answer: d
Clarification: If a line is parallel to y-axis then angle formed by it with x-axis is 90°. So, its inclination is 90°.

14. What is the slope of a line which is parallel to x-axis?
a) -1
b) 0
c) 1
d) Not defined
Answer: b
Clarification: If a line is parallel to x-axis then angle formed by it with x-axis is zero. So, its inclination is zero. Hence slope = tan 0° = 0.

15. What is the slope of a line which is parallel to y-axis?
a) -1
b) 0
c) 1
d) Not defined
Answer: d
Clarification: If a line is parallel to y-axis then angle formed by it with x-axis is zero. So, its inclination is 90°. Hence slope = tan 90° which is not defined.