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Application of Derivatives

JEE Mathematics · 188 questions · Page 16 of 19 · Click an option or "Show Solution" to reveal answer

Q151
If 5f(x)+4f(1x)=x2−2,∀x≠05 f(x)+4 f\left(\dfrac{1}{x}\right)=x^2-2, \forall x \neq 0 and y=9x2f(x)y=9 x^2 f(x), then yy is strictly increasing in :
A (0,15)∪(15,∞)\left(0, \dfrac{1}{\sqrt{5}}\right) \cup\left(\dfrac{1}{\sqrt{5}}, \infty\right)
B (−15,0)∪(15,∞)\left(-\dfrac{1}{\sqrt{5}}, 0\right) \cup\left(\dfrac{1}{\sqrt{5}}, \infty\right)
C (−15,0)∪(0,15)\left(-\dfrac{1}{\sqrt{5}}, 0\right) \cup\left(0, \dfrac{1}{\sqrt{5}}\right)
D (−∞,15)∪(0,15)\left(-\infty, \dfrac{1}{\sqrt{5}}\right) \cup\left(0, \dfrac{1}{\sqrt{5}}\right)
Correct Answer
Option B
Solution
5f(x)+4f(1/x)=x2−25 f(x)+4 f(1 / x)=x^2-2

........(1) Replace xx by 1/x1 / x

5f(1/x)+4f(x)=1x2−25 f(1 / x)+4 f(x)=\frac{1}{x^2}-2

..........(

2) Multiply equation (1) by 5 and multiply equation (2) by 4 and then subtract equation (2) from (1) 25f(x)−16f(x)=5x2−10−4x2+89f(x)=5x2−4x2−29f(x)=5x4−4−2x2x2\begin{aligned} & 25 f(x)-16 f(x)=5 x^2-10-\dfrac{4}{x^2}+8 \\\\ & 9 f(x)=5 x^2-\dfrac{4}{x^2}-2 \\\\ & 9 f(x)=\dfrac{5 x^4-4-2 x^2}{x^2}\end{aligned} y=9x2f(x)y=5x4−2x2−4y′=20x3−4x Put y′>020x3−4x>05x3−x>0x(5x2−1)>0\begin{aligned} & y=9 x^2 f(x) \\\\ & y=5 x^4-2 x^2-4 \\\\ & y^{\prime}=20 x^3-4 x \\\\ & \text{ Put } y^{\prime}>0 \\\\ & 20 x^3-4 x>0 \\\\ & 5 x^3-x>0 \\\\ & x\left(5 x^2-1\right)>0\end{aligned} x∈(−15,0)∪(15,∞)x \in\left(-\dfrac{1}{\sqrt{5}}, 0\right) \cup\left(\dfrac{1}{\sqrt{5}}, \infty\right)

Q152
Let g(x)=3f(x3)+f(3−x)g(x)=3 f\left(\dfrac{x}{3}\right)+f(3-x) and f′′(x)>0f^{\prime \prime}(x)>0 for all x∈(0,3)x \in(0,3). If gg is decreasing in (0,α)(0, \alpha) and increasing in (α,3)(\alpha, 3), then 8α8 \alpha is :
A 0
B 24
C 18
D 20
Correct Answer
Option C
Solution
g(x)=3f(x3)+f(3−x) and f′′(x)>0∀x∈(0,3)g(x)=3 f\left(\frac{x}{3}\right)+f(3-x) \text{ and } f^{\prime \prime}(x)>0 \forall x \in(0,3)
⇒f′(x)\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})

is increasing function

g′(x)=3×13⋅f′(x3)−f′(3−x)=f′(x3)−f′(3−x)\begin{aligned} & g^{\prime}(x)=3 \times \frac{1}{3} \cdot f^{\prime}\left(\frac{x}{3}\right)-f^{\prime}(3-x) \\ & =f^{\prime}\left(\frac{x}{3}\right)-f^{\prime}(3-x) \end{aligned}

