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Differentiation

JEE Mathematics · 66 questions · Page 5 of 7 · Click an option or "Show Solution" to reveal answer

Q41
Let ff and gg be the twice differentiable functions on R\mathbb{R} such that f′′(x)=g′′(x)+6xf''(x)=g''(x)+6x f′(1)=4g′(1)−3=9f'(1)=4g'(1)-3=9 f(2)=3g(2)=12f(2)=3g(2)=12. Then which of the following is NOT true?
A g(−2)−f(−2)=20g(-2)-f(-2)=20
B There exists x0∈(1,3/2)x_0\in(1,3/2) such that f(x0)=g(x0)f(x_0)=g(x_0)
C ∣f′(x)−g′(x)∣<6⇒−1<x<1|f'(x)-g'(x)| < 6\Rightarrow -1 < x < 1
D If −1<x<2-1 < x < 2, then ∣f(x)−g(x)∣<8|f(x)-g(x)| < 8
Correct Answer
Option D
Solution
f′′(x)=g′′(x)+6xf''(x) = g''(x) + 6x
⇒f′(x)=g′(x)+3x2+C\Rightarrow f'(x) = g'(x) + 3{x^2} + C
f′(1)=g′(1)+3+Cf'(1) = g'(1) + 3 + C
⇒g=3+3+C⇒C=3\Rightarrow g = 3 + 3 + C \Rightarrow C = 3
⇒f′(x)=g′(x)+3x2+3\Rightarrow f'(x) = g'(x) + 3{x^2} + 3
⇒f(x)=g(x)+x2+3x+C′\Rightarrow f(x) = g(x) + {x^2} + 3x + C'
x=2x = 2
f(2)=g(2)+14+C′f(2) = g(2) + 14 + C'
12=4+14+C′12 = 4 + 14 + C'
⇒C′=−6\Rightarrow C' = - 6
⇒f(x)=g(2)+x3+3x−6\Rightarrow f(x) = g(2) + {x^3} + 3x - 6
f(−2)=g(−2)−8−6−6f( - 2) = g( - 2) - 8 - 6 - 6
g(−2)−f(−2)=20g( - 2) - f( - 2) = 20
f′(x)−g′(x)=3x2+3f'(x) - g'(x) = 3{x^2} + 3
x∈(−1,1)x \in ( - 1,1)
3x2+3∈(0,6)3{x^2} + 3 \in (0,6)
⇒f′(x)−g′(x)∈(0,6)\Rightarrow f'(x) - g'(x) \in (0,6)
f(x)−g(x)=x3+3x−6f(x) - g(x) = {x^3} + 3x - 6

At

x=−1x = - 1
∣f(−1)−g(−1)∣=10|f( - 1) - g( - 1)| = 10

∴\therefore Option (4) is false.

Q42
Let y(x)=(1+x)(1+x2)(1+x4)(1+x8)(1+x16)y(x) = (1 + x)(1 + {x^2})(1 + {x^4})(1 + {x^8})(1 + {x^{16}}). Then y′−y′′y' - y'' at x=−1x = - 1 is equal to
A 496
B 976
C 464
D 944
Correct Answer
Option A
Solution
y=1−x321−x=1+x+x2+x3+…+x31y′=1+2x+3x2+…+31x30y′(−1)=1−2+3−4+…+31=16y′′(x)=2+6x+12x2+…+31.30x29y′′(−1)=2−6+12…31.30=480y′′(−1)−y′(−1)=−496\begin{aligned} & y=\frac{1-x^{32}}{1-x}=1+x+x^2+x^3+\ldots+x^{31} \\\\ & y^{\prime}=1+2 x+3 x^2+\ldots+31 x^{30} \\\\ & y^{\prime}(-1)=1-2+3-4+\ldots+31=16 \\\\ & y^{\prime \prime}(x)=2+6 x+12 x^2+\ldots+31.30 x^{29} \\\\ & y^{\prime \prime}(-1)=2-6+12 \ldots 31.30=480 \\\\ & y^{\prime \prime}(-1)-y^{\prime}(-1)=-496 \end{aligned}
Q43
If f(x)=x3−x2f′(1)+xf′′(2)−f′′′(3),x∈Rf(x) = {x^3} - {x^2}f'(1) + xf''(2) - f'''(3),x \in \mathbb{R}, then
A 2f(0)−f(1)+f(3)=f(2)2f(0) - f(1) + f(3) = f(2)
B f(1)+f(2)+f(3)=f(0)f(1) + f(2) + f(3) = f(0)
C f(3)−f(2)=f(1)f(3) - f(2) = f(1)
D 3f(1)+f(2)=f(3)3f(1) + f(2) = f(3)
Correct Answer
Option A
Solution
f(x)=x3−x2f′(1)+xf′′(2)−f′′′(3),x∈Rf(x)=x^3-x^2 f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime \prime}(3), x \in R

