JEE Mathematics Application of Derivatives MCQs Set 19

Answer: (b) (i), (iii) and (iv) only

Explanation:
1. The domain is \( x\gt 2 \). \( f'(x)=\frac1{x-2}+\frac1{x^2}\gt 0 \), so f is increasing and (i) is true; (ii) is false because the interval lies outside the domain.
2. \( f''(x)=-\frac1{(x-2)^2}-\frac2{x^3}\lt 0 \) for \( x\gt 2 \), so f is concave downward, and (iii) is true.
3. The inverse of an increasing function is increasing, so (iv) is true.

Teacher's Note:
1. Find the domain first, then check the signs of the first and second derivatives.

Answer: (c) Statement 1 is true, Statement 2 is false

Explanation:
1. Let \( g(x)=\frac{\ln x}x \). \( g'(x)=\frac{1-\ln x}{x^2}\lt 0 \) for \( x\gt e \), so \( x^{1/x} \) is decreasing on \( [e,\infty) \). Statement 2 is false.
2. Since \( \pi\gt e \), \( \pi^{1/\pi}\lt e^{1/e} \). Raising to the power \( e\pi \) gives \( \pi^e\lt e^\pi \). Statement 1 is true.

Teacher's Note:
1. Compare \( \frac{\ln x}x \) at e and at \( \pi \) using its derivative.

Answer: (b) Statement 1 is true, Statement 2 is true and Statement 2 is NOT a correct explanation for Statement 1

Explanation:
1. \( f(0)=f(1)=0 \) and for \( 0\lt x\lt 1 \), \( x^{50}\lt x^{20} \), so \( f\lt 0 \). The global maximum on \([0,1]\) is 0. Statement 1 is true.
2. \( f'(x)=50x^{49}-20x^{19} \) is 0 at \( x=0 \), so Statement 2 is true.
3. The maximum value is attained at both end points and Statement 2 does not explain this (at \( x=1 \) the derivative is not zero). So Statement 2 is not a correct explanation.

Teacher's Note:
1. Compare the values of f at the end points and in between; a stationary point alone does not make it a maximum.

Answer: (b) \( y=4x+4 \)

Explanation:
1. A tangent to \( y=-x^2 \) at \( x=p\lt 0 \) is \( y=-2px+p^2 \). A tangent to \( y=x^2+8 \) at \( x=q\ge0 \) is \( y=2qx+8-q^2 \).
2. Equal slopes give \( p=-q \), and equal intercepts give \( p^2=8-q^2 \). So \( 2q^2=8 \), \( q=2 \), \( p=-2 \).
3. The line is \( y=4x+4 \).

Teacher's Note:
1. Write the tangent to each branch and match both slope and intercept.

Answer: (b) Parabola

Explanation:
1. Differentiating: \( 2yy'=4a\left(1+\cos\frac xa\right) \). The tangent is horizontal when \( \cos\frac xa=-1 \).
2. Then \( \sin\frac xa=0 \), so on the curve \( y^2=4a\cdot x \).
3. These points lie on the parabola \( y^2=4ax \).

Teacher's Note:
1. At the required points the sine term vanishes and the equation reduces to a parabola.

Answer: (d) h is odd and increasing

Explanation:
1. Since \( |x|\operatorname{sgn}x=x \), \( \ln f(x)=\{x\}\ln a \) and \( \ln g(x)=[x]\ln a \).
2. So \( (\ln a)h(x)=\left(\{x\}+[x]\right)\ln a=x\ln a \), giving \( h(x)=x \).
3. \( h(x)=x \) is an odd function and is increasing.

Teacher's Note:
1. Use \( \{x\}+[x]=x \) and \( |x|\operatorname{sgn}x=x \).

Answer: (d) \( \left[-\sqrt{130},-\sqrt2\right)\cup\left(\sqrt2,\sqrt{130}\right] \)

Explanation:
1. For \( x\le1 \), \( f'(x)=3x^2-2x+10\gt 0 \), so f increases and its greatest value there is \( f(1)=5 \).
2. For \( x\gt 1 \), the function \( -2x+\log_2\left(b^2-2\right) \) decreases, and its values stay below \( -2+\log_2\left(b^2-2\right) \). We need \( -2+\log_2\left(b^2-2\right)\le5 \), i.e. \( b^2-2\le128 \), so \( b^2\le130 \).
3. The logarithm needs \( b^2\gt 2 \). So \( b\in\left[-\sqrt{130},-\sqrt2\right)\cup\left(\sqrt2,\sqrt{130}\right] \).

