Integral Calculus for JEE Main & Advanced: Practice Problems

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25 Multiple Choice Questions (MCQs) on Integral Calculus for students of class XI and XII preparing for board examinations or JEE Mains, IIT Advanced WBJEE or any other competitive entrance examination.

👨‍🏫 Author: Vinod Singh

📞 WA: +91-9038126497

Prime Maths

Advanced Integral Calculus - Definite and Indefinite Integration

Test your understanding of core concepts.Master Integral Calculus for JEE Main & Advanced. Practice hand-picked problems with step-by-step solutions, advanced shortcuts, and integration techniques.

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Question 1

Let \(f(x)=\frac{x}{(1+x^n)^{1/n}}\) for \(n \geq 2\) and \(g(x)=f \circ f \circ \dots \circ f\) (\(n\) times), then \(\int x^{n-2} g(x) dx \) is equal to

A\( \frac{1}{n(n+1)} (1+nx^n)^{1-\frac{1}{n}}+c\) B\( \frac{1}{n(n-1)} (1+nx^n)^{1-\frac{1}{n}}+c\) C\( \frac{1}{n(n-1)} (1+nx^n)^{1+\frac{1}{n}}+c \) D\( \frac{1}{n(n-1)} (1-nx^n)^{1-\frac{1}{n}}+c\)

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Question 2

\(\int_{\frac{1}{2026}}^{2026} \frac{tan^{-1} x}{x} dx \quad =\)

A \(\frac{\pi}{4} \ln 2026\) B \(\sqrt{2}\frac{\pi}{8} \ln 2026\) C \(\frac{\pi}{2} \ln 2026\) D \(\frac{\pi}{3} \ln 2026\)

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Question 3

\(\int \frac{dx}{\sqrt[3]{x}+\sqrt[5]{x}} \quad =\)

A \( = 15 \left( \frac{x^{2/3}}{10} - \frac{x^{8/15}}{8} + \frac{x^{2/5}}{6} - \frac{x^{4/15}}{4} + \frac{x^{2/15}}{2} - \frac{1}{2} \ln |x^{2/15} + 1| \right) + c \) B \( = 15 \left( \frac{x^{2/3}}{10} + \frac{x^{8/15}}{8} + \frac{x^{2/5}}{6} + \frac{x^{4/15}}{4} + \frac{x^{2/15}}{2} - \frac{1}{2} \ln |x^{2/15} + 1| \right) + c \) C \( = 15 \left( -\frac{x^{2/3}}{10} - \frac{x^{8/15}}{8} - \frac{x^{2/5}}{6} - \frac{x^{4/15}}{4} - \frac{x^{2/15}}{2} - \frac{1}{2} \ln |x^{2/15} + 1| \right) + c \) D \( = 17 \left( \frac{x^{2/3}}{10} - \frac{x^{8/15}}{8} + \frac{x^{2/5}}{6} - \frac{x^{4/15}}{4} + \frac{x^{2/15}}{2} - \frac{1}{2} \ln |x^{2/15} + 1| \right) + c \)

Put \(x=z^{15}\)

Question 4

\(\int (x^6+x^3) \sqrt[3]{x^3+2} \quad dx \quad =\)

A\(\frac{1}{2}(x^6+2x^3)^{\frac{3}{4}} +c\) B\(\frac{1}{8}(x^6+2x^3)^{\frac{5}{3}} +c\) C\(\frac{1}{8}(x^6+2x^3)^{\frac{4}{3}} +c\) D\(\frac{1}{8}(x^6+x^3)^{\frac{4}{3}} +c\)

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Question 5

\( \int_{\pi/2}^{5\pi/2} \frac{e^{tan^{-1}(\sin x)}}{e^{tan^{-1}(\sin x)}+e^{tan^{-1}(\cos x)}} dx \quad = \)

A\( 2\pi \) B\( \frac{\pi}{8}\) C\( \frac{\pi}{2} \) D\( \pi \)

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Question 6

\(\int \frac{dx}{\tan x + \sec x + \cot x + \csc x} dx \quad = \)

A\(\frac{1}{2}(\cos x - \sin x -x)+c\) B\(\frac{1}{2}(-\cos x + \sin x -x)+c\) C\(\frac{1}{2}(-\cos x + \sin x + x)+c\) D\(\frac{1}{2}(-\cos x - \sin x -x)+c\)

