NCERT Exemplar Solutions Class 11 Maths Relations and Functions — Long Answer Type Q16–23

NCERT Exemplar Problems, Class 11 Mathematics — Relations and Functions, Long Answer Type questions 16 to 23. All 8 questions solved.

Eight questions that mix three skills: finding the domain (where the formula makes sense, Q17, Q21, Q22), finding the range (the set of values actually produced, Q18), and handling modulus functions by splitting into cases (Q19). Q16, Q20 and Q23 are algebraic checks. For domains, look for denominators, even roots and the sign of what is under them. For ranges, either solve $y=f(x)$ for $x$ or bound the expression piece by piece.

Two printed answers are wrong: Question 18(i) (the range is incomplete) and Question 19 (the last branch is $2x$, not $2$). Both are explained where they occur.

Key insight. Domain asks “which $x$ may go in?”, range asks “which $y$ actually come out?”. For a modulus, remove the bars by cases: the expression inside each bar has a fixed sign on each interval, and the breakpoints are where it is zero.

Question 16

Is $g=\{(1,1),(2,3),(3,5),(4,7)\}$ a function? Justify. If it is described by the relation $g(x)=\alpha x+\beta$, what values should be assigned to $\alpha$ and $\beta$?

Solution. The first coordinates $1,2,3,4$ are all different, so every input has exactly one output. It is a function. Now fit $g(x)=\alpha x+\beta$ using two points: $g(1)=1$ gives $\alpha+\beta=1$ and $g(2)=3$ gives $2\alpha+\beta=3$. Subtracting, $\alpha=2$, and then $\beta=-1$. A check on the other two points confirms it: $g(3)=6-1=5$ and $g(4)=8-1=7$.

Question 17

Find the domain of each function: (i) $f(x)=\dfrac1{\sqrt{1-\cos x}}$ (ii) $f(x)=\dfrac1{\sqrt{x+|x|}}$ (iii) $f(x)=x|x|$ (iv) $f(x)=\dfrac{x^3-x+3}{x^2-1}$ (v) $f(x)=\dfrac{3x}{2x-8}$.

Solution.

(i) Under the root we need $1-\cos x>0$ (strictly, since it is also in a denominator). Always $1-\cos x\ge0$, and it is zero exactly when $\cos x=1$, i.e. $x=2n\pi$. So exclude those points: $\mathbf R-\{2n\pi: n\in\mathbf Z\}$.

(ii) For $x\ge0$, $x+|x|=2x$, which is positive when $x>0$. For $x<0$, $x+|x|=0$, which would be a zero denominator. So the domain is $x>0$, i.e. $\mathbf R^+$.

(iii) $x|x|$ is defined for every real $x$: domain $\mathbf R$.

(iv) The denominator vanishes at $x^2=1$, so exclude $\pm1$: $\mathbf R-\{-1,1\}$.

(v) $2x-8=0$ at $x=4$: $\mathbf R-\{4\}$.

Question 18

Find the range of (i) $f(x)=\dfrac3{2-x^2}$ (ii) $f(x)=1-|x-2|$ (iii) $f(x)=|x-3|$ (iv) $f(x)=1+3\cos2x$.

Solution.

(i) The domain is $x\ne\pm\sqrt2$. Let $t=2-x^2$. Then $t$ takes every value in $(-\infty,2]$ except $0$. Split by sign. For $0<t\le2$, $\frac3t$ runs over $\left[\frac32,\infty\right)$ (largest $t=2$ gives the smallest value $\frac32$). For $t<0$, $\frac3t$ is negative and runs over $(-\infty,0)$. So the range is $(-\infty,0)\cup\left[\frac32,\infty\right)$. The printed key gives only $\left[\frac32,\infty\right)$, which forgets the negative values: at $x=2$, $f(2)=\frac3{2-4}=-\frac32$, which is not in the key’s range.

(ii) $|x-2|\ge0$ takes every value in $[0,\infty)$, so $1-|x-2|$ takes every value in $(-\infty,1]$.

(iii) $|x-3|$ takes every value in $[0,\infty)$.

(iv) From the hint, $-1\le\cos2x\le1$, so $-3\le3\cos2x\le3$ and $-2\le1+3\cos2x\le4$. Both ends are attained.

Question 19

Redefine $f(x)=|x-2|+|2+x|$, $-3\le x\le3$.

Solution. The bars change at $x=2$ and at $x=-2$ (where $x+2=0$), giving three intervals.