Teacher's Note:
1. The branch for \( x\gt 1 \) is decreasing, so its supremum is its limiting value at \( x=1 \).

Answer: (c) TFTF

Explanation:
1. S1: \( y=2x^{-2}-x^{-4} \), so \( y'=\frac{4(1-x^2)}{x^5} \), which changes sign at \( x=\pm1 \). So y is neither increasing nor decreasing overall, and S1 is true.
2. S2: a strictly increasing function can have no solution (for example \( e^x=-1 \)), so S2 is false.
3. S3: \( f(x)=x \) on the open interval \( (0,1) \) has no local extremum, so S3 is true.
4. S4: \( \{6\}=0 \) is the minimum value of the fractional part, not the maximum, so S4 is false.

Teacher's Note:
1. Test each statement with a simple counterexample or by checking the derivative.

Answer: (d) (i), (ii), (iii) and (iv)

Explanation:
1. \( f'(x)=4x^3(12\ln x-7)+12x^3=16x^3(3\ln x-1) \), which changes from negative to positive at \( x=e^{1/3} \), so (ii) is true.
2. \( f''(x)=48x^2(3\ln x-1)+48x^2=144x^2\ln x \).
3. \( f''\lt 0 \) for \( 0\lt x\lt 1 \) and \( f''\gt 0 \) for \( x\gt 1 \), so (iii) and (iv) are true, and the concavity changes at \( x=1 \) where \( f(1)=-7 \), so (i) is true.

Teacher's Note:
1. The sign of \( f'' \) decides concavity, and a sign change of \( f'' \) marks an inflection point.

Answer: (c) 1

Explanation:
1. Let \( f(x)=x^7+14x^5+16x^3+30x-560 \). \( f'(x)=7x^6+70x^4+48x^2+30\gt 0 \), so f is strictly increasing.
2. f tends to \(-\infty\) and \(+\infty\) at the two ends, so it crosses the axis exactly once.

Teacher's Note:
1. A strictly increasing continuous function can cross zero at most once.

Answer: (a) (i), (iii) and (iv) only

Explanation:
1. Let \( g=3x^4+8x^3-18x^2+60 \). \( g'=12x(x+3)(x-1) \), so g has local minima at \(-3\) and 1 and a local maximum at 0.
2. \( g(0)=60\gt 0 \) and \( g(1)=53\gt 0 \), while \( g(-3)=-75\lt 0 \). The function \( f=\frac{40}g \) therefore has local minimum at 0 and local maxima at 1 and at \(-3\) (where g\lt 0 and \( |g| \) is largest). So (i) and (iv) are true and (ii) is false.
3. g has real zeros (it is negative at \(-3\) and positive far away), so f is unbounded above near them and has no absolute maximum. (iii) is true.

Teacher's Note:
1. For a reciprocal function, maxima of f correspond to minima of g when g is positive, and also when g is negative the extreme with largest absolute value.

Answer: (b) \( 2\sqrt2 \)

Explanation:
1. \( a=2t \). The curve is \( y=(x-a)^2+a \), so its vertex is \( (a,a)=(2t,2t) \).
2. The distance from the origin is \( 2t\sqrt2 \).
3. Its rate of change is \( 2\sqrt2 \), so \( k=2\sqrt2 \).

Teacher's Note:
1. Write the parabola in vertex form and express the vertex in terms of t.

Answer: (d) \( 16\sqrt2+2 \)

Explanation:
1. With \( a=2t \), at \( x=0 \) the parabola has \( y=a^2+a \) and slope \( -2a \). So the tangent is \( y=-2ax+a^2+a \), giving \( m=-4t \) and \( c=a^2+a=4t^2+2t \).
2. \( \frac{dc}{dt}=8t+2 \).
3. At \( t=2\sqrt2 \): \( \ell=16\sqrt2+2 \).

Teacher's Note:
1. The tangent at \( x=0 \) has intercept equal to the value of the parabola at 0.