If

g\mathrm{g}

is decreasing in

(0,α)(0, \alpha)
\begin{aligned} & \mathrm{g}^{\prime}(\mathrm{x}) Therefore

\alpha=\frac{9}{4}

ThenThen

8 \alpha=8 \times \frac{9}{4}=18$$

Q153
 If f(x)=∣x32x2+11+3x3x2+22xx3+6x3−x4x2−2∣ for all x∈R, then 2f(0)+f′(0) is equal to \text{ If } f(x)=\left|\begin{array}{ccc} x^3 & 2 x^2+1 & 1+3 x \\ 3 x^2+2 & 2 x & x^3+6 \\ x^3-x & 4 & x^2-2 \end{array}\right| \text{ for all } x \in \mathbb{R} \text{, then } 2 f(0)+f^{\prime}(0) \text{ is equal to }
A 24
B 18
C 42
D 48
Correct Answer
Option C
Solution
f(0)=∣01120604−2∣=12f′(x)=∣3x24x33x2+22xx3+6x3−x4x2−2∣+\begin{aligned} & f(0)=\left|\begin{array}{ccc} 0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{array}\right|=12 \\ & f^{\prime}(x)=\left|\begin{array}{ccc} 3 x^2 & 4 x & 3 \\ 3 x^2+2 & 2 x & x^3+6 \\ x^3-x & 4 & x^2-2 \end{array}\right|+ \end{aligned}
∣x32x2+11+3x6x23x2x3−x4x2−2∣+∣x32x2+11+3x3x2+22xx3+63x2−102x∣∴f′(0)=∣00320604−2∣+∣01102004−2∣+∣011206−100∣=24−6=18∴2f(0)+f′(0)=42\begin{aligned} & \left|\begin{array}{ccc} x^3 & 2 x^2+1 & 1+3 x \\ 6 x & 2 & 3 x^2 \\ x^3-x & 4 & x^2-2 \end{array}\right|+ \\ & \left|\begin{array}{ccc} x^3 & 2 x^2+1 & 1+3 x \\ 3 x^2+2 & 2 x & x^3+6 \\ 3 x^2-1 & 0 & 2 x \end{array}\right| \\ & \therefore f^{\prime}(0)=\left|\begin{array}{ccc} 0 & 0 & 3 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{array}\right|+\left|\begin{array}{ccc} 0 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 4 & -2 \end{array}\right|+\left|\begin{array}{ccc} 0 & 1 & 1 \\ 2 & 0 & 6 \\ -1 & 0 & 0 \end{array}\right| \\ & =24-6=18 \\ & \therefore 2 f(0)+f^{\prime}(0)=42 \end{aligned}
Q154
Consider the function f:[12,1]→Rf:\left[\dfrac{1}{2}, 1\right] \rightarrow \mathbb{R} defined by f(x)=42x3−32x−1f(x)=4 \sqrt{2} x^3-3 \sqrt{2} x-1. Consider the statements (I) The curve y=f(x)y=f(x) intersects the xx-axis exactly at one point. (II) The curve y=f(x)y=f(x) intersects the xx-axis at x=cos⁡π12x=\cos \dfrac{\pi}{12}. Then
A Both (I) and (II) are correct.
B Only (I) is correct.
C Both (I) and (II) are incorrect.
D Only (II) is correct.
Correct Answer
Option A
Solution
\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=12 \sqrt{2} \mathrm{x}^2-3 \sqrt{2} \geq 0 \text{ for }\left[\frac{1}{2}, 1\right] \\ & \mathrm{f}\left(\frac{1}{2}\right)

\mathrm{f}(1)>0 \Rightarrow(\mathrm{A})

iscorrect.is correct.

f(x)=\sqrt{2}\left(4 x^3-3 x\right)-1=0

LetLet

\cos \alpha=\mathrm{x}

,,

\cos 3 \alpha=\cos \frac{\pi}{4} \Rightarrow \alpha=\frac{\pi}{12}

\mathrm{x}=\cos \frac{\pi}{12}$$ (4) is correct.

Q155
The function f(x)=2x+3(x)23,x∈Rf(x)=2 x+3(x)^{\dfrac{2}{3}}, x \in \mathbb{R}, has
A exactly one point of local minima and no point of local maxima
B exactly one point of local maxima and exactly one point of local minima
C exactly two points of local maxima and exactly one point of local minima
D exactly one point of local maxima and no point of local minima
Correct Answer
Option B
Solution
f(x)=2x+3(x)23f′(x)=2+2x−13=2(1+1x13)=2(x13+1x13)\begin{aligned} & f(x)=2 x+3(x)^{\frac{2}{3}} \\ & f^{\prime}(x)=2+2 x^{\frac{-1}{3}} \\ & =2\left(1+\frac{1}{x^{\frac{1}{3}}}\right) \\ & =2\left(\frac{x^{\frac{1}{3}}+1}{x^{\frac{1}{3}}}\right) \end{aligned}

