Functions and its properties MCQ Quiz - Objective Question with Answer for Functions and its properties - Download Free PDF

Given, f(x + y) = f(x) + f(y), 

⇒ f(x) = kx

and f(1) = 10 

⇒ f(x) = 10x

So,  \(\displaystyle\sum_{r=1}^{n}\)(f(r))2 =  \(\displaystyle\sum_{r=1}^{n}\)(10r)2 =  \(\displaystyle\sum_{r=1}^{n}\)100(r)2 

⇒  \(\displaystyle\sum_{r=1}^{n}\)(f(r))2 = 100  \(\displaystyle\sum_{r=1}^{n}\)(r)2 

 ⇒  \(\displaystyle\sum_{r=1}^{n}\)(f(r))2 = 100(1² + 2² + ....+ n²)

⇒  \(\displaystyle\sum_{r=1}^{n}\)(f(r))2 = 100(\(n(n+1)(2n+1)\over 6\))

⇒  \(\displaystyle\sum_{r=1}^{n}\)(f(r))2 =  \(​\frac{50}{3}\)n(n + 1)(2n + 1)

∴ The correct answer is option (3).

Given:

\(2f(x) + f(\frac{1}{x}) = logx\)

Concept:

In such questions, we generally try to reduce the expression to a single function f(x) by elimination method.

Formula:

log(ex) = x

log(1/ex) = log(e-x) = -x

Calculation:

\(2f(x) + f(\frac{1}{x}) = logx\) - (i)

Replacing x by 1/x in (i),

⇒ \(2f(\frac{1}{x}) + f({x}) = log(\frac{1}{x})\) - (ii)

Now, 2 × (i) - (ii) :

⇒ 3f(x) = 2log(x) - log(1/x)

⇒ f(x) = \(\frac{1}{3}\)(2log(x) - log(1/x)) - (iii)

Replacing x by ex in (iii) -

⇒ f(ex) =  \(\frac{1}{3}\)(2log(ex) - log(1/ex)) 

⇒ f(ex) =  \(\frac{1}{3}\)(2x + x)

⇒ f(ex) =  x

Given:

f(x) = bx2 + cx + d

f(x + 1) - f(x) = 8x + 3

Calculation:

f(x) = bx2 + cx + d    ___(i)

⇒ f(x + 1) = b(x + 1)2 + c(x + 1) + d      __(ii)

Putting value of (i) and (ii) in the given equation f(x + 1) - f(x) = 8x + 3, we get

⇒ b(x + 1)2 + c(x + 1) + d - (bx2 + cx + d) = 8x + 3

⇒ b(x2 + 2x + 1) + c - bx2 = 8x + 3

⇒ 2bx + b + c = 8x + 3

Equating the coefficients of x and constant terms, we get

⇒ 2b = 8 and b + c = 3

⇒ b = 4 and c = -1

∴ The correct answer is option (2).

Formula used:

• (a - b)² = a² - 2ab + b²
• (a + b)2 = a2 + 2ab + b2
• (a + b)(a - b) = a² - b²

Calculation:

Given that,

\(f(x)=\frac{1+x}{1-x}\)    ----(1)

We have to find the value of

F(x) = \(\frac{f(x).f(x^2)}{1+[f(x)]^2}\) -----(2)

⇒  F(x) = \(\frac{\big(\frac{1+x}{1-x}\big).\big(\frac{1+x^2}{1-x^2}\big)}{1+\big({\frac{1+x}{1-x}}\big)^2}\)

⇒  F(x) =  \(\frac{\frac{(1+x)(1+x^2)}{(1-x)(1-x^2)}}{{\frac{(1-x)^2+(1+x)^2}{(1-x)^2}}}\)

By using above formula, 

⇒  F(x) =  \(\frac{\frac{(1+x)(1+x^2)}{(1-x)(1-x)(1+x)}}{{\frac{1+x^2-2x+1+x^2+2x}{(1-x)^2}}}\)

⇒  F(x) =  \(\frac{\frac{(1+x)(1+x^2)}{(1-x)^2(1+x)}}{{\frac{2(1+x^2)}{(1-x)^2}}}\)

⇒  F(x) =  \(\frac{1}{2}\)

∴ The value of the given function F(x) = \(\frac{1}{2}\).

Concept:

• Reflexive:

Let R be a relation on a non-empty set A, if every element of A is related to itself then R is said to be a reflexive relation.

Thus, R is reflexive ⇔ (a, a) ∈ R, ∀ a ∈ A.

• Symmetric:

Let R be a relation on a non-empty set A, then the relation R is said to be symmetric relation ⇔ (a, b) ∈ R ⇒ (b, a) ∀ a, b ∈ A.

