JEE Main Maths Applications of Derivatives Previous Year Questions With Solutions

JEE main previous year solved questions on Applications of Derivatives give students the opportunity to learn and solve questions in a more effective manner. In calculus, we use derivative to determine the maximum and minimum values of particular functions and many more. There are many important applications of derivatives. In this section, we will solve the problems based on the concepts such as maxima and minima of a function, approximation, Rolle’s theorem and intermediate value theorem, monotonicity, tangent and normal, subtangent and the subnormal, shortest distance between two curves etc. Collection of chapter wise JEE previous year questions with solutions is very useful study material for the JEE aspirants to check their preparation level. Students can expect about 1-2 questions from this chapter in the JEE examination. Click on the below link to get your PDF.

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JEE Main Applications of Derivatives Past Year Questions With Solutions

Question 1: Given function f(x)=(e2x1e2x+1)f(x)=\left( \frac{{{e}^{2x}}-1}{{{e}^{2x}}+1} \right) is

A) Increasing

B) Decreasing

C) Even

D) None of these


f(x)=e2x1e2x+1f(x)=e2x1e2x+1=1e2x1+e2xf(x)=e2x1e2x+1=f(x)f(x)=\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1} \\ f(-x)=\frac{{{e}^{-2x}}-1}{{{e}^{-2x}}+1}=\frac{1-{{e}^{2x}}}{1+{{e}^{2x}}}\\ f(x)=-\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1}=-f(x)\\

f(x) is an odd function.


f(x)=e2x1e2x+1f(x)=4e2x(1+e2x)2>0nRf(x)=\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1} \Rightarrow {f}'(x)=\frac{4{{e}^{2x}}}{{{(1+{{e}^{2x}})}^{2}}}>0\,\forall \,n\in R

f(x) is an increasing function.

Question 2: If f(x)=2x33x212x+5f(x)=2{{x}^{3}}-3{{x}^{2}}-12x+5 and x[2,4],x\in [-2,\,4], then the maximum value of function is at the following value of x

A) 2    B) 1    C) 2     D) 4


f(x)=6x26x12f(x)=0(x2)(x+1)=0x=1,2f'(x)=6{{x}^{2}}-6x-12 \\ f'(x)=0\Rightarrow (x-2)(x+1)=0\Rightarrow x=-1,\,2

Here, f(4)=1284848+5=37f(1)=23+12+5=12f(2)=161224+5=15f(2)=1612+24+5=1f(4)=128-48-48+5=37\\ f(-1)=-2-3+12+5=12\\ f(2)=16-12-24+5=-15\\ f(-2)=-16-12+24+5=1

Therefore, the maximum value of function is 37 at x = 4.

Question 3: The ratio of the height of cone of maximum volume inscribed in a sphere to its radius is

A)34    B)43    C)12    D)23A) \frac{3}{4}\; \; B) \frac{4}{3} \; \; C) \frac{1}{2} \; \; D) \frac{2}{3}


Diameter of the sphere = 2r

The radius of the cone is x and height is y.

Volume of cone is given by

V=13πx2y=13πy(2ry)y=13π(2ry2y3)dVdy=13π(4ry3y2)dVdy=013π(4ry3y2)=0y(4r3y)=0y=43r,0V=\frac{1}{3}\pi {{x}^{2}}y=\frac{1}{3}\pi y(2r-y)y=\frac{1}{3}\pi (2r{{y}^{2}}-{{y}^{3}})\\ \frac{dV}{dy}=\frac{1}{3}\pi (4ry-3{{y}^{2}})\\ \frac{dV}{dy}=0\\ \frac{1}{3}\pi (4ry-3{{y}^{2}})=0\\ y(4r-3y)=0\\ y=\frac{4}{3}r,\,0


d2Vdy2=13π(4r6y),\frac{{{d}^{2}}V}{d{{y}^{2}}}=\frac{1}{3}\pi (4r-6y),

Put y=43ry=\frac{4}{3}r d2Vdy2=13π(4r6×43r)\frac{{{d}^{2}}V}{d{{y}^{2}}}=\frac{1}{3}\pi \,\left( 4r-6\times \frac{4}{3}r \right) = negative value

So, volume of cone is maximum at y=43ry=\frac{4}{3}r

so, HeightRadius=yr=43.\frac{\text{Height}}{\text{Radius}}=\frac{y}{r}=\frac{4}{3}.