Let f′(1)=a,f′′(2)=b,f′′′(3)=c\mathrm{f}^{\prime}(1)=\mathrm{a}, \mathrm{f}^{\prime \prime}(2)=\mathrm{b}, \mathrm{f}^{\prime \prime \prime}(3)=\mathrm{c}

f(x)=x3−ax2+bx−cf′(x)=3x2−2ax+bf′′(x)=6x−2af′′′(x)=6c=6,a=3,b=6f(x)=x3−3x2+6x−6f(1)=−2,f(2)=2,f(3)=12,f(0)=−62f(0)−f(1)+f(3)=2=f(2)\begin{aligned} & f(x)=x^3-a x^2+b x-c \\\\ & f^{\prime}(x)=3 x^2-2 a x+b \\\\ & f^{\prime \prime}(x)=6 x-2 a \\\\ & f^{\prime \prime \prime}(x)=6 \\\\ & c=6, a=3, b=6 \\\\ & f(x)=x^3-3 x^2+6 x-6 \\\\ & f(1)=-2, f(2)=2, f(3)=12, f(0)=-6 \\\\ & 2 f(0)-f(1)+f(3)=2=f(2) \end{aligned}
Q44
For the differentiable function f:R−{0}→Rf: \mathbb{R}-\{0\} \rightarrow \mathbb{R}, let 3f(x)+2f(1x)=1x−103 f(x)+2 f\left(\dfrac{1}{x}\right)=\dfrac{1}{x}-10, then ∣f(3)+f′(14)∣\left|f(3)+f^{\prime}\left(\dfrac{1}{4}\right)\right| is equal to
A 13
B 295\dfrac{29}{5}
C 335\dfrac{33}{5}
D 7
Correct Answer
Option A
Solution

Given the equation:

3f(x)+2f(1x)=1x−103f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10

Replace

xx

with

1x\frac{1}{x}

in the original equation:

3f(1x)+2f(x)=x−103f\left(\frac{1}{x}\right) + 2f(x) = x - 10

Now, we have two equations:

3f(x)+2f(1x)=1x−103f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10
3f(1x)+2f(x)=x−103f\left(\frac{1}{x}\right) + 2f(x) = x - 10

By adding the two equations, we can find

f(x)f(x)

:

5f(x)=3x−2x−105f(x) = \frac{3}{x} - 2x - 10

Now, let's differentiate both sides with respect to

xx

:

5f′(x)=−3x2−25f'(x) = -\frac{3}{x^2} - 2

Now, we can find the values for

f(3)f(3)

and

f′(14)f'\left(\frac{1}{4}\right)

:

f(3)=15(1−6−10)=−3f(3) = \frac{1}{5}(1 - 6 - 10) = -3
f′(14)=15(−48−2)=−10f'\left(\frac{1}{4}\right) = \frac{1}{5}(-48 - 2) = -10

Finally, calculate the expression we are interested in :

