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, Rolleβs 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.
JEE Main Applications of Derivatives Past Year Questions With Solutions
Question 1:
\(\begin{array}{l}\text{Given function}\ f(x)=\left( \frac{{{e}^{2x}}-1}{{{e}^{2x}}+1} \right)\ \text{is}\end{array} \)
A) Increasing
B) Decreasing
C) Even
D) None of these
Solution:
\(\begin{array}{l}f(x)=\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1} \\ f(-x)=\frac{{{e}^{-2x}}-1}{{{e}^{-2x}}+1}=\frac{1-{{e}^{2x}}}{1+{{e}^{2x}}}\\ f(x)=-\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1}=-f(x)\\\end{array} \)
f(x) is an odd function.
Again,
\(\begin{array}{l}f(x)=\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1} \Rightarrow {f}'(x)=\frac{4{{e}^{2x}}}{{{(1+{{e}^{2x}})}^{2}}}>0\,\forall \,n\in R\end{array} \)
f(x) is an increasing function.
Question 2: If f(x) = 2x3 – 3x2 – 12x + 5 and x belongs to [-2, 4], then the maximum value of function is at the following value of x
A) 2
B) 1
C) 2
D) 4
Solution:
\(\begin{array}{l}f'(x)=6{{x}^{2}}-6x-12 \\ f'(x)=0\Rightarrow (x-2)(x+1)=0\Rightarrow x=-1,\,2\end{array} \)
Here,
\(\begin{array}{l}f(4)=128-48-48+5=37\\ f(-1)=-2-3+12+5=12\\ f(2)=16-12-24+5=-15\\ f(-2)=-16-12+24+5=1\end{array} \)
Therefore, the maximum value of function is 37 at x = 4.
Hence option D is the answer.
Question 3: The ratio of the height of cone of maximum volume inscribed in a sphere to its radius is
A) 3/4
B) 4/3
C) 1/2
D) 2/3
Solution:
Diameter of the sphere = 2r
The radius of the cone is x and height is y.
Volume of cone is given by
\(\begin{array}{l}V=\frac{1}{3}\pi {{x}^{2}}y=\frac{1}{3}\pi y(2r-y)y=\frac{1}{3}\pi (2r{{y}^{2}}-{{y}^{3}})\\ \frac{dV}{dy}=\frac{1}{3}\pi (4ry-3{{y}^{2}})\\ \frac{dV}{dy}=0\\ \frac{1}{3}\pi (4ry-3{{y}^{2}})=0\\ y(4r-3y)=0\\ y=\frac{4}{3}r,\,0\end{array} \)
Now,
\(\begin{array}{l}\frac{{{d}^{2}}V}{d{{y}^{2}}}=\frac{1}{3}\pi (4r-6y),\end{array} \)
Put
\(\begin{array}{l}y=\frac{4}{3}r\end{array} \)
\(\begin{array}{l}\frac{{{d}^{2}}V}{d{{y}^{2}}}=\frac{1}{3}\pi \,\left( 4r-6\times \frac{4}{3}r \right)\end{array} \)
= negative value
So, volume of cone is maximum at
\(\begin{array}{l}y=\frac{4}{3}r\end{array} \)
so,
\(\begin{array}{l}\frac{\text{Height}}{\text{Radius}}=\frac{y}{r}=\frac{4}{3}.\end{array} \)
Question 4: If ab = 2a + 3b, a > 0, b > 0 then the minimum value of ab is
A) 12
B) 24
C) 1/4
D) None of these
Solution:
\(\begin{array}{l}ab=2a+3b\\ (a-3)b=2a\\ b=\frac{2a}{a-3}\end{array} \)
Now
\(\begin{array}{l}z=ab=\frac{2{{a}^{2}}}{a-3}\\ \frac{dz}{da}=\frac{2[(a-3)2a-{{a}^{2}}]}{{{(a-3)}^{2}}}=\frac{2[{{a}^{2}}-6a]}{{{(a-3)}^{2}}}\end{array} \)
Put
\(\begin{array}{l}\frac{dz}{da}=0, \\ {{a}^{2}}-6a=0, \\ a=0,\,6\end{array} \)
Now at
\(\begin{array}{l}a=6,\frac{{{d}^{2}}z}{d{{a}^{2}}}=+ve\end{array} \)
When a = 6, b = 4;
(ab)min. = 6 Γ 4 = 24.
