As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background.

Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced .

Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced .

**Q3.**The function f:R→R defined by f(x)=(x-1)(x-2)(x-3) is

Solution

(b) Given, f(x)=(x-1)(x-2)(x-3) ⇒ f(1)=f(2)=f(3)=0 ⇒ f(x) is not one-one. For each y∈R, there exists x∈R such that fx=y. Therefore, f is onto.

(b) Given, f(x)=(x-1)(x-2)(x-3) ⇒ f(1)=f(2)=f(3)=0 ⇒ f(x) is not one-one. For each y∈R, there exists x∈R such that fx=y. Therefore, f is onto.

**Q7.**In a function f(x) is defined for x∈[0,1], then the function f(2x+3) is defined for

Solution

(b) Since f(x) is defined for x∈[0, 1]. Therefore, f(2x+3) exists if 0 ≤ 2x+3 ≤ 1 ⇒ -3/2 ≤ x ≤ -1 ⇒ x∈[-3/2, -1]

(b) Since f(x) is defined for x∈[0, 1]. Therefore, f(2x+3) exists if 0 ≤ 2x+3 ≤ 1 ⇒ -3/2 ≤ x ≤ -1 ⇒ x∈[-3/2, -1]

**Q9.**Which of the following functions has period Ï€ ?

Solution

(a) We know that tan x has period Ï€. Therefore, |tan(tanx)| has period Ï€/2. Also,cos2x has period Ï€. Therefore, period of |tan(tanx)| +cos(cos2x) is Ï€. Clearly, 2sin Ï€ x3+3cos 2Ï€ x3 has its period equal to the LCM of 6 and 3 i.e., 6 6cos 2 Ï€ x+Ï€/4+5sin (Ï€ x+3Ï€/4) has period 2 The function tantan4x +|sinsin4x| has period 2

(a) We know that tan x has period Ï€. Therefore, |tan(tanx)| has period Ï€/2. Also,cos2x has period Ï€. Therefore, period of |tan(tanx)| +cos(cos2x) is Ï€. Clearly, 2sin Ï€ x3+3cos 2Ï€ x3 has its period equal to the LCM of 6 and 3 i.e., 6 6cos 2 Ï€ x+Ï€/4+5sin (Ï€ x+3Ï€/4) has period 2 The function tantan4x +|sinsin4x| has period 2

**Q10.**The function f:C→C defined by f(x)=(ax+b)/(cx+d) for x∈C where bd≠0 reduces to a constant function, if

Solution

(c) Given, f(x)=(ax+b)/(cx+d) It reduces the constant function if a/c=b/d ⇒ ad=bc

(c) Given, f(x)=(ax+b)/(cx+d) It reduces the constant function if a/c=b/d ⇒ ad=bc