Category: Class 12 Mathematics Objective Questions

Mathematics written test Questions & Answers on “Composition of Functions and Invertible Function”.

1. The composition of functions is both commutative and associative.
a) True
b) False
Answer: b
Clarification: The given statement is false. The composition of functions is associative i.e. fο(g ο h)=(f ο g)οh. The composition of functions is not commutative i.e. g ο f ≠ f ο g.

2. If f:R→R, g(x)=3x²+7 and f(x)=√x, then gοf(x) is equal to _______
a) 3x-7
b) 3x-9
c) 3x+7
d) 3x-8
Answer: c
Clarification: Given that, g(x)=3x²+7 and f(x)=√x
∴ gοf(x)=g(f(x))=g(√x)=3(√x)²+7=3x+7.
Hence, gοf(x)=3x+7.

3. If f:R→R is given by f(x)=(5+x⁴)^1/4, then fοf(x) is _______
a) x
b) 10+x⁴
c) 5+x⁴
d) (10+x⁴)^1/4
Answer: d
Clarification: Given that f(x)=(5+x⁴)^1/4
∴ fοf(x)=f(f(x))=(5+{(5+x⁴)^1/4}⁴)^1/4
=(5+(5+x⁴))^1/4=(10+x⁴)^1/4.

4. If f:R→R f(x)=cos⁡x and g(x)=7x³+6, then fοg(x) is ______
a) cos⁡(7x³+6)
b) cos⁡x
c) cos⁡(x³)
d) (cos(frac{x^3+6}{7}))
Answer: a
Clarification: Given that, f:R→R, f(x)=cos⁡x and g(x)=7x³+6
Then, fοg(x) = f(g(x))=cos⁡(g(x))=cos⁡(7x³+6).

5. A function is invertible if it is ____________
a) surjective
b) bijective
c) injective
d) neither surjective nor injective
Answer: b
Clarification: A function is invertible if and only if it is bijective i.e. the function is both injective and surjective. If a function f:A→B is bijective, then there exists a function g:B→A such that f(x)=y⇔g(y)=x, then g is called the inverse of the function.

6. The function f:R→R defined by f(x)=5x+9 is invertible.
a) True
b) False
Answer: a
Clarification: The given statement is true. A function is invertible if it is bijective.
For one – one: Consider f(x₁)=f(x₂)
∴ 5x₁+9=5x₂+9
⇒x₁=x₂. Hence, the function is one – one.
For onto: For any real number y in the co-domain R, there exists an element x=(frac{y-9}{5}) such that f(x)=(f(frac{y-9}{5})=5(frac{y-9}{5}))+9=y.
Therefore, the function is onto.

7. If f:N→N, g:N→N and h:N→R is defined f(x)=3x-5, g(y)=6y² and h(z)=tan⁡z, find ho(gof).
a) tan⁡(6(3x-5))
b) tan⁡(6(3x-5)²)
c) tan⁡(3x-5)
d) 6 tan⁡(3x-5)²
Answer: b
Clarification: Given that, f(x)=3x-5, g(y)=6y² and h(z)=tan⁡z,
Then, ho(gof)=hο(g(f(x))=h(6(3x-5)²)=tan⁡(6(3x-5)²)
∴ ho(gof)=tan⁡(6(3x-5)²)

8. Let M={7,8,9}. Determine which of the following functions is invertible for f:M→M.
a) f = {(7,7),(8,8),(9,9)}
b) f = {(7,8),(7,9),(8,9)}
c) f = {(8,8),(8,7),(9,8)}
d) f = {(9,7),(9,8),(9,9)}
Answer: a
Clarification: The function f = {(7,7),(8,8),(9,9)} is invertible as it is both one – one and onto. The function is one – one as every element in the domain has a distinct image in the co – domain. The function is onto because every element in the codomain M = {7,8,9} has a pre – image in the domain.

9. Let f:R₊→[9,∞) given by f(x)=x²+9. Find the inverse of f.
a) (sqrt{x-9})
b) (sqrt{9-x})
c) (sqrt{x^2-9})
d) x²+9
Answer: a
Clarification: The function f(x)=x²+9 is bijective.
Therefore, f(x)=x²+9
i.e.y=x²+9
x=(sqrt{y-9})
⇒f^-1 (x)=(sqrt{x-9}).