Question 5: If y = cot-1 (cos 2x)1/2, then find the value of dy/dx at x = Ο/6.
Solution:
\(\begin{array}{l}\frac{dy}{dx}=\frac{-1}{1+\cos 2x}\times \frac{1}{2\sqrt{\cos 2x}}\times -2\sin 2x\\=\frac{\tan x}{\sqrt{\cos 2x}}\\ \Rightarrow {{\left( \frac{dy}{dx} \right)}_{x=\pi /6}}=\frac{1/\sqrt{3}}{\sqrt{1/2}}=\sqrt{\frac{2}{3}}.\end{array} \)
Question 6: If f(x + y) = f(x).f(y) for all x and y and f(5) = 2, f'(0) = 3, then find f'(5).
Solution:
Let x = 5, y = 0,
βΒ f(5+0) = f(5) f(0)
βΒ f(5) = f(5) f(0)
βΒ f(0) = 1
Therefore,
\(\begin{array}{l}f'(5)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(5+h)-f(5)}{h}\\ =\underset{h\to 0}{\mathop{\lim }}\,\frac{f(5)f(h)-f(5)}{h}=\underset{h\to 0}{\mathop{\lim 2}}\,\left[ \frac{f(h)-1}{h} \right] \left\{\text \ because \ f(5)=2 \right\}\\ =2\underset{h\to 0}{\mathop{\lim }}\,.\left[ \frac{f(h)-f(0)}{h} \right]=2\times f'(0)=2\times 3=6.\end{array} \)
Question 7: If xexy = y + sin2x, then at x = 0, dy/dx =
Solution:
We are given that
\(\begin{array}{l}x{{e}^{xy}}=y+{{\sin }^{2}}x\end{array} \)
When x = 0, we get y = 0
Differentiating both sides with respect to x, we get;
\(\begin{array}{l}{{e}^{xy}}+x{{e}^{xy}}\left[ x\frac{dy}{dx}+y \right]=\frac{dy}{dx}+2\sin x\cos x\end{array} \)
Putting x = 0, y = 0, we get;
dy/dx = 1.
Question 8: If y = f(2x – 1)/(x2 + 1) and f ‘(x) = sin x2 then find dy/dx.
Solution:
Let t = (2x – 1)/(x2 + 1) then
\(\begin{array}{l}\frac{dy}{dx}=f'(t).\frac{dt}{dx}\\ =\sin {{t}^{2}}\frac{d}{dx}\left( \frac{2x-1}{{{x}^{2}}+1} \right)=\frac{2(1+x-{{x}^{2}})}{{{(1+{{x}^{2}})}^{2}}}.\sin {{\left( \frac{2x-1}{{{x}^{2}}+1} \right)}^{2}}.\end{array} \)
Question 9: If x = sec ΞΈ – cos ΞΈ and y = secn ΞΈ – cosn ΞΈ, then which of the following option is correct?