See that \(\frac{1}{\tan x + \sec x + \cot x + \csc x} = \frac{\sin x \cos x}{1+\sin x +\cos x}=\frac{1}{2}\frac{(\sin x +\cos x)^2-1^2}{1+\sin x +\cos x}\)

Question 7

If \( f \) is an even function and \(I= \int_{0}^{\pi/2}f(\cos 2x) \cos x \quad dx \), then

A \( I= \sqrt{2}\int_{0}^{\pi/4}f(\cos 2x) \cos x \quad dx \) B \( I= \sqrt{2}\int_{0}^{\pi/4}f(\cos 2x) \sin x \quad dx \) C \( I= \sqrt{2}\int_{0}^{\pi/4}f(\sin 2x) \cos x \quad dx \) D \( I= \sqrt{2}\int_{0}^{\pi/4}f(\sin 2x) \sin x \quad dx \)

Use King's Rule

Question 8

Let \( f \) be a polynomial function such that \( f(x^2+1) = x^4+5x^2+2\), for all \( x \in \mathbb{R}.\) Then \(\int_{0}^{3} f(x) dx \) is equal to

A \( \frac{41}{3} \) B \( \frac{33}{2} \) C \( \frac{27}{2} \) D \( \frac{5}{3} \)

See that \(f(x^2+1)=(x^2+1)^2+3(x^2+1)-2\)

Question 9

Evaluate \(\int_{-\pi/3}^{\pi/3} \frac{\pi+4x^3}{2-cos\big(|x|+\frac{\pi}{3}\big)} dx\)

A\(\frac{4\pi}{\sqrt{3}}tan^{-1}(\frac{1}{3})\) B\(\frac{4\pi}{\sqrt{2}}tan^{-1}(\frac{1}{2})\) C\(\frac{4\pi}{\sqrt{3}}tan^{-1}(\frac{1}{2})\) D\(\frac{4\pi}{\sqrt{2}}tan^{-1}(\frac{1}{3})\)

Express \(\frac{\pi+4x^3}{2-cos\big(|x|+\frac{\pi}{3}\big)}\) as a sum of an odd and an even function, then split

Question 10

Lrt \(f: (0,\infty) \rightarrow \mathbb{R}\) and \(F(x)=\int_{0}^{x} f(t) dt.\) If \(F(x^2)=x^2(1+x),\) then \(f(4)\) equals

A\(\frac{5}{4}\) B\(7\) C\(4\) D\(2\)

See that \(F'(x)=f(x)\) by fundamental theorem of integral calculus and \(f(x^2)=F'(x^2)= \frac{3x+2}{2}\)

Question 11

Evaluate \( \int sin^{-1} \big(\frac{2x+2}{\sqrt{4x^2+8x+13}} \big) dx.\)

A\((x+1)tan^{-1}\big(\frac{2x+2}{3}\big)+\frac{3}{4}ln|4x^2+8x+13|+c\) B\((x+1)tan^{-1}\big(\frac{2x+2}{3}\big)-\frac{1}{4}ln|4x^2+8x+13|+c\) C\((x+1)tan^{-1}\big(\frac{2x+2}{3}\big)-\frac{3}{2}ln|4x^2+8x+13|+c\) D\((x+1)tan^{-1}\big(\frac{2x+2}{3}\big)-\frac{3}{4}ln|4x^2+8x+13|+c\)

\(4x^2+8x+13=(2x+2)^2+3^2\) now put \( 2x+2 = t\) and then subsutitute \( t=3 \tan \theta \)

Question 12

Let \(T > 0\) be a fixed real number. Suppose \(f\) is a continuous function such that for all \(x \in \mathbb{R}, f(x+T)=f(x).\) If \(I=\int_{0}^{T} f(x) \\dx\) then the value of \(\int_{3}^{3+3T} f(2x) \\dx\) is

A\(3I\) B\(2I \) C\(\frac{3}{2}I\) D\(6I\)

\(\int_{3}^{3+3T} f(2x) \\dx\) = \(\frac{1}{2}\int_{6}^{6+6T} f(t) \\dt\) on substituting \(2x=t\) now split the integral form 6 to T, T to 2T, ...., 5T to 6T and 6T to 6+6T and think!

