Skip to main content

Physics Library

Free Physics Study Material for CSIR NET, IIT JAM, GATE, JEST & TIFR

Physics Library
by Pravegaa Education
Visit Pravegaa.com
Login / Register
Menu
Menu

Chapter 14  Calculus, Series and Integrals Previous Year Questions with Solutions

Pravegaa PYQ Series · Volume 1: Mathematical Physics

MP14: Calculus, Series and Integrals

58 previous-year questions with complete solutions, from IIT JAM, GATE, CSIR NET-JRF, JEST and TIFR, arranged from IIT JAM and GATE through to TIFR and from the most recent paper backwards. Click Show solution under any question.

  • IIT JAM: 15
  • CSIR NET: 6
  • JEST: 15
  • TIFR GS: 22
MP14.1IIT JAM 2026, Q25MCQ | 2 marks

Using Taylor series, expand \(f(x)=x^3-\dfrac18\) around \(x_0=1\) up to second order in \(x\). The coefficient of \(x\) is:

  1. \(-2\)

  2. 3

  3. 2

  4. \(-3\)

Show solution

\(f(1)=\tfrac78\), \(f'(x)=3x^2\Rightarrow f'(1)=3\), \(f''(x)=6x\Rightarrow f''(1)=6\). To second order, \[ f(x)\approx\frac78+3(x-1)+\frac{6}{2!}(x-1)^2=\frac78+3x-3+3x^2-6x+3=3x^2-3x+\frac78 . \] \[ \boxed{\text{Coefficient of } x=-3} \]

Answer: (D) \(-3\)

MP14.2IIT JAM 2025, Q12MCQ | 2 marks

Given a function \(f(x,y)=\dfrac{x}{a}e^{y}+\dfrac{y}{b}e^{x}\), where \(x=at\) and \(y=bt\) (\(a\) and \(b\) are non-zero constants), the value of \(\dfrac{df}{dt}\) at \(t=0\) is

  1. \(-1\)

  2. 0

  3. 1

  4. 2

Show solution

By the chain rule, \[\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt} =\left(\frac{e^{y}}{a}+\frac{y}{b}e^{x}\right)a+\left(\frac{x}{a}e^{y}+\frac{e^{x}}{b}\right)b .\] At \(t=0\), \(x=y=0\): \[\left.\frac{df}{dt}\right|_{t=0}=\frac{1}{a}\cdot a+\frac{1}{b}\cdot b=\boxed{2}\]

Answer: (D) 2

MP14.3IIT JAM 2024, Q53NAT | 2 marks

In the Taylor expansion of function, \(F(x)=e^x\sin x\), around \(x=0\), the coefficient of \(x^5\) is . (Rounded off to three decimal places)

Show solution

Write \(F(x)=\operatorname{Im}\,e^{x}e^{ix}=\operatorname{Im}\,e^{(1+i)x}=\operatorname{Im}\displaystyle\sum_{n=0}^\infty\frac{(1+i)^n}{n!}x^n\). The coefficient of \(x^5\) is \(\operatorname{Im}(1+i)^5/5!\): \[ (1+i)^5=\left(\sqrt2\,e^{i\pi/4}\right)^5=4\sqrt2\,e^{i5\pi/4}=4\sqrt2\left(-\frac{1}{\sqrt2}-\frac{i}{\sqrt2}\right)=-4-4i . \] \[ \text{Coefficient of }x^5=\frac{-4}{120}=-\frac{1}{30}=\boxed{-0.033} \]

Answer: \(-0.033\) (range \(-0.034\) to \(-0.032\))

MP14.4IIT JAM 2023, Q4MCQ | 1 mark

The plot of the function \(f(x)=\big||x|-1\big|\) is

  1. Figure

  2. Figure

  3. Figure

  4. Figure

Show solution

The function is even and never negative. At \(x=0\): \(f(0)=|0-1|=1\); at \(x=\pm1\): \(f(\pm1)=|1-1|=0\); for \(|x|>1\), \(f=|x|-1\) rises with unit slope. So the graph is a “W”: value \(1\) at the origin, zeros at \(x=\pm1\), and straight lines of slope \(\pm1\) outside.

(A) takes negative values, (C) never reaches zero and (D) is negative at \(x=0\). Only (B) satisfies all these conditions. \[\boxed{\text{(B)}}\]

Answer: (B)

MP14.5IIT JAM 2023, Q43NAT | 1 mark

The absolute error in the value of \(\sin\theta\) if approximated up to two terms in the Taylor's series for \(\theta=60^\circ\) is (rounded off to three decimal places).

