NCERT Exemplar Solutions Class 11 Maths Limits and Derivatives — Objective Type and Fill in the Blanks Q65–80

NCERT Exemplar Problems, Class 11 Mathematics — Limits and Derivatives, Objective Type questions 65 to 76 and Fill in the Blanks questions 77 to 80. All 16 questions solved.

This page finishes the unit. Questions 65 to 67 are about one-sided limits and the fractional part $x-[x]$; Questions 68 to 76 are derivatives evaluated at a point (and one trap in Question 74); Questions 77 to 80 are the blanks, with a trigonometric limit, a limit that fixes $m$, the derivative of the exponential series, and a limit with the greatest-integer function. All sixteen printed answers agree with the working below.

Key insight. Before differentiating a “messy” function at a point, simplify it (Questions 69, 70, 71), and before taking a limit where a greatest-integer function or a piecewise definition appears, split into left and right (Questions 65 and 80). The two habits together settle most of this page.

Choose the correct answer out of the 4 options given against each Exercise 54 to 76 (M.C.Q.). This page covers 65 to 76.

Question 65

Let $f(x)=\begin{cases}x^2-1, & 0<x<2\\ 2x+3, & 2\le x<3\end{cases}$. The quadratic equation whose roots are $\displaystyle\lim_{x\to2^-}f(x)$ and $\displaystyle\lim_{x\to2^+}f(x)$ is

Solution. Just left of $2$ the rule is $x^2-1$, so the left-hand limit is $4-1=3$. Just right of $2$ the rule is $2x+3$, so the right-hand limit is $4+3=7$. A quadratic with roots $3$ and $7$ has sum $10$ and product $21$:

$$x^2-10x+21=0.$$

Question 66

$\displaystyle\lim_{x\to0}\dfrac{\tan2x-x}{3x-\sin x}$ is

Solution. Divide numerator and denominator by $x$:

$$\frac{2\cdot\frac{\tan2x}{2x}-1}{3-\frac{\sin x}{x}}\to\frac{2-1}{3-1}=\frac12.$$

Question 67

Let $f(x)=x-[x]$, $x\in\mathbf R$. Then $f’\left(\dfrac12\right)$ is

Solution. The book omits the variable before the membership sign in “$\in\mathbf R$”; it means $x\in\mathbf R$. Near $x=\frac12$, say for $0<x<1$, the greatest integer $[x]$ is $0$, so $f(x)=x$ there. The function is locally just $x$, so $f’\left(\frac12\right)=1$. (The function is not differentiable at integers, but $\frac12$ is not one.)

Question 68

If $y=\sqrt x+\dfrac1{\sqrt x}$, then $\dfrac{dy}{dx}$ at $x=1$ is

Solution. $y=x^{1/2}+x^{-1/2}$, so $\dfrac{dy}{dx}=\dfrac12x^{-1/2}-\dfrac12x^{-3/2}$. At $x=1$ this is $\frac12-\frac12=0$.

Question 69

If $f(x)=\dfrac{x-4}{2\sqrt x}$, then $f'(1)$ is

Solution. Split the fraction: $f(x)=\dfrac12x^{1/2}-2x^{-1/2}$. Then $f'(x)=\dfrac14x^{-1/2}+x^{-3/2}$, and $f'(1)=\frac14+1=\frac54$.

Question 70

If $y=\dfrac{1+\frac1{x^2}}{1-\frac1{x^2}}$, then $\dfrac{dy}{dx}$ is

Solution. Multiply top and bottom by $x^2$: $y=\dfrac{x^2+1}{x^2-1}$. Quotient rule:

$$\frac{dy}{dx}=\frac{2x(x^2-1)-(x^2+1)\cdot2x}{(x^2-1)^2}=\frac{-4x}{(x^2-1)^2}.$$

Question 71

If $y=\dfrac{\sin x+\cos x}{\sin x-\cos x}$, then $\dfrac{dy}{dx}$ at $x=0$ is

Solution. Quotient rule with $u=\sin x+\cos x$, $u’=\cos x-\sin x$, $v=\sin x-\cos x$, $v’=\cos x+\sin x$:

$$y’=\frac{(\cos x-\sin x)(\sin x-\cos x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x-\cos x)^2}=\frac{-(\sin x-\cos x)^2-(\sin x+\cos x)^2}{(\sin x-\cos x)^2}=\frac{-2}{(\sin x-\cos x)^2}.$$

At $x=0$: $\dfrac{-2}{(0-1)^2}=-2$.

Question 72

If $y=\dfrac{\sin(x+9)}{\cos x}$, then $\dfrac{dy}{dx}$ at $x=0$ is

Solution. Quotient rule:

$$y’=\frac{\cos(x+9)\cos x+\sin(x+9)\sin x}{\cos^2x}=\frac{\cos\big((x+9)-x\big)}{\cos^2x}=\frac{\cos9}{\cos^2x}.$$

At $x=0$ this is $\cos9$.