Question 4: If ab=2a+3b,a>0,b>0ab=2a+3b,\,a>0,\,\,b>0 then the minimum value of ab is A) 12

B) 24

C) 1 / 4

D) None of these


ab=2a+3b(a3)b=2ab=2aa3ab=2a+3b\\ (a-3)b=2a\\ b=\frac{2a}{a-3}


z=ab=2a2a3dzda=2[(a3)2aa2](a3)2=2[a26a](a3)2z=ab=\frac{2{{a}^{2}}}{a-3}\\ \frac{dz}{da}=\frac{2[(a-3)2a-{{a}^{2}}]}{{{(a-3)}^{2}}}=\frac{2[{{a}^{2}}-6a]}{{{(a-3)}^{2}}}

Put dzda=0,a26a=0,a=0,6\frac{dz}{da}=0, \\ {{a}^{2}}-6a=0, \\ a=0,\,6

Now at a=6,d2zda2=+vea=6,\frac{{{d}^{2}}z}{d{{a}^{2}}}=+ve

When a = 6, b = 4;

(ab)min. = 6 × 4 = 24.

Question 5: If y=cot1(cos2x)1/2,y={{\cot }^{-1}}{{(\cos 2x)}^{1/2}}, then find the value of dydx\frac{dy}{dx} at x=π6x=\frac{\pi }{6}.


dydx=11+cos2x×12cos2x×2sin2x=tanxcos2x(dydx)x=π/6=1/31/2=23.\frac{dy}{dx}=\frac{-1}{1+\cos 2x}\times \frac{1}{2\sqrt{\cos 2x}}\times -2\sin 2x\\=\frac{\tan x}{\sqrt{\cos 2x}}\\ \Rightarrow {{\left( \frac{dy}{dx} \right)}_{x=\pi /6}}=\frac{1/\sqrt{3}}{\sqrt{1/2}}=\sqrt{\frac{2}{3}}.

Question 6: If f(x+y)=f(x).f(y)f(x+y)=f(x).f(y) for all x and y and f(5)=2,f(0)=3,f(5)=2, f'(0)=3, then find f(5)f'(5).


Let x=5,y=0f(5+0)=f(5).f(0)f(5)=f(5)f(0)f(0)=1x=5,\,\,\,y=0\Rightarrow f(5+0)=f(5).f(0) \\ f(5)=f(5)f(0)\Rightarrow f(0)=1


f(5)=limh0f(5+h)f(5)h=limh0f(5)f(h)f(5)h=lim2h0[f(h)1h]{ because f(5)=2}=2limh0.[f(h)f(0)h]=2×f(0)=2×3=6.f'(5)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(5+h)-f(5)}{h}\\ =\underset{h\to 0}{\mathop{\lim }}\,\frac{f(5)f(h)-f(5)}{h}=\underset{h\to 0}{\mathop{\lim 2}}\,\left[ \frac{f(h)-1}{h} \right] \left\{\text \ because \ f(5)=2 \right\}\\ =2\underset{h\to 0}{\mathop{\lim }}\,.\left[ \frac{f(h)-f(0)}{h} \right]=2\times f'(0)=2\times 3=6.