Show solution

With \(\theta=60^\circ=\pi/3=1.047198\) rad, the two-term Taylor approximation is \[ \sin\theta\approx\theta-\frac{\theta^3}{3!}=1.047198-\frac{(1.047198)^3}{6}=1.047198-0.191397=0.855801 . \] The exact value is \(\sin60^\circ=\dfrac{\sqrt3}{2}=0.866025\), so the absolute error is \[ \Delta=0.866025-0.855801=0.010225\approx0.010 . \] \[\boxed{\Delta\approx0.010}\]

Answer: 0.010 (range 0.009 to 0.011)

MP14.6IIT JAM 2022, Q11MCQ | 2 marks

The function \(f(x)=e^{\sin x}\) is expanded as a Taylor series in \(x\), around \(x=0\), in the form \(f(x)=\sum_{n=0}^{\infty}a_nx^n\). The value of \(a_0+a_1+a_2\) is

  1. \(0\)

  2. \(\dfrac32\)

  3. \(\dfrac52\)

  4. \(5\)

Show solution

\begin{align*} f(x)&=e^{\sin x} &&\Rightarrow f(0)=1\\ f'(x)&=e^{\sin x}\cos x &&\Rightarrow f'(0)=1\\ f''(x)&=e^{\sin x}\cos^2x-e^{\sin x}\sin x &&\Rightarrow f''(0)=1 \end{align*} So \(f(x)=1+x+\dfrac{x^2}{2}+\cdots\), giving \(a_0=1\), \(a_1=1\), \(a_2=\tfrac12\) and \[\boxed{a_0+a_1+a_2=\frac52}\]

Answer: (C) \(5/2\)

MP14.7IIT JAM 2022, Q41NAT | 1 mark

The integral \(\iint(x^2+y^2)\,dx\,dy\) over the area of a disk of radius 2 in the \(xy\) plane is \(\pi\).

Show solution

In plane polar coordinates \(x^2+y^2=r^2\) and \(dx\,dy=r\,dr\,d\theta\): \[ I=\int_0^{2\pi}\!\!\int_0^2 r^2\,r\,dr\,d\theta=2\pi\left[\frac{r^4}{4}\right]_0^2=2\pi\times4=8\pi . \] \[\boxed{I=8\pi\;\Rightarrow\;8}\]

Answer: 8

MP14.8IIT JAM 2021, Q1MCQ | 1 mark

The function \(e^{\cos x}\) is Taylor expanded about \(x=0\). The coefficient of \(x^2\) is

  1. \(-\dfrac{1}{2}\)

  2. \(-\dfrac{e}{2}\)

  3. \(\dfrac{e}{2}\)

  4. Zero

Show solution

The coefficient of \(x^2\) in the Taylor series about \(x=0\) is \(f''(0)/2!\). With \(f(x)=e^{\cos x}\), \[ f'(x)=-\sin x\,e^{\cos x},\qquad f''(x)=\left(\sin^2x-\cos x\right)e^{\cos x} \;\Rightarrow\; f''(0)=(0-1)e=-e . \] Hence \[ \boxed{\text{Coefficient of }x^2=\frac{f''(0)}{2!}=-\frac{e}{2}} \] (Check: \(e^{\cos x}=e\cdot e^{-x^2/2+\cdots}=e\left(1-\tfrac{x^2}{2}+\cdots\right)\).)

Answer: (B) \(-\dfrac{e}{2}\)

MP14.9IIT JAM 2020, Q1MCQ | 1 mark

Which one of the following functions has a discontinuity in the second derivative at \(x=0\), where \(x\) is a real variable?

  1. \(f(x)=|x|^3\)

  2. \(f(x)=x|x|\)

  3. \(f(x)=\cos(|x|)\)

  4. \(f(x)=|x|^2\)

Show solution

For \(f(x)=x|x|\): \(f(x)=x^2\) for \(x\ge 0\) and \(f(x)=-x^2\) for \(x<0\). Hence \[ f'(x)=2|x|,\qquad f''(x)=\begin{cases}+2, & x>0\\ -2, & x<0\end{cases} \] so \(f''\) jumps from \(-2\) to \(+2\) at \(x=0\).

For the others: \(|x|^3\) gives \(f''=6|x|\) (continuous); \(\cos|x|=\cos x\) and \(|x|^2=x^2\) are smooth everywhere. \[ \boxed{f(x)=x|x|} \]

Answer: (B) \(f(x)=x|x|\)

MP14.10IIT JAM 2020, Q13MCQ | 2 marks

\(\displaystyle\lim_{x\to0^+}x^x\) is equal to

  1. \(0\)

  2. \(\infty\)

  3. \(e\)

  4. \(1\)

Show solution

Let \(y=x^x\), so \(\ln y=x\ln x=\dfrac{\ln x}{1/x}\). By L'Hospital's rule, \[ \lim_{x\to0^+}\ln y=\lim_{x\to0^+}\frac{1/x}{-1/x^2}=\lim_{x\to0^+}(-x)=0 \] \[ \Rightarrow\ \lim_{x\to0^+}y=e^0=\boxed{1} \]