• Transitive:

Let R be a relation on a non-empty set A, then the relation R is said to be transitive relation ⇔ (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, ∀ a, b, c ∈ A.

• Equivalence:

Let R be a relation on a non-empty set A, then the relation R is said to be equivalence relation if R is reflexive, symmetric and transitive.

Calculation:

Given: R is a relation on the set of integers Z such that R = {(a, b) : 2 divides a - b where a, b ∈ Z}

Reflexive:

As we can see that a - a = 0 ∀ a ∈ Z ⇒ (a, a) ∈ R ∀ a ∈ Z

So, the given relation R is reflexive.

Symmetric:

Let (a, b) ∈ R ⇒ 2 divides a - b i.e a - b = 2k where k is some integer

So, if a - b = 2k ⇒ b - a = 2l where l = - k ∈ Z

So, 2 also divides b - a ⇒ (b, a) ∈ R

So, the given relation R is  also symmetric.

Transitive:

Let (a, b) and (b, c) ∈ R i.e 2 divides both a - b and b - c also.

⇒ a - b = 2k where k is some integer and similarly b - c = 2l where l is some integer.

⇒ (a - b) + (b - c) = 2k + 2l

⇒ a - c = 2m where m = l + k ∈ Z

So, 2 also divides a - c ⇒ (a, c) ∈ R

So, the given relation R is also transitive.

As we know that if a relation R is reflexive, symmetric and transitive then R is said to be an equivalence relation

Hence, option 4 is the correct answer.

Calculation:

Given, \(\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{x\cos x - \log (1 + x)}}{{{x^2}}}} \right]\)

which is in the form \(\left(\dfrac{0}{0}\right)\)

using L-Hospital Rule:

on differentiating N^r and ω^r

\(= lim_{x\to 0}\left[\dfrac{cosx -x sinx - \frac{1}{1+x}}{2x}\right]\)

again differentiating w.r.to x

\(= lim_{x\to 0}\left[\dfrac{-sinx -( sinx + xcos x)+\frac{1}{(1+x)^2}}{2}\right]\)

put x → 0

\(= \dfrac{-sin 0^\circ - ( sin 0^\circ + 0. cos 0^ \circ)+ \frac{1}{1+0}}{2}\)

= 1/2

Explanation:

We have f(x) = \(\frac{1}{x}\) ∀ x ∈ R.

For x=0, f(x) is not defined.

Hence, the function f(x) is not defined function.

Given:

f(a) = 2, f'(a) = 1, g(a) = -1, g'(a) = 2

Concept:

L'Hospital's Rule: It tells us that if we have an indeterminate form '0/0' or '∞/∞' all we need to do is differentiate the numerator and the denominator separately and then take the limit.

Calculation: 

\(Let\ \ y\ = \ \mathop {\lim }\limits_{x \to a} .\frac{{g(x)f(a) - g(a)f(x)}}{{x - a}}\)

when x = a, it gives 0/0 form. Hence, we can solve the limit by simply using L'Hospital's rule. 

Hence, differentiate the numerator and the denominator separately with respect to x, we will get

\(⇒ \ y\ = \ \mathop {\lim }\limits_{x \to a} .\frac{{g'(x)f(a) - g(a)f'(x)}}{{ 1}}\)

Put x = a

 \(⇒ \ y\ = \ \mathop {{g'(a)f(a) - g(a)f'(a)}}\)

But according to question,

f(a) = 2, f'(a) = 1, g(a) = -1, g'(a) = 2

⇒ y = (2)(2) - (-1)(1) 

⇒ y = 5

Hence, the value of above limit is 5.

Concept:

• Reflexive:

Let R be a relation on a non-empty set A, if every element of A is related to itself then R is said to be a reflexive relation.

Thus, R is reflexive ⇔ (a, a) ∈ R, ∀ a ∈ A.

• Symmetric:

Let R be a relation on a non-empty set A, then the relation R is said to be symmetric relation ⇔ (a, b) ∈ R ⇒ (b, a) ∀ a, b ∈ A.

• Transitive:

Let R be a relation on a non-empty set A, then the relation R is said to be transitive relation ⇔ (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, ∀ a, b, c ∈ A.

• Equivalence:

Let R be a relation on a non-empty set A, then the relation R is said to be equivalence relation if R is reflexive, symmetric and transitive.

Calculation:

Given: R is a relation on the set of integers Z such that R = {(a, b) : 2 divides a - b where a, b ∈ Z}

Reflexive:

As we can see that a - a = 0 ∀ a ∈ Z ⇒ (a, a) ∈ R ∀ a ∈ Z

So, the given relation R is reflexive.