Question 7: If xexy=y+sin2x,x{{e}^{xy}}=y+{{\sin }^{2}}x, then at x=0,dydx=x=0,\frac{dy}{dx}=


We are given that xexy=y+sin2xx{{e}^{xy}}=y+{{\sin }^{2}}x

When x=0, we get y=0

Differentiating both sides with respect to x, we get

exy+xexy[xdydx+y]=dydx+2sinxcosx{{e}^{xy}}+x{{e}^{xy}}\left[ x\frac{dy}{dx}+y \right]=\frac{dy}{dx}+2\sin x\cos x

Putting x = 0, y = 0, we get


Question 8: If y=f(2x1x2+1) and f(x)=sinx2,y=f\left( \frac{2x-1}{{{x}^{2}}+1} \right) \text \ and \ {f}'(x)=\sin {{x}^{2}}, then find dydx=\frac{dy}{dx}=.


Let t=2x1x2+1,t=\frac{2x-1}{{{x}^{2}}+1}, then

dydx=f(t).dtdx=sint2ddx(2x1x2+1)=2(1+xx2)(1+x2)2.sin(2x1x2+1)2.\frac{dy}{dx}=f'(t).\frac{dt}{dx}\\ =\sin {{t}^{2}}\frac{d}{dx}\left( \frac{2x-1}{{{x}^{2}}+1} \right)=\frac{2(1+x-{{x}^{2}})}{{{(1+{{x}^{2}})}^{2}}}.\sin {{\left( \frac{2x-1}{{{x}^{2}}+1} \right)}^{2}}.

Question 9: If x=secθcosθx=\sec \theta -\cos \theta and y=secnθcosnθ,y={{\sec }^{n}}\theta -{{\cos }^{n}}\theta , then which of the following option is correct.

A)(x2+4) (dydx)2=n2(y2+4)B)(x2+4) (dydx)2=x2(y2+4)C)(x2+4) (dydx)2=(y2+4)D) None of theseA) ({{x}^{2}}+4)\text{ }{{\left( \frac{dy}{dx} \right)}^{2}}={{n}^{2}}({{y}^{2}}+4)\\ B) ({{x}^{2}}+4)\text{ }{{\left( \frac{dy}{dx} \right)}^{2}}={{x}^{2}}({{y}^{2}}+4)\\ C) ({{x}^{2}}+4)\text{ }{{\left( \frac{dy}{dx} \right)}^{2}}=({{y}^{2}}+4) \\ D) \text \ None \ of \ these


dydx=dy/dθdx/dθ=nsecnθtanθ+ncosn1θsinθsecθtanθ+sinθ=n(secnθ+cosnθ)secθ+cosθ(dydx)2=n2(secnθ+cosnθ)2(secθ+cosθ)2=n2[(secnθcosnθ)2+4secnθcosnθ](secθcosθ)2+4secθ.cosθ=n2(y2+4)x2+4(x2+4) (dydx)2=n2(y2+4).\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }\\ =\frac{n{{\sec }^{n}}\theta \tan \theta +n{{\cos }^{n-1}}\theta \sin \theta }{\sec \theta \tan \theta +\sin \theta }\\ =\frac{n({{\sec }^{n}}\theta +{{\cos }^{n}}\theta )}{\sec \theta +\cos \theta }\\ {{\left( \frac{dy}{dx} \right)}^{2}}=\frac{{{n}^{2}}{{({{\sec }^{n}}\theta +{{\cos }^{n}}\theta )}^{2}}}{{{(\sec \theta +\cos \theta )}^{2}}}\\ =\frac{{{n}^{2}}[{{({{\sec }^{n}}\theta -{{\cos }^{n}}\theta )}^{2}}+4{{\sec }^{n}}\theta {{\cos }^{n}}\theta ]}{{{(\sec \theta -\cos \theta )}^{2}}+4\sec \theta .\cos \theta }=\frac{{{n}^{2}}({{y}^{2}}+4)}{{{x}^{2}}+4}\\ ({{x}^{2}}+4)\text{ }{{\left( \frac{dy}{dx} \right)}^{2}}={{n}^{2}}({{y}^{2}}+4).