• For $-3\le x<-2$: both $x-2$ and $x+2$ are negative, so $f(x)=-(x-2)-(x+2)=-2x$.
• For $-2\le x<2$: $x+2\ge0$ and $x-2<0$, so $f(x)=-(x-2)+(x+2)=4$.
• For $2\le x\le3$: both are non-negative, so $f(x)=(x-2)+(x+2)=2x$.

The printed key shows $2$ for the last interval, which is wrong: at $x=3$, $f(3)=1+5=6=2(3)$, not $2$. The branches also join up (at $x=-2$ both give $4$, at $x=2$ both give $4$), which the key’s “$2$” would break.

Question 20

If $f(x)=\dfrac{x-1}{x+1}$, show that (i) $f\left(\dfrac1x\right)=-f(x)$ (ii) $f\left(-\dfrac1x\right)=\dfrac{-1}{f(x)}$.

Solution. Substitute and clear the small fractions by multiplying top and bottom by $x$.

(i) $f\left(\frac1x\right)=\dfrac{\frac1x-1}{\frac1x+1}=\dfrac{1-x}{1+x}=-\dfrac{x-1}{x+1}=-f(x)$.

(ii) $f\left(-\frac1x\right)=\dfrac{-\frac1x-1}{-\frac1x+1}=\dfrac{-1-x}{-1+x}=-\dfrac{x+1}{x-1}$. Since $f(x)=\frac{x-1}{x+1}$, we have $\frac{-1}{f(x)}=-\frac{x+1}{x-1}$, the same expression.

Question 21

Let $f(x)=\sqrt x$ and $g(x)=x$ be defined on $\mathbf R^+\cup\{0\}$. Find (i) $(f+g)(x)$ (ii) $(f-g)(x)$ (iii) $(fg)(x)$ (iv) $\left(\dfrac fg\right)(x)$.

Solution. Combine pointwise on the common domain $[0,\infty)$.

(i) $\sqrt x+x$. (ii) $\sqrt x-x$. (iii) $\sqrt x\cdot x=x^{3/2}$. (iv) $\dfrac{\sqrt x}{x}=\dfrac1{\sqrt x}$, but this needs $g(x)\ne0$, so $x=0$ is removed and the domain is $x>0$.

Question 22

Find the domain and range of $f(x)=\dfrac1{\sqrt{x-5}}$.

Solution. The root needs $x-5\ge0$ and the denominator cannot be zero, so $x-5>0$: domain $(5,\infty)$. As $x$ runs over $(5,\infty)$, $\sqrt{x-5}$ runs over all positive numbers $(0,\infty)$, and the reciprocal of every positive number is a positive number, so the range is $(0,\infty)=\mathbf R^+$.

Question 23

If $f(x)=y=\dfrac{ax-b}{cx-a}$, prove that $f(y)=x$.

Solution. Substitute $y$ into $f$: $f(y)=\dfrac{ay-b}{cy-a}$. First the numerator: $$ay-b=\frac{a(ax-b)-b(cx-a)}{cx-a}=\frac{a^2x-ab-bcx+ab}{cx-a}=\frac{x(a^2-bc)}{cx-a}.$$ Now the denominator: $$cy-a=\frac{c(ax-b)-a(cx-a)}{cx-a}=\frac{acx-bc-acx+a^2}{cx-a}=\frac{a^2-bc}{cx-a}.$$ Dividing, the common factor $\frac{a^2-bc}{cx-a}$ cancels (it is non-zero when $f$ is not a constant function, i.e. $a^2\ne bc$), leaving $f(y)=x$.

Common mistakes

• Q18(i): stopping at the positive branch. $2-x^2$ is negative for $|x|>\sqrt2$, which gives negative function values; check at least one $x$ beyond $\sqrt2$.
• Q19: writing the last branch as a constant. Both bars are positive there, so the two $x$’s add and the answer depends on $x$.
• Q17(i): including $x=2n\pi$ because $\sqrt{1-\cos x}$ is “fine” there. It is zero, and zero cannot sit in a denominator.
• Q17(ii): admitting $x\le0$ since $x+|x|\ge0$. At $x\le0$ the sum is exactly $0$, so the strict condition $x+|x|>0$ leaves $x>0$ only.
• Q21(iv): leaving $x=0$ in the domain of $f/g$ because it is in the domain of both $f$ and $g$. A quotient also needs $g(x)\ne0$.
• Q16: fitting a line through two points and not checking the rest. Verify $g(3)$ and $g(4)$ to be sure the whole set really is linear.

Practise next

• This unit’s other pages: Short Answer Type, Q1-15 and Objective, Fill in the Blanks and True or False, Q24-42.
• NCERT textbook: Exercise 2.3 for domain and range of functions, Exercise 2.2 for relations from rules.
• Miscellaneous Exercise, Chapter 2.