∣f(3)+f′(14)∣=∣−3−10∣=13\left|f(3) + f'\left(\frac{1}{4}\right)\right| = \left|-3 - 10\right| = 13
Q45
Let f(x)=sin⁡x+cos⁡x−2sin⁡x−cos⁡x,x∈[0,π]−{π4}f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}, x \in[0, \pi]-\left\{\frac{\pi}{4}\right\}. Then f(7π12)f′′(7π12)f\left(\dfrac{7 \pi}{12}\right) f^{\prime \prime}\left(\dfrac{7 \pi}{12}\right) is equal to
A 233\dfrac{2}{3 \sqrt{3}}
B 29\dfrac{2}{9}
C −133\dfrac{-1}{3 \sqrt{3}}
D −23\dfrac{-2}{3}
Correct Answer
Option B
Solution
f(x)=sin⁡x+cos⁡x−2sin⁡x−cos⁡xf(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}
=12sin⁡x+12cos⁡x−112sin⁡x−12cos⁡x=cos⁡(x−π4)−1sin⁡(x−π4)=−2sin⁡2(x2−π8)2sin⁡(x2−π8)cos⁡(x2−π8)\begin{aligned} & =\frac{\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x-1}{\frac{1}{\sqrt{2}} \sin x-\frac{1}{\sqrt{2}} \cos x} \\\\ & =\frac{\cos \left(x-\frac{\pi}{4}\right)-1}{\sin \left(x-\frac{\pi}{4}\right)} \\\\ & =\frac{-2 \sin ^2\left(\frac{x}{2}-\frac{\pi}{8}\right)}{2 \sin \left(\frac{x}{2}-\frac{\pi}{8}\right) \cos \left(\frac{x}{2}-\frac{\pi}{8}\right)} \end{aligned}
⇒f(x)=−tan⁡(x2−π8)⇒f′(x)=−12sec⁡2(x2−π8)\begin{aligned} & \Rightarrow f(x)=-\tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \\\\ & \Rightarrow f^{\prime}(x)=-\frac{1}{2} \sec ^2\left(\frac{x}{2}-\frac{\pi}{8}\right) \end{aligned}
⇒f′′(x)=−12⋅2sec⁡(x2−π8)⋅sec⁡(x2−π8)tan⁡(x2−π8)×12=−12sec⁡2(x2−π8)⋅tan⁡(x2−π8)\begin{aligned} & \Rightarrow f^{\prime \prime}(x)=-\frac{1}{2} \cdot 2 \sec \left(\frac{x}{2}-\frac{\pi}{8}\right) \cdot \sec \left(\frac{x}{2}-\frac{\pi}{8}\right)\tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \times \frac{1}{2} \\\\ & =-\frac{1}{2} \sec ^2\left(\frac{x}{2}-\frac{\pi}{8}\right) \cdot \tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \end{aligned}
 Now, f(7π12)f′′(7π12)=−tan⁡(7π24−π8)×−12sec⁡2(7π24−π8)×tan⁡(7π24−π8)=12tan⁡2(π6)×sec⁡2π6=12×13×43=29\begin{aligned} & \text{ Now, } f\left(\frac{7 \pi}{12}\right) f^{\prime \prime}\left(\frac{7 \pi}{12}\right) \\\\ & =-\tan \left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \times \frac{-1}{2} \sec ^2\left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \times \tan \left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \\\\ & =\frac{1}{2} \tan ^2\left(\frac{\pi}{6}\right) \times \sec ^2 \frac{\pi}{6} \\\\ & =\frac{1}{2} \times \frac{1}{3} \times \frac{4}{3}=\frac{2}{9} \end{aligned}
Q46
Let f:(0,∞)→Rf:(0, \infty) \rightarrow \mathbf{R} be a function which is differentiable at all points of its domain and satisfies the condition x2f′(x)=2xf(x)+3x^2 f^{\prime}(x)=2 x f(x)+3, with f(1)=4f(1)=4. Then 2f(2)2 f(2) is equal to :
A 19
B 23
C 29
D 39
Correct Answer
Option D
Solution
x2f′(x)−2xf(x)=3(x2f′(x)−2xf(x)(x2)2)=3(x2)2⇒ddx(f(x)x2)=3x4\begin{aligned} & x^2 f^{\prime}(x)-2 x f(x)=3 \\ & \left(\frac{x^2 f^{\prime}(x)-2 x f(x)}{\left(x^2\right)^2}\right)=\frac{3}{\left(x^2\right)^2} \\ & \Rightarrow \frac{d}{d x}\left(\frac{f(x)}{x^2}\right)=\frac{3}{x^4} \end{aligned}

Integrating both sides

f(x)x2=−1x3+Cf(x)=−1x+Cx2 put x=14=−1+C⇒C=5f(x)=−1x+5x2\begin{aligned} & \frac{\mathrm{f}(\mathrm{x})}{\mathrm{x}^2}=-\frac{1}{\mathrm{x}^3}+\mathrm{C} \\ & \mathrm{f}(\mathrm{x})=-\frac{1}{\mathrm{x}}+\mathrm{Cx}^2 \\ & \text{ put } \mathrm{x}=1 \\ & 4=-1+\mathrm{C} \Rightarrow \mathrm{C}=5 \\ & \mathrm{f}(\mathrm{x})=-\frac{1}{\mathrm{x}}+5 \mathrm{x}^2 \end{aligned}