Question 13

The integral \(\int_{-1/a}^{1/a} \big( [x]+\ln (\frac{1+x}{1-x})\big) dx\) where \(a>1\) equals

A\(0\) B\(\frac{-1}{a}\) C\(1\) D\(a \ln \frac{1}{a}\)

Find out the odd function!

Question 14

For any natural number \(m\), evaluate \(\int (x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^m+6)^{1/m} dx, x>0\)

A \( \frac{1}{6(m+1)} (2x^{3m}+3x^{2m}+6x^m)^{\frac{m-1}{m}}+c\) B \( \frac{1}{6(m-1)} (2x^{3m}+3x^{2m}+6x^m)^{\frac{m+1}{m}}+c\) C \( \frac{1}{(m+1)} (2x^{3m}+3x^{2m}+6x^m)^{\frac{m+1}{m}}+c\) D \( \frac{1}{6(m+1)} (2x^{3m}+3x^{2m}+6x^m)^{\frac{m+1}{m}}+c\)

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Question 15

Let \(f(x)= \frac{e^x}{1+e^x} \quad I_1 = \int_{f(-a)}^{f(a)}xg(x(1-x)) dx \) and \(I_2 = \int_{f(-a)}^{f(a)} g(x(1-x)) dx, \) then the value of \(\frac{I_2}{I_1}\) is

A\(3\) B\(4\) C\(2\) D\(-1\)

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Question 16

Evaluate \(\int_{0}^{2\pi} |1+2\sin x| dx \)

A\(\frac{2\pi}{3} (4+\sqrt{3})\) B\(\frac{4\pi}{3} (2+\sqrt{3})\) C\(\frac{\pi}{3} (4+\sqrt{3})\) D\(\frac{\pi}{3} (2+\sqrt{3})\)

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Question 17

The integral \(\int \frac{2x^{12}+5x^9}{(x^5+x^3+1)^3} dx \)

A\(\frac{-x^5}{(x^5+x^3+1)^{2}}+c\) B\(\frac{x^{10}}{2(x^5+x^3+1)^{2}}+c\) C\(\frac{x^5}{2(x^5+x^3+1)^{2}}+c\) D\(\frac{-x^{10}}{2(x^5+x^3+1)^{2}}+c\)

substitute \(\x=\frac{1}{t})

Question 18

If \(f\) and \(g\) be continuous functions on \([0,a]\) such that \(f(x)=f(a-x)\) and \(g(x)+g(a-x)=4,\) then \(\int_{0}^{a} f(x)g(x) dx \) is equal to

A\(4\int_{0}^{a} f(x)dx \) B\(4\int_{0}^{a} f(x)dx \) C\(2\int_{0}^{a} f(x)dx \) D\(-3\int_{0}^{a} f(x)dx \)

Apply King's Theorem

Question 19

The value of \(\int_{-\pi/2}^{\pi/2}\frac{x^2 \cos x}{1+e^x}dx\) is equal to

A\(\frac{\pi^2}{4}-2\) B\(\frac{\pi^2}{4}+2\) C\(\pi^2-e^{-\pi/2}\) D\(\pi^2-e^{-\pi/2}\)

Use \(\int_{a}^{b} f(x) dx = \) \(\int_{a}^{b} f(a+b-x) dx\) then add the resulting integral with the given one and then apply integration by parts

Question 20

The value of \(\int_{\sqrt{\log 2}}^{\sqrt{\log 3} }\frac{x \sin x^2}{\sin x^2+\sin (\log 6 -x^2)}dx\) is equal to

A\(\frac{1}{4} \log \frac{3}{2}\) B\(\frac{1}{2} \log \frac{3}{2}\) C\(\log \frac{3}{2}\) D\(\frac{1}{6} \log \frac{3}{2}\)

put \(\log 6 - x^2 = t\) and then apply \(\int_{a}^{b} f(x) dx = \int_{a}^{0b} f(a+b-x) dx\)

Question 21

The value of \(\int_{-2}^{0} (x^3+3x^2+3x+3+(x+1) \cos (x+1)) dx\) is equal to

A\( 0 \) B\( 3 \) C\( 4 \) D\( 1 \)