10. Let the function f be defined by f(x)=(frac{9+3x}{7-2x}), then f^-1(x) is ______
a) (frac{9-3x}{7+2x})
b) (frac{7x-9}{2x+3})
c) (frac{2x-7}{3x+9})
d) (frac{2x-3}{7x+9})
Answer: b
Clarification: The function f(x)=(frac{9+3x}{7-2x}) is bijective.
∴ f(x)=(frac{9+3x}{7-2x})
i.e.y=(frac{9+3x}{7-2x})
7y-2xy=9+3x
7y-9=x(2y+3)
x=(frac{7y-9}{2y+3})
⇒f^-1 (x)=(frac{7y-9}{2x+3}).

Mathematics Multiple Choice Questions on “Application of Determinants – 1”.

1. What is the value of (begin{vmatrix}-bc & ca + ab & ca + ab \ab + bc & -ca & ab + bc \bc + ca & bc + ca & -ab end {vmatrix}) ?
a) Σab
b) (Σab)²
c) (Σab)³
d) (Σab)⁴
Answer: c
Clarification: Applying R₁ –> R₁ + R₂ + R₃
(begin{vmatrix}Sigma ab & Sigma ab & Sigma ab \ab + bc & -ca & ab + bc \bc + ca & bc + ca & -ab end {vmatrix})
This is equal to,
= Σab (begin{vmatrix}1 & 1 & 1 \ab + bc & -ca & ab + bc \bc + ca & bc + ca & -ab end {vmatrix})
Applying C₁ –> C₁ – C₂ and C₂ –> C₂ – C₃
= Σab (begin{vmatrix}0 & 0 & 1 \ Sigma ab & -Sigma ab & ab + bc \0 & Sigma ca & -ab end {vmatrix})
= (Σab)³

2. If the system of equation 2x + 5y + 8z = 0, x + 4y + 7z = 0, 6x + 9y – αz = 0 has a non trivial solution then what is the value of α?
a) -12
b) 0
c) 12
d) 2
Answer: c
Clarification: Here, in L.H.S we have,
(begin{vmatrix}2 & 5 & 8 \1 & 4 & 7 \6 & 4 & alpha end {vmatrix})
So, for trivial roots the above value is = 0
=>(begin{vmatrix}2 & 5 & 8 \1 & 4 & 7 \6 & 4 & alpha end {vmatrix}) = 0
Solving it further we get α = 12

3. What is the value of x if, (begin{vmatrix}x & 3 & 6 \3 & 6 & x \6 & x & 3 end {vmatrix}) = (begin{vmatrix}2 & x & 7 \x & 7 & 2 \7 & 2 & x end {vmatrix}) = (begin{vmatrix}4 & 5 & x \5 & x & 4 \x & 4 & 6 end {vmatrix})?
a) 9
b) -9
c) 0
d) Can’t be predicted
Answer: b
Clarification: Given that,
(begin{vmatrix}x & 3 & 6 \3 & 6 & x \6 & x & 3 end {vmatrix}) = (begin{vmatrix}2 & x & 7 \x & 7 & 2\7 & 2 & x end {vmatrix}) = (begin{vmatrix}4 & 5 & x \ 5 & x & 4 \x & 4 & 6 end {vmatrix})
So, by circular determinant property,
Sum of the elements of a row = 0
So, x + 3 + 6 = 2 + x + 7 = 4 + 5 + x = 0
=> x = -9

4. Which one of the following is correct if a, b and c are the sides of a triangle ABC and (begin{vmatrix}a^2 & b^2 & c^2 \(a + 1)^2 & (b + 1)^2 & (c + 1)^2 \ (a – 1)^2 & (b – 1)^2 & (c – 1)^2 end {vmatrix}) ?
a) ABC is an equilateral triangle
b) ABC is an isosceles triangle
c) ABC is a right angled triangle
d) ABC is a scalene triangle
Answer: b
Clarification: When a = b or b = c or c = a the determinant reduces to 0
It is not necessary that a = b = c for determinant to be 0
Therefore, the triangle is isosceles.