So, maxima (M) at x = −-1 & minima (m) at x = 0

Q156
Let f(x)=(x+3)2(x−2)3,x∈[−4,4]f(x)=(x+3)^2(x-2)^3, x \in[-4,4]. If MM and mm are the maximum and minimum values of ff, respectively in [−4,4][-4,4], then the value of M−mM-m is
A 108
B 392
C 608
D 600
Correct Answer
Option C
Solution
f′(x)=(x+3)2⋅3(x−2)2+(x−2)32(x+3)=5(x+3)(x−2)2(x+1)f′(x)=0,x=−3,−1,2\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=(\mathrm{x}+3)^2 \cdot 3(\mathrm{x}-2)^2+(\mathrm{x}-2)^3 2(\mathrm{x}+3) \\ & =5(\mathrm{x}+3)(\mathrm{x}-2)^2(\mathrm{x}+1) \\ & \mathrm{f}^{\prime}(\mathrm{x})=0, \mathrm{x}=-3,-1,2 \end{aligned}
f(−4)=−216f(−3)=0,f(4)=49×8=392M=392,m=−216M−m=392+216=608 Ans = ’3’ \begin{aligned} & f(-4)=-216 \\ & f(-3)=0, f(4)=49 \times 8=392 \\ & M=392, m=-216 \\ & M-m=392+216=608 \\ & \text{ Ans }=\text{ '3' } \end{aligned}
Q157
The maximum area of a triangle whose one vertex is at (0,0)(0,0) and the other two vertices lie on the curve y=−2x2+54y=-2 x^2+54 at points (x,y)(x, y) and (−x,y)(-x, y), where y>0y>0, is :
A 108
B 122
C 88
D 92
Correct Answer
Option A
Solution

Area of

Δ\Delta
=12∣001xy1−xy1∣⇒∣12(xy+xy)∣=∣xy∣Area⁡(Δ)=∣xy∣=∣x(−2x2+54)∣d(Δ)dx=∣(−6x2+54)∣⇒dΔdx=0 at x=3 Area =3(−2×9+54)=108\begin{aligned} & =\frac{1}{2}\left|\begin{array}{ccc} 0 & 0 & 1 \\ x & y & 1 \\ -x & y & 1 \end{array}\right| \\ & \Rightarrow\left|\frac{1}{2}(x y+x y)\right|=|x y| \\ & \operatorname{Area}(\Delta)=|x y|=\left|x\left(-2 x^2+54\right)\right| \\ & \frac{d(\Delta)}{d x}=\left|\left(-6 x^2+54\right)\right| \Rightarrow \frac{d \Delta}{d x}=0 \text{ at } x=3 \\ & \text{ Area }=3(-2 \times 9+54)=108 \end{aligned}
Q158
Let the sum of the maximum and the minimum values of the function f(x)=2x2−3x+82x2+3x+8f(x)=\dfrac{2 x^2-3 x+8}{2 x^2+3 x+8} be mn\dfrac{m}{n}, where gcd⁡(m,n)=1\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1. Then m+n\mathrm{m}+\mathrm{n} is equal to :
A 217
B 182
C 201
D 195
Correct Answer
Option C
Solution
f(x)=2x2−3x+82x2+3x+8=y,2x2+3x+8>0∀x∈R⇒x2(2y−2)+x(3y+3)+8y−8=0\begin{aligned} & f(x)=\frac{2 x^2-3 x+8}{2 x^2+3 x+8}=y, 2 x^2+3 x+8>0 \quad \forall x \in \mathbb{R} \\ & \Rightarrow \quad x^2(2 y-2)+x(3 y+3)+8 y-8=0 \end{aligned}

Since

x∈Rx \in \mathbb{R}

, the equation has real roots

⇒D≥0⇒(3y+3)2−4(2y−2)(8y−8)≥0⇒9(y+1)2−64y(y−1)2≥0⇒(3y+3)2−(8y−8)2≥0⇒(11y−5)(−5y+11)≥0⇒(y−511)(y−115)≤0⇒y∈[511,115]\begin{aligned} & \Rightarrow \quad D \geq 0 \\ & \Rightarrow(3 y+3)^2-4(2 y-2)(8 y-8) \geq 0 \\ & \Rightarrow 9(y+1)^2-64 y(y-1)^2 \geq 0 \\ & \Rightarrow(3 y+3)^2-(8 y-8)^2 \geq 0 \\ & \Rightarrow(11 y-5)(-5 y+11) \geq 0 \\ & \Rightarrow\left(y-\frac{5}{11}\right)\left(y-\frac{11}{5}\right) \leq 0 \\ & \Rightarrow y \in\left[\frac{5}{11}, \frac{11}{5}\right] \end{aligned}