Use \((x+1)^3\) to reduce the expression then substitute \(x+1=z\)

Question 22

Let \( f(x) = 7\tan^8 x + 7 \tan^6 x -3 \tan^4 x -3 \tan^2 x \), for all \(x \in \big(\frac{-\pi}{2},\frac{\pi}{2} \big)\). Then, the correct expression is

A \(\int_{0}^{\pi/4} x f(x) dx = \frac{1}{12}\) B \(\int_{0}^{\pi/4} x f(x) dx = \frac{1}{6}\) C \(\int_{0}^{\pi/4} x f(x) dx = \frac{5}{12}\) D \(\int_{0}^{\pi/4} x f(x) dx = \frac{3}{4}\)

Easy!

Question 23

The value of \(\int_{0}^{1/2} \frac{1+\sqrt{3}}{((x+1)^2(1-x)^6)^{1/4}} dx\) is equal to

A\( 2 \) B\( \frac{1}{\sqrt{3}} \) C\( 3 \) D\( 4 \)

After simplifying the integrand substitute \(x= \cos \theta\)

Question 24

The value of \(\,(5050)\frac{\int_{0}^{1} (1 - x^{50})^{100} \, dx}{\int_{0}^{1} (1 - x^{50})^{101} \, dx}\) is

A\( 5050 \) B\( 5051 \) C\( 1 \) D\( 5500 \)

\(If I_n \int_{0}^{1} (1-x^{50})^n dx\) then by use of integration by parts it is easy to see thet \((1+50n)I_n=50nI_{n-1}\). Take \((1-x^{50})^n\) as the first function

Question 25

The value of \(\int_{-2}^{2} |1-x^2| dx\) is equal to

A\(2\) B\(3\) C\(\frac{3}{4}\) D\(4\)

\(\int_{-2}^{2} |1-x^2| dx=\)\(\int_{-2}^{-1} -(1-x^2) dx\) \(+\int_{-1}^{1} (1-x^2) dx\)\(+\int_{1}^{2} -(1-x^2) dx\)

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Relations and Functions Advanced Problems for JEE Main and WBJEE

Prime Maths

25 Multiple Choice Questions (MCQs) on Relations and Functions for students of class XI and XII preparing for board examinations or JEE Mains, IIT Advanced WBJEE or any other competitive entrance examination.

👨‍🏫 Author: Singh

📞 WA: +91-9038126497

Prime Maths

Advanced Algebra - Relations and Functions

Test your understanding of core concepts.

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Question 1

The number of reflexive relations on a set \(A\) of \(n\) elements is equal to

A\( 2^{n^2}\) B\( n^2\) C\( 2^{n(n-1)} \) D\( n^2-n\)

For the relation to be reflexive, \((a_i,a_i) \in A \times A \quad \forall a_i \in A\) So the from the remaining \(n^2 - n\) elements, any number of elements can be choosen to form a subset of \(A \times A\)

Question 2

Let \(X\) be a non-void set. If \(\rho_1\) and \(\rho_2\) be the transitive relations on \(X\), then [ \( \circ\) denotes composition ]

A \(\rho_1 \circ \rho_2\) is transitive relation B \(\rho_1 \circ \rho_2\) is not transitive relation C \(\rho_1 \circ \rho_2\) is equivalence relation D \(\rho_1 \circ \rho_2\) is not any relation on \(X\)

Let's define a set \(X = \{1, 2, 3, 4, 5\}\). Now, let's create two relations on \(X\), \(\rho_1\) and \(\rho_2\), both of which are transitive:Let \(\rho_1 = \{(1, 2), (3, 4)\}\) Let \(\rho_2 = \{(2, 3), (4, 5)\}\) (Note: These are trivially transitive). Now, let's find the composition \(\rho_1 \circ \rho_2\). By definition of composition, \((x, z) \in \rho_1 \circ \rho_2\) if there is some intermediate element \(y\) such that \((x, y) \in \rho_1\) and \((y, z) \in \rho_2\). So, the resulting composition is:\(\rho_1 \circ \rho_2 = \{(1, 3), (3, 5)\}\) which is clearly not transitive.