Question 73

If $f(x)=1+x+\dfrac{x^2}2+\dots+\dfrac{x^{100}}{100}$, then $f'(1)$ is equal to

Solution. Differentiate term by term: $\dfrac{d}{dx}\dfrac{x^k}{k}=x^{k-1}$, so

$$f'(x)=1+x+x^2+\dots+x^{99}.$$

At $x=1$ each of the $100$ terms is $1$, so $f'(1)=100$.

Question 74

If $f(x)=\dfrac{x^n-a^n}{x-a}$ for some constant $a$, then $f'(a)$ is

Solution. The function is not defined at $x=a$ (the denominator is zero), so there is no value $f(a)$ and no derivative at $a$. This is the trap: the limit of $f(x)$ at $a$ is $na^{n-1}$, which is not what $f'(a)$ means. (If the hole were filled in with that value, $f'(a)$ would be $\frac{n(n-1)}2a^{n-2}$, which depends on $n$ and $a$ and is not among the options either.)

Question 75

If $f(x)=x^{100}+x^{99}+\dots+x+1$, then $f'(1)$ is equal to

Solution. $f'(x)=100x^{99}+99x^{98}+\dots+2x+1$. At $x=1$ this is $100+99+\dots+2+1=\dfrac{100\cdot101}2=5050$.

Question 76

If $f(x)=1-x+x^2-x^3+\dots-x^{99}+x^{100}$, then $f'(1)$ is equal to

Solution. $f'(x)=-1+2x-3x^2+\dots-99x^{98}+100x^{99}$. At $x=1$ the terms alternate in sign, so pair them: $(-1+2)+(-3+4)+\dots+(-99+100)$ is $50$ pairs, each equal to $1$. The total is $50$.

Fill in the blanks in Questions 77 to 80.

Question 77

If $f(x)=\dfrac{\tan x}{x-\pi}$, then $\displaystyle\lim_{x\to\pi}f(x)=\_\_\_\_$.

Solution. Put $t=x-\pi\to0$. Because $\tan$ has period $\pi$, $\tan x=\tan(\pi+t)=\tan t$. So the limit is $\dfrac{\tan t}{t}\to1$.

Question 78

$\displaystyle\lim_{x\to0}\left(\sin mx\,\cot\dfrac x{\sqrt3}\right)=2$, then $m=\_\_\_\_$.

Solution. Write $\cot\dfrac x{\sqrt3}=\dfrac{\cos\frac x{\sqrt3}}{\sin\frac x{\sqrt3}}$ and match each sine with its argument:

$$\sin mx\cot\frac x{\sqrt3}=\frac{\sin mx}{mx}\cdot\frac{\frac x{\sqrt3}}{\sin\frac x{\sqrt3}}\cdot\cos\frac x{\sqrt3}\cdot\frac{mx}{x/\sqrt3}\to1\cdot1\cdot1\cdot m\sqrt3.$$

Setting $m\sqrt3=2$ gives $m=\dfrac2{\sqrt3}=\dfrac{2\sqrt3}3$.

Question 79

If $y=1+\dfrac x{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dots$, then $\dfrac{dy}{dx}=\_\_\_\_$.

Solution. Differentiate term by term: $\dfrac{d}{dx}\dfrac{x^k}{k!}=\dfrac{x^{k-1}}{(k-1)!}$. The terms become $0+1+\dfrac x{1!}+\dfrac{x^2}{2!}+\dots$, which is the original series again. So $\dfrac{dy}{dx}=y$. (The series is that of $e^x$, so $y=e^x$ and $\dfrac{dy}{dx}=e^x=y$.)

Question 80

$\displaystyle\lim_{x\to3^+}\dfrac{x}{[x]}=\_\_\_\_$.

Solution. For $x$ slightly more than $3$ (say $3<x<4$) the greatest integer is $[x]=3$, so $\dfrac x{[x]}=\dfrac x3\to\dfrac33=1$. (From the left the integer part would be $2$ and the limit $\frac32$; the “$+$” in the question matters.)

Common mistakes

• Question 65. Using the same piece for both one-sided limits. The left-hand limit at $2$ uses $x^2-1$ (the piece valid for $x<2$), the right-hand limit uses $2x+3$.
• Question 67. Thinking $f(x)=x-[x]$ is not differentiable because the greatest-integer function is not continuous. It is not differentiable at integers, but at $\frac12$ it is locally $x$ and has derivative $1$.
• Question 69 and 70. Applying the quotient rule to the unsimplified forms and getting tangled in nested fractions. Simplify first.
• Question 71. Concluding “does not exist” because the denominator $\sin x-\cos x$ seems to vanish. It vanishes at $x=\frac\pi4$, not at $0$.
• Question 74. Reading $f'(a)$ as the limit of $f(x)$ at $a$ and writing $na^{n-1}$. The function is not even defined at $a$.
• Question 76. Evaluating $f'(1)$ by substituting $1$ into $f$ itself (which gives $1$) instead of into the derivative; the derivative’s terms $-1+2-3+4-\dots$ pair up to $50$.

Practise next

• Other pages in this unit: Short Answer Type Q1–14, Short Answer Type Q15–28, Short and Long Answer Type Q29–46 and Long Answer and Objective Type Q47–64.
• NCERT textbook: Exercise 12.1, Exercise 12.2 and the Miscellaneous Exercise.