Important Solved Functions problems for JEE Maths




Functions Chapter in mathematics is quite important for JEE Preparation. Fundamentals of this chapter is required to clear to move ahead.I am trying to give few Important Solved Functions problems for JEE Maths in this article .

You can refer below link for below for the detailed study material on Relations and Function

Relation and function Study material

So Here we go.I hope you will like it

Question 1

If 1 lies between the roots of equation {{y}^{2}}-my+1=0 and [x] denote the greatest integer’s value of , then find the value of \left[ {{\left( \frac{4|x|}{|x{{|}^{2}}+16} \right)}^{m}} \right]

Solution

{{y}^{2}}-my+1=0

Let f(x) = {{y}^{2}}-my+1

Since 1 lies between the roots

(1)   \begin{align*} & \therefore f(1)<0 \\ & 2-m<0 \\ \end{align*}

This implies that

m>2

Now we know that Arithmetic Mean between two number is always greater than or equal to the Geometric mean

A.M\ge G.M

So choosing two number as |x|^{2} and 16

(2)   \begin{align*} & \Rightarrow \frac{{{\left| x \right|}^{2}}+16}{2}\ge \left| x \right|\cdot 4 \\ & \Rightarrow 0\le \frac{4\left| x \right|}{{{\left| x \right|}^{2}}+16}\le \frac{1}{2} \\ & \Rightarrow 0<{{\left( \frac{4\left| x \right|}{{{\left| x \right|}^{2}}+16} \right)}^{m}}<\frac{1}{4} \\ \end{align*}

Hence as per definition of function [x]

\left[ {{\left( \frac{4|x|}{|x{{|}^{2}}+16} \right)}^{m}} \right]=0

Question 2

Find the number of real roots of  {{2}^{x}}={{x}^{2}}+1

Solution

Let

(3)   \begin{align*} & f(x)={{2}^{x}}-({{x}^{2}}+1) \\ & f(0)=0 \\ & f'(x)={{2}^{x}}ln2-2x \\ & f''(x)={{2}^{x}}{{(ln2)}^{2}}-2 \\ \end{align*}

Clearly f''(x)=0 has only the real root

This implies that f'(x) has a local minima of negative value.

f(x) has a local minima and local maxima

This implies that f(x)=0 has three real roots.

Question 3




If f(x) and g(x) be periodic and non-periodic functions respectively, then f(g(x)) is

(a) always periodic

(b) never periodic

(c) periodic, when g(x) is a linear function of x

(d) can’t say

Solution

(c) Since we know if f(x) is periodic then f(ax+b) is also periodic function. Hence g(x) should be linear function of x.

Question 4

Let f(x) and g(x) be bijective functions where f:\{a,b,c,d\}\to \{1,2,3,4\} and g:\{3,4,5,6\}\to \{w,x,y,z\} respectively. The number of elements in the range set of g(f(x)) is

(a) 1

(b) 2

(c) 3

(d) 4

Solution

(b) Range of f(x) for which g(f(x)) is defined is {3,4}

Hence domain of g(f(x)) has two elements.

Therefore range of g(f(x))  also has two elements.

Question 5

Domain (D) and Range (R) of f(x)={{\sin }^{-1}}\left( {{\cos }^{-1}}\left[ x \right] \right) is respectively, where [.] denotes the greatest integer function.

(a) D=x\in [1,2),R\equiv \{0\}

(b) D=x\in [0,1],R\equiv \{-1,0,1\}

(c) D=x\in [-1,1],R\equiv \left\{ 0,{{\sin }^{-1}}\left( \frac{\pi }{2} \right),si{{n}^{-1}}\left( \pi  \right) \right\}

(d) D=x\in [-1,1],R\equiv \left\{ -\frac{\pi }{2},0,\frac{\pi }{2} \right\}

Solution (a)

f(x)={{\sin }^{-1}}\left( {{\cos }^{-1}}\left[ x \right] \right)

f(x) is defined when \frac{x\in [1,2)}{{{\cos }^{-1}}[x]}=\frac{\pi }{2}>1 as, when x\in [1,2);

Range of f(x) is 0

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