Question 3

If one root of \(x^2 + px - q^2 = 0\), \(p\) and \(q\) are real, be less than \(2\) and other be greater than \(2\), then

A \( 4+2p+q^2 > 0 \) B \( 4+2p+q^2 \\< 0 \) C \( 4+2p-q^2 > 0 \) D \( 4+2p-q^2 \\< 0 \)

Let \(y=x^2+px-q^2\) as \(x \rightarrow \pm \infty, y \rightarrow \infty\) thus the resulting parabola is open in the positive direction of \(y\)-axis. As \(2\), lies between the roots, \(y\) is less than \(0\) at \(x=2\).

Question 4

If \(R\) and \(Q\) are equivalence relations on set \(A\), then which of the following is not an equivalence relation

A\(R^{-1} \cap Q^{-1}\) B\(Q^{-1}\) C\(R \cup Q\) D\(R \cap Q\)

Let \( A=\{1,2,3\}, P =\{(1,1),(2,2),(3,3),(2,3),(3,2)\}\) and \(Q =\{(1,1),(2,2),(3,3),(1,2),(2,1)\}\). It is easy to see that both \(P \text{ and } Q\) are equivalence relation on \(A\) but \(P \cup Q\) is not an equivalence relation.

Question 5

Let \( \rho \) be a relation defined on set of natural numbers \( \mathbb{N} \), as \( \rho = \{(x, y) \in \mathbb{N} \times \mathbb{N} : 2x + y = 41\} \). Then domain A and range B are

A\( A \subset \{x \in \mathbb{N} : 1 \le x \le 20\} \) and \( B \subset \{y \in \mathbb{N} : 1 \le y \le 39\} \) B\( A = \{x \in \mathbb{N} : 1 \le x \le 15\} \) and \( B = \{y \in \mathbb{N} : 2 \le y \le 30\} \) C\( A \equiv \mathbb{N}, B \equiv \mathbb{Q} \) D\( A \equiv \mathbb{Q}, B \equiv \mathbb{Q} \)

See that \(y= 41 - 2x\) and for \(x \in \{1,2,...,20 \}\), \( y \in \{1,3,5,...,39\}\)

Question 6

In \(\mathbb{R}\), a relation \(\rho\) is defined as follows: \(\forall a,b \in \mathbb{R}\), \(a\rho b \quad \text{holds iff} \quad a^2 -4ab+3b^2=0 \)

A\(\rho\) is equivalence relation B\(\rho\) is only symmetric C\(\rho\) is only reflexive D\(\rho\) is only transitive

See that \(a^2-4ab+3b^2 =0 \implies a=b\) or \(a=3b\)

Question 7

Let \( f : \mathbb{R} \to \mathbb{R} \) be a function defined by \( f(x) = \frac{e^{|x|} - e^{-x}}{e^x + e^{-x}} \), then

A \( f \) is both one-one and onto B \( f \) is one-one but not onto C \( f \) is onto but not one-one D \( f \) is neither one-one nor onto

\Injectivity (One-One): For \( x < 0 \), \( |x| = -x \). $$ f(x) = \frac{e^{-x} - e^{-x}}{e^x + e^{-x}} = 0 $$ Since \( f(-1) = f(-2) = 0 \), multiple inputs yield the same output. Therefore, \( f \) is not one-one. 2. Surjectivity (Onto): For \( x \ge 0 \), \( |x| = x \). $$ f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} $$ The range for \( x \ge 0 \) is \( [0, 1) \) ( use limits), the total range of \( f \) is \( [0, 1) \). Since the range \( [0, 1) \neq \) codomain \( \mathbb{R} \), \( f \) is not onto.

Question 8

Let \( f : \mathbb{R} \to \mathbb{R} \) be a function defined by \( f(x) = 2026x^3+2024x+2023 \), then

A \( f \) is both one-one and onto B \( f \) is one-one but not onto C \( f \) is onto but not one-one D \( f \) is neither one-one nor onto

It is easy to see that \(f\) is injective. For surjectivity, think about the nature of roots of a cubic equation.