Symmetric:

Let (a, b) ∈ R ⇒ 2 divides a - b i.e a - b = 2k where k is some integer

So, if a - b = 2k ⇒ b - a = 2l where l = - k ∈ Z

So, 2 also divides b - a ⇒ (b, a) ∈ R

So, the given relation R is  also symmetric.

Transitive:

Let (a, b) and (b, c) ∈ R i.e 2 divides both a - b and b - c also.

⇒ a - b = 2k where k is some integer and similarly b - c = 2l where l is some integer.

⇒ (a - b) + (b - c) = 2k + 2l

⇒ a - c = 2m where m = l + k ∈ Z

So, 2 also divides a - c ⇒ (a, c) ∈ R

So, the given relation R is also transitive.

As we know that if a relation R is reflexive, symmetric and transitive then R is said to be an equivalence relation

Hence, option 4 is the correct answer.

Given, f(x + y) = f(x) + f(y), 

⇒ f(x) = kx

and f(1) = 10 

⇒ f(x) = 10x

So,  \(\displaystyle\sum_{r=1}^{n}\)(f(r))2 =  \(\displaystyle\sum_{r=1}^{n}\)(10r)2 =  \(\displaystyle\sum_{r=1}^{n}\)100(r)2 

⇒  \(\displaystyle\sum_{r=1}^{n}\)(f(r))2 = 100  \(\displaystyle\sum_{r=1}^{n}\)(r)2 

 ⇒  \(\displaystyle\sum_{r=1}^{n}\)(f(r))2 = 100(1² + 2² + ....+ n²)

⇒  \(\displaystyle\sum_{r=1}^{n}\)(f(r))2 = 100(\(n(n+1)(2n+1)\over 6\))

⇒  \(\displaystyle\sum_{r=1}^{n}\)(f(r))2 =  \(​\frac{50}{3}\)n(n + 1)(2n + 1)

∴ The correct answer is option (3).

Calculation:

ϕ(x) = ax          -----(1)

⇒ [ϕ(p)]3 = (a^p)³ 

⇒ [ϕ(p)]3 = a^3p 

⇒ [ϕ(p)]3  = ϕ(3p)    [Using the equation (1)]

∴  Required value is  ϕ(3p). 

Given:

f(x) = bx2 + cx + d

f(x + 1) - f(x) = 8x + 3

Calculation:

f(x) = bx2 + cx + d    ___(i)

⇒ f(x + 1) = b(x + 1)2 + c(x + 1) + d      __(ii)

Putting value of (i) and (ii) in the given equation f(x + 1) - f(x) = 8x + 3, we get

⇒ b(x + 1)2 + c(x + 1) + d - (bx2 + cx + d) = 8x + 3

⇒ b(x2 + 2x + 1) + c - bx2 = 8x + 3

⇒ 2bx + b + c = 8x + 3

Equating the coefficients of x and constant terms, we get

⇒ 2b = 8 and b + c = 3

⇒ b = 4 and c = -1

∴ The correct answer is option (2).

Calculation:

Given, \(\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{x\cos x - \log (1 + x)}}{{{x^2}}}} \right]\)

which is in the form \(\left(\dfrac{0}{0}\right)\)

using L-Hospital Rule:

on differentiating N^r and ω^r

\(= lim_{x\to 0}\left[\dfrac{cosx -x sinx - \frac{1}{1+x}}{2x}\right]\)

again differentiating w.r.to x

\(= lim_{x\to 0}\left[\dfrac{-sinx -( sinx + xcos x)+\frac{1}{(1+x)^2}}{2}\right]\)

put x → 0

\(= \dfrac{-sin 0^\circ - ( sin 0^\circ + 0. cos 0^ \circ)+ \frac{1}{1+0}}{2}\)

= 1/2

Explanation:

We have f(x) = \(\frac{1}{x}\) ∀ x ∈ R.

For x=0, f(x) is not defined.

Hence, the function f(x) is not defined function.

Given:

\(2f(x) + f(\frac{1}{x}) = logx\)

Concept:

In such questions, we generally try to reduce the expression to a single function f(x) by elimination method.

Formula:

log(ex) = x

log(1/ex) = log(e-x) = -x

Calculation:

\(2f(x) + f(\frac{1}{x}) = logx\) - (i)

Replacing x by 1/x in (i),

⇒ \(2f(\frac{1}{x}) + f({x}) = log(\frac{1}{x})\) - (ii)

Now, 2 × (i) - (ii) :

⇒ 3f(x) = 2log(x) - log(1/x)

⇒ f(x) = \(\frac{1}{3}\)(2log(x) - log(1/x)) - (iii)

Replacing x by ex in (iii) -

⇒ f(ex) =  \(\frac{1}{3}\)(2log(ex) - log(1/ex)) 

⇒ f(ex) =  \(\frac{1}{3}\)(2x + x)

⇒ f(ex) =  x