Sum of maximum and minimum value

ymax⁡+ymin⁡=511+115=25+12155=14655=mn⇒m+n=201\begin{aligned} & y_{\max }+y_{\min }=\frac{5}{11}+\frac{11}{5}=\frac{25+121}{55} \\ & =\frac{146}{55}=\frac{m}{n} \Rightarrow m+n=201 \end{aligned}
Q159
Let f(x)=3x−2+4−xf(x)=3 \sqrt{x-2}+\sqrt{4-x} be a real valued function. If α\alpha and β\beta are respectively the minimum and the maximum values of ff, then α2+2β2\alpha^2+2 \beta^2 is equal to
A 42
B 38
C 24
D 44
Correct Answer
Option A
Solution
f(x)=3x−2+4−x Let x=2sin⁡2θ+4cos⁡2θ=32sin⁡2θ+4cos⁡2θ−2+4−2sin⁡2θ−4cos⁡2θ=32cos⁡2θ+2sin⁡2θ=32∣cos⁡θ∣+2∣sin⁡θ∣\begin{aligned} & f(x)=3 \sqrt{x-2}+\sqrt{4-x} \\ & \text{ Let } x=2 \sin ^2 \theta+4 \cos ^2 \theta \\ & =3 \sqrt{2 \sin ^2 \theta+4 \cos ^2 \theta-2}+\sqrt{4-2 \sin ^2 \theta-4 \cos ^2 \theta} \\ & =3 \sqrt{2 \cos ^2 \theta}+\sqrt{2 \sin ^2 \theta} \\ & =3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \end{aligned}
⇒32cos⁡θ+2sin⁡θ≤18+2⇒32cos⁡θ+2sin⁡θ≤20\begin{aligned} & \Rightarrow 3 \sqrt{2} \cos \theta+\sqrt{2} \sin \theta \leq \sqrt{18+2} \\ & \Rightarrow 3 \sqrt{2} \cos \theta+\sqrt{2} \sin \theta \leq \sqrt{20} \end{aligned}

Minimum value exists when

θ=π2\theta=\frac{\pi}{2}
 So, minimum value =2⇒α=2 and β=20⇒α2+2β2=2+40=42\begin{aligned} & \text{ So, minimum value }=\sqrt{2} \\ & \Rightarrow \alpha=\sqrt{2} \text{ and } \beta=\sqrt{20} \\ & \Rightarrow \alpha^2+2 \beta^2=2+40 \\ & =42 \end{aligned}
Q160
If the function f(x)=2x3−9ax2+12a2x+1,a>0f(x)=2 x^3-9 \mathrm{ax}^2+12 \mathrm{a}^2 x+1, \mathrm{a}> 0 has a local maximum at x=αx=\alpha and a local minimum at x=α2x=\alpha^2, then α\alpha and α2\alpha^2 are the roots of the equation :
A x2−6x+8=0x^2-6 x+8=0
B 8x2−6x+1=08 x^2-6 x+1=0
C 8x2+6x−1=08 x^2+6 x-1=0
D x2+6x+8=0x^2+6 x+8=0
Correct Answer
Option A
Solution
f(x)=6x2−18ax+12a2=6(x2−3a+2a2)=6(x−a)(x−2a)=0x=a,2a\begin{aligned} & f(x)=6 x^2-18 a x+12 a^2 \\ & =6\left(x^2-3 a+2 a^2\right) \\ & =6(x-a)(x-2 a)=0 \\ & x=a, 2 a \end{aligned}
a=α,2a=α2⇒α=0,2a=\alpha, \quad 2 a=\alpha^2 \quad \Rightarrow \alpha=0,2
a>0∴α=2α2=4\begin{array}{lll} a>0 & \therefore & \alpha=2 \\ & & \alpha^2=4 \end{array}
∴x2−6x+8=0\therefore x^2-6 x+8=0

is the required quadratic equation.

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