Now 2×f(2)=2×[−12+5×22]2 \times f(2)=2 \times\left[-\dfrac{1}{2}+5 \times 2^2\right]

=39=39
Q47
If 2xy+3yx=202 x^{y}+3 y^{x}=20, then dydx\dfrac{d y}{d x} at (2,2)(2,2) is equal to :
A −(3+log⁡e164+log⁡e8)-\left(\dfrac{3+\log _{e} 16}{4+\log _{e} 8}\right)
B −(2+log⁡e83+log⁡e4)-\left(\dfrac{2+\log _{e} 8}{3+\log _{e} 4}\right)
C −(3+log⁡e82+log⁡e4)-\left(\dfrac{3+\log _{e} 8}{2+\log _{e} 4}\right)
D −(3+log⁡e42+log⁡e8)-\left(\dfrac{3+\log _{e} 4}{2+\log _{e} 8}\right)
Correct Answer
Option B
Solution

Given, 2xy+3yx=202 x^y+3 y^x=20 ..........(i) Let u=xyu=x^y On taking log both sides, we get log⁡u=ylog⁡x\log u=y \log x On differentiating both sides with respect to xx, we get

1ududx=y1x+log⁡xdydx⇒dudx=u(yx+log⁡xdydx)⇒dudx=xy(yx+log⁡xdydx).........(ii)\begin{array}{rlrl} & \frac{1}{u} \frac{d u}{d x} =y \frac{1}{x}+\log x \frac{d y}{d x} \\\\ & \Rightarrow \frac{d u}{d x} =u\left(\frac{y}{x}+\log x \frac{d y}{d x}\right) \\\\ & \Rightarrow \frac{d u}{d x} =x^y\left(\frac{y}{x}+\log x \frac{d y}{d x}\right) .........(ii) \end{array}

Also, let v=yxv=y^x On taking log both sides, we get log⁡v=xlog⁡y\log v=x \log y On differentiating both sides, we get

1vdvdx=x1ydydx+log⁡y⋅1⇒dvdx=v(xydydx+log⁡y)⇒dvdx=yx(xydydx+log⁡y)..........(iii)\begin{array}{rlrl} & \frac{1}{v} \frac{d v}{d x} =x \frac{1}{y} \frac{d y}{d x}+\log y \cdot 1 \\\\ &\Rightarrow \frac{d v}{d x} =v\left(\frac{x}{y} \frac{d y}{d x}+\log y\right) \\\\ &\Rightarrow \frac{d v}{d x} =y^x\left(\frac{x}{y} \frac{d y}{d x}+\log y\right) ..........(iii) \end{array}

Now, from Equation (i), 2u+3v=202 u+3 v=20

⇒2dudx+3dvdx=0⇒2xy(yx+log⁡xdydx)+3yx(xydydx+log⁡y)=0 [Using Eqs. (ii) and (iii)] \begin{aligned} & \Rightarrow 2 \frac{d u}{d x}+3 \frac{d v}{d x}=0 \\\\ & \Rightarrow 2 x^y\left(\frac{y}{x}+\log x \frac{d y}{d x}\right)+3 y^x\left(\frac{x}{y} \frac{d y}{d x}+\log y\right)=0 \text{ [Using Eqs. (ii) and (iii)] } \end{aligned}