Answer: (c) \( 2\pi \)

Explanation:
1. On \( [0,\pi] \), \( \frac{x+\sin x}2\in\left[0,\frac\pi2\right] \), so \( f(x)=\frac{x+\sin x}2 \), which is increasing since \( f'=\frac{1+\cos x}2\ge0 \).
2. On \( [\pi,2\pi] \): \( f(x)=\pi-f(2\pi-x)=\pi-\frac{2\pi-x-\sin x}2=\frac{x+\sin x}2 \), again increasing. So f increases on \( [0,2\pi] \).
3. On \( [2\pi,4\pi] \), \( f(x)=f(4\pi-x) \) is a mirror image, so f decreases there. Hence \( \alpha=2\pi \).

Teacher's Note:
1. Find the formula for f on each interval and check where it increases.

Answer: (b) \( \alpha \)

Explanation:
1. On \( [0,2\pi] \), \( f(x)=\frac{x+\sin x}2 \) is increasing and on \( [2\pi,4\pi] \), \( f(x)=f(4\pi-x) \).
2. So the graph is symmetric about the line \( x=2\pi \), and \( \beta=2\pi \).
3. Since \( \alpha=2\pi \), \( \beta=\alpha \).

Teacher's Note:
1. The relation \( f(x)=f(4\pi-x) \) shows the axis of symmetry is the midpoint of 0 and \( 4\pi \).

Answer: (a) \( \frac\beta2 \)

Explanation:
1. f increases on \( [0,2\pi] \) with \( f(2\pi)=\frac{2\pi+\sin2\pi}2=\pi \).
2. On \( [2\pi,4\pi] \) it decreases symmetrically.
3. So the maximum value is \( \pi=\frac\beta2 \).

Teacher's Note:
1. The maximum occurs at the axis of symmetry where f stops increasing.

Answer: (a) \( \frac\pi2 \)

Explanation:
1. \( \frac{dy}{dx}=2x-5 \). At \( (2,0) \) the slope is \(-1\) and at \( (3,0) \) it is 1.
2. The product of the slopes is \(-1\).
3. So the tangents are perpendicular and the angle is \( \frac\pi2 \).

Teacher's Note:
1. Two lines are perpendicular when the product of their slopes is \(-1\).

Answer: (c) \( f(6)\ge8 \)

Explanation:
1. By the mean value theorem, \( f(6)-f(1)=5f'(c) \) for some \( c\in(1,6) \).
2. Since \( f'(c)\ge2 \), \( f(6)\ge f(1)+10 \).
3. So \( f(6)\ge-2+10=8 \).

Teacher's Note:
1. The mean value theorem converts a bound on the derivative into a bound on the change in the function.

Answer: (a) \( 2\log_3e \)

Explanation:
1. \( f'(c)=\frac{f(3)-f(1)}{3-1}=\frac{\ln3}2 \).
2. \( f'(c)=\frac1c \), so \( \frac1c=\frac{\ln3}2 \).
3. \( c=\frac2{\ln3}=2\log_3e \), which lies in \( (1,3) \).

Teacher's Note:
1. Equate \( f'(c) \) with the average rate of change.

Answer: (b) \( \left(-\frac\pi2,\frac\pi4\right) \)

Explanation:
1. \( \tan^{-1} \) is increasing, so f increases exactly where \( \sin x+\cos x \) increases.
2. \( \frac d{dx}(\sin x+\cos x)=\cos x-\sin x\gt 0 \) when \( x\lt \frac\pi4 \) (within \( \left(-\frac{3\pi}4,\frac\pi4\right) \)).
3. Among the options only \( \left(-\frac\pi2,\frac\pi4\right) \) lies entirely in this region.

Teacher's Note:
1. Composition with an increasing function preserves monotonicity.

Answer: (b) \( P(-1) \) is not minimum but \( P(1) \) is the maximum of P

Explanation:
1. \( P'(x)=4x^3+3ax^2+2bx+c \) has 0 as its only real root, so \( c=0 \) and \( P'(x)=x\left(4x^2+3ax+2b\right) \) with a quadratic factor that has no real root and is positive.
2. Hence \( P'\lt 0 \) for \( x\lt 0 \) and \( P'\gt 0 \) for \( x\gt 0 \): P decreases on \( [-1,0] \) and increases on \( [0,1] \). The minimum is at \( x=0 \), so \( P(-1) \) is not the minimum.
3. The maximum on \( [-1,1] \) is at an end point, and \( P(-1)\lt P(1) \) makes \( P(1) \) the maximum.

Teacher's Note:
1. Show that the derivative has the sign of x, so the function has its minimum at 0.