\(\begin{array}{l}A) ({{x}^{2}}+4)\text{ }{{\left( \frac{dy}{dx} \right)}^{2}}={{n}^{2}}({{y}^{2}}+4)\\ B) ({{x}^{2}}+4)\text{ }{{\left( \frac{dy}{dx} \right)}^{2}}={{x}^{2}}({{y}^{2}}+4)\\ C) ({{x}^{2}}+4)\text{ }{{\left( \frac{dy}{dx} \right)}^{2}}=({{y}^{2}}+4) \\ D)\ \text{None of these}\end{array} \)
Solution:
\(\begin{array}{l}\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }\\ =\frac{n{{\sec }^{n}}\theta \tan \theta +n{{\cos }^{n-1}}\theta \sin \theta }{\sec \theta \tan \theta +\sin \theta }\\ =\frac{n({{\sec }^{n}}\theta +{{\cos }^{n}}\theta )}{\sec \theta +\cos \theta }\\ {{\left( \frac{dy}{dx} \right)}^{2}}=\frac{{{n}^{2}}{{({{\sec }^{n}}\theta +{{\cos }^{n}}\theta )}^{2}}}{{{(\sec \theta +\cos \theta )}^{2}}}\\ =\frac{{{n}^{2}}[{{({{\sec }^{n}}\theta -{{\cos }^{n}}\theta )}^{2}}+4{{\sec }^{n}}\theta {{\cos }^{n}}\theta ]}{{{(\sec \theta -\cos \theta )}^{2}}+4\sec \theta .\cos \theta }=\frac{{{n}^{2}}({{y}^{2}}+4)}{{{x}^{2}}+4}\\ ({{x}^{2}}+4)\text{ }{{\left( \frac{dy}{dx} \right)}^{2}}={{n}^{2}}({{y}^{2}}+4).\end{array} \)
Option (A) is correct.
Question 10: If y2 = p(x) is a polynomial of degree three, then 2(d/dx)(y3. d2y/dx2) is
Solution:
\(\begin{array}{l}2y\frac{dy}{dx}=p'(x)\Rightarrow 2\frac{dy}{dx}=\frac{{p}'(x)}{y}\\ \Rightarrow 2\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{y{p}”(x)-{p}'(x){y}’}{{{y}^{2}}}\\ 2{{y}^{3}}\frac{{{d}^{2}}y}{d{{x}^{2}}}={{y}^{2}}{p}”(x)-y\frac{dy}{dx}{p}'(x)\\ =p(x)\,{p}”(x)-\frac{1}{2}{{\{{p}'(x)\}}^{2}} \\ 2\frac{d}{dx}\left( {{y}^{3}}\frac{{{d}^{2}}y}{d{{x}^{2}}} \right)={p}'(x){p}”(x)+p(x){p}”'(x)-{p}'(x){p}”(x)\\ =p(x){p}”'(x).\end{array} \)
Question 11: At what points of the curve y = (2/3)x3 + (1/2)x2, tangent makes the equal angle with axis?
Solution:
\(\begin{array}{l}y=\frac{2}{3}{{x}^{3}}+\frac{1}{2}{{x}^{2}}\\ \frac{dy}{dx}=2{{x}^{2}}+x \rightarrow (i)\end{array} \)
Now tangent makes equal angle with axis
y = 450 and y = -450
\(\begin{array}{l}\frac{dy}{dx}=\tan (\pm {{45}^{o}})=\pm \tan ({{45}^{o}})=\pm 1\end{array} \)
From equation (i),
\(\begin{array}{l}2{{x}^{2}}+x=1\end{array} \)
[taking positive sign]
\(\begin{array}{l}2{{x}^{2}}+x-1=0\\ (2x-1)\,(x+1)=0 \\ x=\frac{1}{2},-1\end{array} \)
From the given curve, when
\(\begin{array}{l}x=\frac{1}{2}, y=\frac{2}{3}.\frac{1}{8}+\frac{1}{2}.\frac{1}{4}=\frac{1}{12}+\frac{1}{8}=\frac{5}{24}\end{array} \)
and when x = -1,
\(\begin{array}{l}y=\frac{2}{3}(-1)+\frac{1}{2}.1 \\ =-\frac{2}{3}+\frac{1}{2}=-\frac{1}{6}\end{array} \)
Therefore, the required points are (1/2, 5/24) and (-1, -1/6).