On putting x=2x=2 and y=2y=2, we get

2(4)(1+log⁡2dydx)+3(4)(dydx+log⁡2)=0⇒dydx(8log⁡2+12)+(8+12log⁡2)=0⇒dydx=−(2+3log⁡23+2log⁡2)⇒dydx=−(2+log⁡83+log⁡4)\begin{aligned} & 2(4)\left(1+\log 2 \frac{d y}{d x}\right)+3(4)\left(\frac{d y}{d x}+\log 2\right)=0 \\\\ & \Rightarrow \frac{d y}{d x}(8 \log 2+12)+(8+12 \log 2)=0 \\\\ & \Rightarrow \frac{d y}{d x}=-\left(\frac{2+3 \log 2}{3+2 \log 2}\right) \\\\ & \Rightarrow \frac{d y}{d x}=-\left(\frac{2+\log 8}{3+\log 4}\right) \end{aligned}
Q48
Suppose f(x)=(2x+2−x)tan⁡xtan⁡−1(x2−x+1)(7x2+3x+1)3f(x)=\dfrac{\left(2^x+2^{-x}\right) \tan x \sqrt{\tan ^{-1}\left(x^2-x+1\right)}}{\left(7 x^2+3 x+1\right)^3}. Then the value of f′(0)f^{\prime}(0) is equal to
A π\pi
B π\sqrt{\pi}
C 0
D π2\dfrac{\pi}{2}
Correct Answer
Option B
Solution
f′(0)=lim⁡h→0f(h)−f(0)h=lim⁡h→0(2h+2−h)tan⁡htan⁡−1(h2−h+1)−0(7h2+3h+1)3h=π\begin{aligned} & f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\ & =\lim _{h \rightarrow 0} \frac{\left(2^h+2^{-h}\right) \tan h \sqrt{\tan ^{-1}\left(h^2-h+1\right)}-0}{\left(7 h^2+3 h+1\right)^3 h} \\ & =\sqrt{\pi} \end{aligned}
Q49
$$\text { Let } y=\log _e\left(\frac{1-x^2}{1+x^2}\right),-1
A 732
B 736
C 742
D 746
Correct Answer
Option B
Solution
y=log⁡e(1−x21+x2)dydx=y′=−4x1−x4\begin{aligned} & y=\log _e\left(\frac{1-x^2}{1+x^2}\right) \\ & \frac{d y}{d x}=y^{\prime}=\frac{-4 x}{1-x^4} \end{aligned}

Again,

d2ydx2=y′′=−4(1+3x4)(1−x4)2\frac{d^2 y}{d x^2}=y^{\prime \prime}=\frac{-4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}

Again

y′−y′′=−4x1−x4+4(1+3x4)(1−x4)2y^{\prime}-y^{\prime \prime}=\frac{-4 x}{1-x^4}+\frac{4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}

at

x=12\mathrm{x}=\frac{1}{2}

,

y′−y′′=736225y^{\prime}-y^{\prime \prime}=\frac{736}{225}

Thus

225(y′−y′′)=225×736225=736225\left(y^{\prime}-y^{\prime \prime}\right)=225 \times \frac{736}{225}=736
Q50
Let f:R−{0}→Rf: \mathbb{R}-\{0\} \rightarrow \mathbb{R} be a function satisfying f(xy)=f(x)f(y)f\left(\dfrac{x}{y}\right)=\dfrac{f(x)}{f(y)} for all x,y,f(y)≠0x, y, f(y) \neq 0. If f′(1)=2024f^{\prime}(1)=2024, then
A xf′(x)+2024f(x)=0x f^{\prime}(x)+2024 f(x)=0
B xf′(x)−2023f(x)=0x f^{\prime}(x)-2023 f(x)=0
C xf′(x)−2024f(x)=0x f^{\prime}(x)-2024 f(x)=0
D xf′(x)+f(x)=2024x f^{\prime}(x)+f(x)=2024
Correct Answer
Option C
Solution
f(xy)=f(x)f(y)f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}
f′(1)=2024f(1)=1\begin{aligned} & \mathrm{f}^{\prime}(1)=2024 \\ & \mathrm{f}(1)=1 \end{aligned}

Partially differentiating w. r. t. x

f′(xy)⋅1y=1f(y)f′(x)y→xf′(1)⋅1x=f′(x)f(x)2024f(x)=xf′(x)⇒xf′(x)−2024 f(x)=0\begin{aligned} & \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{\mathrm{y}}\right) \cdot \frac{1}{\mathrm{y}}=\frac{1}{\mathrm{f}(\mathrm{y})} \mathrm{f}^{\prime}(\mathrm{x}) \\ & \mathrm{y} \rightarrow \mathrm{x} \\ & \mathrm{f}^{\prime}(1) \cdot \frac{1}{\mathrm{x}}=\frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \\ & 2024 \mathrm{f}(\mathrm{x})=\mathrm{xf}^{\prime}(\mathrm{x}) \Rightarrow \mathrm{xf}^{\prime}(\mathrm{x})-2024 \mathrm{~f}(\mathrm{x})=0 \end{aligned}
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