Answer: (D) \(1\)

MP14.11IIT JAM 2019, Q1MCQ | 1 mark

The function \(f(x)=\dfrac{8x}{x^2+9}\) is continuous everywhere except at

  1. \(x=0\)

  2. \(x=\pm 9\)

  3. \(x=\pm 9i\)

  4. \(x=\pm 3i\)

Show solution

A rational function is discontinuous only where its denominator vanishes: \[ x^2+9=0 \;\Rightarrow\; \boxed{x=\pm 3i} \]

Answer: (D) \(x=\pm 3i\)

MP14.12IIT JAM 2018, Q41NAT | 1 mark

The coefficient of \(x^3\) in the Taylor expansion of \(\sin(\sin x)\) around \(x=0\) is .

(Specify your answer upto two digits after the decimal point)

Show solution

Let \(f(x)=\sin(\sin x)\). Then \begin{align*} f'(x)&=\cos(\sin x)\cos x,\\ f''(x)&=-\sin(\sin x)\cos^2x-\cos(\sin x)\sin x,\\ f'''(x)&=-\cos(\sin x)\cos^3x+3\sin(\sin x)\sin x\cos x-\cos(\sin x)\cos x . \end{align*} At \(x=0\): \(f(0)=0\), \(f'(0)=1\), \(f''(0)=0\), \(f'''(0)=-1-1=-2\). The coefficient of \(x^3\) is \[ \frac{f'''(0)}{3!}=\frac{-2}{6}=-\frac13 . \] (Check: \(\sin x=x-\tfrac{x^3}{6}+\dots\) and \(\sin u=u-\tfrac{u^3}{6}+\dots\) give \(\sin(\sin x)=x-\tfrac{x^3}{6}-\tfrac{x^3}{6}+\dots=x-\tfrac{x^3}{3}+\dots\)) \[ \boxed{-0.33} \]

Answer: \(-0.33\) (range \(-0.35\) to \(-0.30\))

MP14.13IIT JAM 2017, Q22MCQ | 2 marks

Which one of the following graphs represents the derivative \(f'(x)=\dfrac{df}{dx}\) of the function \(f(x)=\dfrac{1}{1+x^2}\) most closely (graphs are schematic and not drawn to scale)?Figure

Show solution

\[ f'(x)=-\frac{2x}{\left(1+x^2\right)^2}. \] This is an odd function: \(f'>0\) for \(x<0\), \(f'=0\) at \(x=0\), \(f'<0\) for \(x>0\), with extrema at \(x=\mp1/\sqrt3\) and \(f'\to0\) as \(|x|\to\infty\). \[ \boxed{\text{positive lobe for }x<0,\ \text{negative lobe for }x>0} \] This is graph (A).

Answer: (A)

MP14.14IIT JAM 2016, Q12MCQ | 2 marks

The tangent line to the curve \(x^2+xy+5=0\) at \((1,1)\) is represented by

  1. \(y=3x-2\)

  2. \(y=-3x+4\)

  3. \(x=3y-2\)

  4. \(x=-3y+4\)

Show solution

Substituting \((1,1)\) into the curve gives \(1+1+5=7\neq0\), so the point \((1,1)\) does not lie on the curve and no tangent to the curve passes through it. The question is therefore ill-posed.

If one formally uses the implicit derivative, \[ 2x+y+x\frac{dy}{dx}=0\;\Rightarrow\;\frac{dy}{dx}=-\frac{2x+y}{x}\Big|_{(1,1)}=-3, \] the line through \((1,1)\) with this slope is \(y-1=-3(x-1)\), i.e. \(y=-3x+4\). Since the point is not on the curve, full marks were awarded to all candidates. \[ \boxed{\text{Marks to All}} \]

Answer: Marks to All the point \((1,1)\) does not lie on the curve

MP14.15IIT JAM 2016, Q42NAT | 1 mark

Consider a function \(f(x,y)=x^3+y^3\), where \(y\) represents a parabolic curve \(x^2+1\). The total derivative of \(f\) with respect to \(x\), at \(x=1\) is .

Show solution

\[ \frac{df}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{dy}{dx}=3x^2+3y^2(2x). \] At \(x=1\), \(y=1^2+1=2\): \[ \frac{df}{dx}=3(1)+3(4)(2)=3+24=\boxed{27} \]

Answer: 27 (range 27.0 to 27.0)

Continue reading — it's free

Register with your mobile number to unlock the full chapter and all 100+ chapters of the Physics Library.

Register / Login with OTP

Already registered? Use the same button — just verify your number.