Option (A) is correct.

Question 10: If y2=p(x){{y}^{2}}=p(x) is a polynomial of degree three, then 2ddx{y3.d2ydx2}2\frac{d}{dx}\left\{ {{y}^{3}}.\frac{{{d}^{2}}y}{d{{x}^{2}}} \right\} is


2ydydx=p(x)2dydx=p(x)y2d2ydx2=yp(x)p(x)yy22y3d2ydx2=y2p(x)ydydxp(x)=p(x)p(x)12{p(x)}22ddx(y3d2ydx2)=p(x)p(x)+p(x)p(x)p(x)p(x)=p(x)p(x).2y\frac{dy}{dx}=p'(x)\Rightarrow 2\frac{dy}{dx}=\frac{{p}'(x)}{y}\\ \Rightarrow 2\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{y{p}”(x)-{p}'(x){y}’}{{{y}^{2}}}\\ 2{{y}^{3}}\frac{{{d}^{2}}y}{d{{x}^{2}}}={{y}^{2}}{p}”(x)-y\frac{dy}{dx}{p}'(x)\\ =p(x)\,{p}”(x)-\frac{1}{2}{{\{{p}'(x)\}}^{2}} \\ 2\frac{d}{dx}\left( {{y}^{3}}\frac{{{d}^{2}}y}{d{{x}^{2}}} \right)={p}'(x){p}”(x)+p(x){p}”'(x)-{p}'(x){p}”(x)\\ =p(x){p}”'(x).

Question 11: At what points of the curve y=23x3+12x2,y=\frac{2}{3}{{x}^{3}}+\frac{1}{2}{{x}^{2}}, tangent makes the equal angle with axis?


y=23x3+12x2dydx=2x2+x(i)y=\frac{2}{3}{{x}^{3}}+\frac{1}{2}{{x}^{2}}\\ \frac{dy}{dx}=2{{x}^{2}}+x \rightarrow (i)

Now tangent makes equal angle with axis

y = 450 and y = -450

dydx=tan(±45o)=±tan(45o)=±1\frac{dy}{dx}=\tan (\pm {{45}^{o}})=\pm \tan ({{45}^{o}})=\pm 1

From equation (i), 2x2+x=12{{x}^{2}}+x=1 [taking positive sign] 2x2+x1=0(2x1)(x+1)=0x=12,12{{x}^{2}}+x-1=0\\ (2x-1)\,(x+1)=0 \\ x=\frac{1}{2},-1

From the given curve, when x=12,y=23.18+12.14=112+18=524x=\frac{1}{2}, y=\frac{2}{3}.\frac{1}{8}+\frac{1}{2}.\frac{1}{4}=\frac{1}{12}+\frac{1}{8}=\frac{5}{24}

and when x = -1,

y=23(1)+12.1=23+12=16y=\frac{2}{3}(-1)+\frac{1}{2}.1 \\ =-\frac{2}{3}+\frac{1}{2}=-\frac{1}{6}

Therefore, required points are

(12,524) and (1,16).\left( \frac{1}{2},\,\frac{5}{24} \right) \text \ and \ \left( -1,\,-\frac{1}{6} \right).

Question 12: The function f(x)=1xt(et1)(t1)(t2)3(t3)5dtf(x)=\int\limits_{-1}^{x}{t({{e}^{t}}-1)(t-1){{(t-2)}^{3}}{{(t-3)}^{5}}}dt has a local minimum at x =


f(x)=1xt(et1)(t1)(t2)3(t3)5dt   therefore  f(x)=x(ex1)(x1)(x2)3(x3)5f(x)=\int_{-1}^{x}{t({{e}^{t}}-1)(t-1){{(t-2)}^{3}}{{(t-3)}^{5}}}dt \; \text \ therefore \; f'(x)=x({{e}^{x}}-1)(x-1){{(x-2)}^{3}}{{(x-3)}^{5}}

For local minima, slope i.e. f′ (x) should change sign from −ve to +ve.

f(x)=0x=0,1,2,3f’ (x)=0\Rightarrow x=0,\,1,\,2,\,3

If x = 0 + h; f′ (x) = (+) (+) (−) (−1) (−1) = −ve

Hence at x = 0 neither maxima or minima.