Question 9

Let \(u + v + w = 3\), \(u, v, w \in \mathbb{R}\) and \(f(x) = ux^2 + vx + w\) be such that \(f(x + y) = f(x) + f(y) + xy\), \(\forall x, y \in \mathbb{R}\). Then \(v\) is equal to

A\(\frac{5}{2}\) B\(\frac{1}{2}\) C\(\frac{1}{\sqrt{2}}\) D\(3\)

Start by substituting \(f(x) = ux^2 + vx + w\) directly into the given equation \(f(x + y) = f(x) + f(y) + xy\). Expand the left side, and then compare the coefficients of the \(xy\) terms and the constant terms on both sides. This will reveal the values of \(u\) and \(w\) first, which you can then use with the condition \(u + v + w = 3\) to find \(v\).

Question 10

Let \(X\) and \(Y\) be two non-empty sets. Let \(f:X \rightarrow Y\) be a function. For \(A \subset X\) and \(B \subset Y\) , define \(f(A)=\{f(x):x \in A\}\) and \(f^{-1}(B)=\{x \in X : f(x) \in B \}\), then

A\(f(f^{-1}(B))=B\) B\(f(f^{-1}(B)) \subset B\) C\(f^{-1}(f(A))=A\) D\(f^{-1}(f(A)) \subset A\)

Collection of all pre-images will be mapped into \(B\). Not every element in \(Y\) has a pre-image. Additional exercise, what will happen if the function is surfective?

Question 11

Range of the function \(f(x)= \frac{x^2+x+2}{x^2+x+1}\ \: x \in \mathbb{R}\)

A\((1,\infty)\) B\((1, 1.5]\) C\((1,\frac{7}{3}]\) D\([1,\frac{7}{3}]\)

Rewrite the function as \(f(x)=1+\frac{1}{x^2+x+1}\)

Question 12

Let \(X=\{v,i,n,o,d\}\) and \(Y=\{p,m\}\). The number of onto ( surjective) functions from \(X\) to \(Y\) is

A\(32\) B\(28 \) C\(30\) D\(36\)

To find the number of onto functions, first calculate the total number of unrestricted mappings from \(\{v, i, n, o, d\}\) to \(\{p, m\}\), which is \(2^5\). Then, simply subtract the cases where the mapping is not onto—specifically, the exact number of scenarios where all \(5\) elements map exclusively to just a single output in the codomain.

Question 13

Find the natural number \(a\) for which \(\sum_{k=1}^{n} f(a+k) = 16(2^n-1)\) where the function \(f\) satisfies the relation \(f(x+y)=f(x)f(y)\) for all natural numbers \(x,y\) and further , \(f(1)=2\).

A\(11\) B\(5\) C\(7\) D\(3\)

Hint not yet added.

Question 14

The domain of the definition of the function \(f(x)=\frac{1}{4-x^2} + log_{10} (x^3-x)\) is

A \( (-1,0) \cup (1,2) \cup (3,\infty) \) B \( (-2,-1) \cup (-1,0) \cup (2,\infty) \) C \( (-1,0) \cup (1,2) \cup (2,\infty) \) D \( (1,2) \cup (2,\infty)\)

Critical points of \(x^3-x\) are \(-1,0\) and \(1\)

Question 15

Let \(f(x)= a^x \quad (a > 0) \) be written as \(f(x)=f_1(x)+f_2(x)\), where \(f_1(x)\) is an even function and \(f_2(x)\) is an odd function. Then \(f_1(x+y)+f_1(x-y)\) equals

A\(2f_1(x+y)f_2(x-y)\) B\(2f_1(x+y)f_1(x-y)\) C\(2f_1(x)f_2(x)\) D\(2f_1(x)f_1(x)\)

see that \(f(x)=a^x=f_1(x)+f_2(x)\) and \(\f(-x)=a^{-x}=f_1(x)-f_2(x))

Question 16

Let \(g(x) = 1 + x - [x]\) and \(f(x) = \begin{cases} -1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{cases}\), then for all \(x\), \(f[g(x)]\) is equal to

A\(x\) B\(1\) C\(f(x)\) D\(g(x)\)

Note that \(x-[x] = \{x\}\) ( fractional part of \(x\) )

Question 17

Let \(\mathbb{N}\) be the set of natural numbers and two functions \(f\) and \(g\) defined as \(f,g:\mathbb{N} \rightarrow \mathbb{N}\) such that \(f(x) = \begin{cases} \frac{n+1}{2}, & \text{ if n is odd} \\ \frac{n}{2}, & \text{ if n is even} \end{cases}\) and \(g(n)=n-(-1)^n\). Then \(f\circ g\) is

Aone-one but not onto Bonto but not one-one Cboth one-one and onto Dneither one-one nor onto

Start by evaluating the composite function for the first couple of natural numbers. What exactly are the outputs for \(n = 1\) and \(n = 2\) when you calculate \((f \circ g)(n)\)? If you choose an even natural number for your initial input, say \(n = 2k\), what does the composite function \((f \circ g)(2k)\) ultimately simplify to?