Question 12: The function
\(\begin{array}{l}f(x)=\int\limits_{-1}^{x}{t({{e}^{t}}-1)(t-1){{(t-2)}^{3}}{{(t-3)}^{5}}}dt\end{array} \)
has a local minimum at x =
Solution:
\(\begin{array}{l}f(x)=\int_{-1}^{x}{t({{e}^{t}}-1)(t-1){{(t-2)}^{3}}{{(t-3)}^{5}}}dt\end{array} \)
\(\begin{array}{l}f'(x)=x({{e}^{x}}-1)(x-1){{(x-2)}^{3}}{{(x-3)}^{5}}\end{array} \)
For local minima, slope, i.e. fβ² (x) should change the sign from βve to +ve.
\(\begin{array}{l}f’ (x)=0\Rightarrow x=0,\,1,\,2,\,3\end{array} \)
If x = 0 + h; fβ² (x) = (+) (+) (β) (β1) (β1) = βve
Hence at x = 0, neither maxima or minima.
If x = 1 β h; fβ² (x) = (+) (+) (β) (β1) (β1) = βve
If x = 1 + h; fβ² (x) = (+) (+) (+) (β1) (β1) = +ve
Hence at x = 1, there is a local minima.
If x = 2 β h; fβ² (x) = (+) (+1) (+) (β) (β) = +ve
If x = 2 + h; fβ² (x) = (+) (+) (+) (+) (β1) = βve
Hence at x = 2, there is a local maxima.
If x = 3 β h; fβ² (x) = (+) (+) (+) (+) (β) = βve
If x = 3 + h; fβ² (x) = (+) (+) (+) (+) (+) = +ve
Hence at x = 3, there is a local minima.
Question 13: Check whether the function
\(\begin{array}{l}f(x)=\frac{\text{ln}(\pi +x)}{\text{ln}(e+x)}\end{array} \)
is increasing or decreasing.
Solution:
Let
\(\begin{array}{l}f(x)=\frac{\ln (\pi +x)}{\ln (e+x)}\\ \therefore \ f'(x)=\frac{\ln (e+x)\times \frac{1}{\pi +x}-\ln (\pi +x)\frac{1}{e+x}}{{{\ln }^{2}}(e+x)}\\ =\frac{(e+x)\ln (e+x)-(\pi +x)\ln (\pi +x)}{{{\ln }^{2}}(e+x)\times (e+x)(\pi +x)}\\ \Rightarrow f'(x)<0 \ \forall \ x\ge 0\ ,\ \{\because \ \pi >e\}\end{array} \)
Hence, f (x) is decreasing in [0, β).
Question 14: Let
\(\begin{array}{l}f(x)=\left\{\begin{Bmatrix} {x}^{\alpha}lnx, &x>0 \\ 0,& x=0 \end{Bmatrix}\right.\end{array} \)
Rolle’s theorem is applicable to f for x β [0, 1], if Ξ± =
A) -2
B) -1
C) 0
D) 1 / 2
Solution:
For Rolle’s theorem to be applicable to f, for x β [0, 1] we should have
(i) f (1) = f (0),
(ii) f is continuous for x β [0, 1] and f is differentiable for x β [0, 1]
From (i), f (1) = 0, which is true.
From (ii),
\(\begin{array}{l}0=f(0)=f({{0}_{+}})=\underset{x\to {{0}_{+}}}{\mathop{\lim }}\,{{x}^{\alpha }}\ln x\end{array} \)
which is true only for positive values of Ξ± > 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. What is the speed of the particle after 1 second?
Solution:
\(\begin{array}{l}s={{t}^{3}}+2{{t}^{2}}+t\\ v=\frac{ds}{dt}=3{{t}^{2}}+4t+1\end{array} \)
Speed of the particle after 1 second is given by
\(\begin{array}{l}{{v}_{(t=1)}}={{\left( \frac{ds}{dt} \right)}_{(t=1)}}=3\times {{1}^{2}}+4\times 1+1=8\,cm/sec.\end{array} \)
Application of Derivatives Important JEE Main Questions 1
Application of Derivatives Important JEE Main Questions 2
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JEE Advanced Maths Method of Differentiation Previous Year Questions with Solutions
Applications of Derivatives – Top 10 Most Important and Expected JEE Questions
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