5. What is the value of k if (begin{vmatrix}y + z & x & x \y & z + x & y \z & z & x + yend {vmatrix}) ?
a) 4
b) -4
c) 1
d) 0
Answer: a
Clarification: Put the value of x, y, z = 1
Thus, putting the value of x = 1, y = 1 and z = 1 on both sides, we get
(begin{vmatrix}2 & 1 & 1 \1 & 2 & 1 \1 & 1 & 2 end {vmatrix}) = k
So, solving the determinant we get k = 4.

6. What will be the value of the given determinant (begin{vmatrix}109 & 102 & 95 \6 & 13 & 20 \1 & -6 & 13 end {vmatrix})?
a) constant other than 0
b) 0
c) 100
d) -1997
Answer: b
Clarification: In the given determinant form the elements are in A.P
Also the common difference of this A.P is 7
Thus the value of the given determinant = 0

7. What is the value of _{r = 1}Σⁿ f(x) if f(r) = (begin{vmatrix}2r & x & n(n + 1) \(6r^2 – 1) & y & n^2 (2n + 3) \(4r^3 – 2nr) & z & n^3(n + 1) end {vmatrix}) where n € N?
a) 1
b) -1
c) 0
d) 2
Answer: c
Clarification: The given determinant is f(r) = (begin{vmatrix}2r & x & n(n + 1) \(6r^2 – 1) & y & n^2(2n + 3) \(4r^3 – 2nr) & z & n^3(n + 1) end {vmatrix})
Now, _{r = 1}Σⁿ (2r) = 2[(n(n + 1))/2] ……….(1)
= n² + n
_{r = 1}Σⁿ(6r² – 1) = 6[((n + 1)(2n + 1))/6] – n ……….(2)
= n(2n² + 2n + n + 1) – n
= 2n³ + 2n² + n² + n – n
= 2n³ + 3n²
= _{r = 1}Σⁿ(4r³ – 2nr) = n³ (n + 1) ……….(3)
From (1), (2) and (3) we get
_{r = 1}Σⁿ f(x) = 0

8. If f(x) = (begin{vmatrix}sec⁡ x & cos ⁡x & sec^2⁡ x + cot ⁡x, cosec x \cos^2 ⁡x & cos^2 ⁡x & cosec^2 x \1 & cos^2 ⁡x & cos^2 ⁡x end {vmatrix}) then what is the value of ₀∫^π/2 f(x) dx = (π/4 + 8/15)?
a) (π/4 + 8/15)
b) (π/4 – 8/15)
c) (π/4 + 8/15)
d) (-π/4 + 8/15)
Answer: c
Clarification: (dy/dx) = (dx/dy)^-1
So, d²y/dx² = -(dx/dy)^-2 d/dx(dx/dy)
= -(dy/dx)²(d²x/dy²)(dy/dx)
= d²y/dx² + (dy/dx)³ d²y/dx² = 0

9. What will be the value of (begin{vmatrix}0 & i – 100 & i – 500 \100 – i & 0 & 1000 – i \500 – i & i – 1000 & 0 end {vmatrix})?
a) 100
b) 500
c) 1000
d) 0
Answer: d
Clarification: The above matrix is a skew symmetric matrix and its order is odd
And we know that for any skew symmetric matrix with odd order has determinant = 0
Therefore, the value of the given determinant = 0

10. If f(x) = (begin{vmatrix}x^n & x^{n+2} & x^{2n} \1 & x^p & p \x^{n+5} & x^{p+6} & x^{2n+5} end {vmatrix}) = 0,then what will be the value of p?
a) xⁿ
b) (n + 1)
c) Either xⁿ or (n + 1)
d) Both xⁿ and (n + 1)
Answer: d
Clarification: Here, C₁ and C₃ becomes equal when we put p = xⁿ
And R₁ and R₃ becomes equal when we put p = n + 1
As both of the conditions are satisfied so d is the correct one.

Mathematics Multiple Choice Questions on “Derivatives Application – Increasing and Decreasing Functions”.

1. What is the nature of function f(x) = 7x-4 on R?
a) Increasing
b) Decreasing
c) Strictly Increasing
d) Increasing and Decreasing
Answer: c
Clarification: Let x1 and x2 be any two numbers in R.
Then x1 7×1 < 7×2.
=> 7×1 – 4 < 7×2 – 4.
As f(x1) < f(x2), thus the function f is strictly increasing on R.