Question 18

If \(f\) is an even function defined on the interval \((-5, 5)\), the four real values of \(x\) satisfying the equation: \(f(x) = f\left(\frac{x+1}{x+2}\right)\) are

A\(x = \frac{-1 \pm \sqrt{7}}{2} \quad \text{and} \quad x = \frac{-3 \pm \sqrt{5}}{2}\) B\(x = \frac{-1 \pm \sqrt{5}}{2} \quad \text{and} \quad x = \frac{-1 \pm \sqrt{5}}{2}\) C\(x = \frac{-2 \pm \sqrt{5}}{2} \quad \text{and} \quad x = \frac{-3 \pm \sqrt{5}}{2}\) D\(x = \frac{-1 \pm \sqrt{5}}{2} \quad \text{and} \quad x = \frac{-3 \pm \sqrt{5}}{2}\)

Since the function \(f\) is an even function, the condition \(f(a) = f(b)\) implies that either \(a = b\) or \(a = -b\).

Question 19

Let \(f:X \rightarrow X\) be such that \(f(f(x))=x\), for all \(x \in X\) and \( X \subset R \), then

A\(f\) is one-to-one B\(f\) is onto C\(f\) is one-to-one not onto D\(f\) is onto but not one-to-one

Hint not yet added.

Question 20

Let \( f : R \rightarrow R \) be such that \( f \) is injective and \( f(x)f(y) = f(x+y) \) for \( \forall x, y \in R \). If \( f(x) \), \( f(y) \), \( f(z) \) are in G.P., then \( x, y, z \) are in

AAP always and \(f(0)=1\) BGP always and \(f(0)=0\) CAP depending on the value of \( x, y, z \) DGP depending on the value of \( x, y, z \)

Hint not yet added.

Question 21

The domain of definition of \( f(x) = \sqrt{\frac{1 - |x|}{2 - |x|}} \) is

A\( (-\infty, -1) \cup (2, \infty) \) B\( [-1, 1] \cup (2, \infty) \cup (-\infty, -2) \) C\( (-\infty, 1) \cup (2, \infty) \) D\( [-1, 1] \cup (2, \infty) \)

Hint not yet added.

Question 22

Let \( f(x) = ax^2 + bx + c \), \( g(x) = px^2 + qx + r \) such that \( f(1) = g(1) \), \( f(2) = g(2) \) and \( f(3) - g(3) = 2 \). Then \( f(4) - g(4) \) is

A4 B5 C6 D7

Hint not yet added.

Question 23

Let \( R \) be the set of real numbers and the functions \( f : R \rightarrow R \) and \( g : R \rightarrow R \) be defined by \( f(x) = x^2 + 2x - 3 \) and \( g(x) = x + 1 \). Then, the value of \( x \) for which \( f(g(x)) = g(f(x)) \) is

A\( -1 \) B\( 0 \) C\( 1 \) D\( 2 \)

Hint not yet added.

Question 24

The minimum value of the function \(f(x)=2|x-1|+|x-2|\) is

A\( 0 \) B\( 1 \) C\( 2 \) D\( 3 \)

Sketch the graph.

Question 25

Let \(\mathrm{A}=\{-3,-2,-1,0,1,2,3\}\) and \(R\) be a relation on \(A\) defined by \(x R y\) if and only if \(2 x-y \in\{0,1\}\). Let \(l\) be the number of elements in \(R\). Let \(m\) and \(n\) be the minimum number of elements required to be added to \(R\) to make it a reflexive and symmetric relation, respectively. Then \(l+\mathrm{m}+ \mathrm{n}\) is equal to :-

A\(18\) B\(15\) C\(16\) D\(17\)

Hint not yet added.

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