If x = 1 − h; f′ (x) = (+) (+) (−) (−1) (−1) = −ve

If x = 1 + h; f′ (x) = (+) (+) (+) (−1) (−1) = +ve

Hence at x = 1 there is a local minima.

If x = 2 − h; f′ (x) = (+) (+1) (+) (−) (−) = +ve

If x = 2 + h; f′ (x) = (+) (+) (+) (+) (−1) = −ve

Hence at x = 2 there is a local maxima.

If x = 3 − h; f′ (x) = (+) (+) (+) (+) (−) = −ve

If x = 3 + h; f′ (x) = (+) (+) (+) (+) (+) = +ve

Hence at x = 3 there is a local minima.

Question 13: Check whether the function f(x)=ln(π+x)ln(e+x)f(x)=\frac{\text{ln}(\pi +x)}{\text{ln}(e+x)} is increasing or decreasing.


Let f(x)=ln(π+x)ln(e+x) therefore f(x)=ln(e+x)×1π+xln(π+x)1e+xln2(e+x)=(e+x)ln(e+x)(π+x)ln(π+x)ln2(e+x)×(e+x)(π+x)f(x)<0  x0 ,  { because π>e}f(x)=\frac{\ln (\pi +x)}{\ln (e+x)}\\ \text \ therefore \ f'(x)=\frac{\ln (e+x)\times \frac{1}{\pi +x}-\ln (\pi +x)\frac{1}{e+x}}{{{\ln }^{2}}(e+x)}\\ =\frac{(e+x)\ln (e+x)-(\pi +x)\ln (\pi +x)}{{{\ln }^{2}}(e+x)\times (e+x)(\pi +x)}\\ \Rightarrow f'(x)<0 \ \forall \ x\ge 0\ ,\ \ \{\text \ because \ \pi >e\}

Hence, f (x) is decreasing in [0, ∞).

Question 14: Let f(x)={{xαlnx,x>00,x=0}f(x)=\left\{\begin{Bmatrix} {x}^{\alpha}lnx, &x>0 \\ 0,& x=0 \end{Bmatrix}\right. Rolle’s theorem is applicable to f for x[0,1],x\in [0,1], if α=\alpha = is

A) 2

B) 1

C) 0

D) 1 / 2


For Rolle’s theorem to be applicable to f, for x[0,1],x\in [0,1], we should have

(i) f (1) = f (0),

(ii) f is continuous for x[0,1],x\in [0,1], and f is differentiable for x[0,1],x\in [0,1],

From (i), f (1) = 0, which is true.

From (ii), 0=f(0)=f(0+)=limx0+xαlnx0=f(0)=f({{0}_{+}})=\underset{x\to {{0}_{+}}}{\mathop{\lim }}\,{{x}^{\alpha }}\ln x which is true only for positive values of α\alpha > 0, thus option (4) is correct.

Question 15: The distance travelled s (in metre) by a particle in t seconds is given by, s=t3+2t2+t.s={{t}^{3}}+2{{t}^{2}}+t. What is the speed of the particle after 1 second?


s=t3+2t2+tv=dsdt=3t2+4t+1s={{t}^{3}}+2{{t}^{2}}+t\\ v=\frac{ds}{dt}=3{{t}^{2}}+4t+1

Speed of the particle after 1 second is given by

v(t=1)=(dsdt)(t=1)=3×12+4×1+1=8cm/sec.{{v}_{(t=1)}}={{\left( \frac{ds}{dt} \right)}_{(t=1)}}=3\times {{1}^{2}}+4\times 1+1=8\,cm/sec.