2. What is the nature of function f(x) = x³ – 3x² + 4x on R?
a) Increasing
b) Decreasing
c) Constant
d) Increasing and Decreasing
Answer: a
Clarification: f(x) = x³ – 3x² + 4x
f’(x) = 3x² – 6x + 4.
f’(x) = 3(x² – 2x + 1) + 1.
=> 3(x-1)² + 1>0, in every interval of R. Therefore the function f is increasing on R.

3. Find the interval in which function f(x) = x² – 4x + 5 is increasing.
a) (2, ∞)
b) (-∞, 2)
c) (3, ∞)
d) (-∞, ∞)
Answer: a
Clarification: f(x) = x² – 4x + 5.
f’(x) = 2x – 4. Therefore f’(x) = 0 gives x = 2.
Now this point x=2 divides the line into two disjoint intervals and the interval namely (2, ∞) is increasing on f(x).

4. Find the interval in which function f(x) = x² – 4x + 5 is decreasing.
a) (2, ∞)
b) (-∞, 2)
c) (3, ∞)
d) (-∞, ∞)
Answer: b
Clarification: f(x) = x² – 4x + 5.
f’(x) = 2x – 4. Therefore f’(x) = 0 gives x = 2.
Now this point x=2 divides the line into two disjoint intervals and the interval namely (-∞, 2) is decreasing on f(x).

5. Find the interval in which function f(x) = sinx+cosx, 0 ≤ x ≤ 2π is decreasing.
a) (π/4, 5π/4)
b) (-π/4, 5π/4)
c) (π/4, -5π/4)
d) (-π/4, π/4)
Answer: a
Clarification: f(x) = sinx+cosx.
f’(x) = cosx – sinx. Now f’(x) = 0 gives sinx = cosx which gives that x=π/4, 5π/4 as 0 ≤ x ≤ 2π.
Therefore on checking the values we get f is decreasing in (π/4, 5π/4).

6. Find the interval in which function f(x) = sinx+cosx is increasing.
a) (5π/4, 2π)
b) [0, π/4) and (5π/4, 2π]
c) (π/4, -5π/4)
d) (-π/4, π/4)
Answer: b
Clarification: f(x) = sinx+cosx.
f’(x) = cosx – sinx. Now f’(x) = 0 gives sinx = cosx which gives that x= π/4, 5π/4 as 0 ≤ x ≤ 2π.
The points x = π/4 and x = 5π/4 divide the interval [0, 2π] into three disjoint intervals which are
[0, π/4), (π/4, 5π/4) and (5π/4, 2π].
Therefore on checking the values we get f is increasing in [0, π/4) and (5π/4, 2π].

7. Is the function f(x) = 3x+10 is increasing on R?
a) True
b) False
Answer: a
Clarification: f(x) = 3x+10.
f’(x) = 3, which shows 3 > 0 for all x ∈ R.
Thus function f(x) is increasing.

8. Find the intervals in which f(x) = 2x² – 3x is increasing.
a) (-1/4, ∞)
b) (-3/4, ∞)
c) (1/4, ∞)
d) (3/4, ∞)
Answer: d
Clarification: f(x) = 2x²-3x.
f’(x) = 4x – 3.
As we know f’(x) = 0, x=-3/4. This shows that function f is increasing in interval (-3/4, ∞) for all x ∈ R.

9. Find the intervals in which f(x) = x² + 2x – 5 is strictly increasing.
a) x>1
b) x<-1
c) x>-1
d) x>2
Answer: c
Clarification: f(x) = x²+2x -5.
f’(x) = 4x – 3.
As we know f’(x) = 0, x=-3/4, which shows that function f is increasing in interval (-3/4, ∞) for all x ∈ R.

10. Nature of the function f(x) = e^2x is _______
a) increasing
b) decreasing
c) constant
d) increasing and decreasing
Answer: a
Clarification: f(x) = e^2x.
f’(x) = 2e^2x.
As we know 2e^2x > 0, so it always has a value greater than zero.
Which shows that function f is increasing for all x ∈ R.