JEE Mathematics Application of Derivatives MCQs Set 28

Answer: (c) \(3^{1/3}\)

Explanation:
1. Consider \(g(x)=\dfrac{\ln x}x\); \(g'(x)=\dfrac{1-\ln x}{x^2}\), so \(g\) increases up to \(x=e\) and decreases afterwards.
2. Among integers, compare \(g(2)=0.3466\), \(g(3)=0.3662\): the maximum is at \(3\), so \(3^{1/3}\) is greatest.

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (d) \((0,\sqrt6-2)\)

Explanation:
1. With \(c=\cos x\in[-1,1]\): \(f'(x)=4c^2-8(a+1)c-4(a^2+2a-3)\).
2. For \(a>0\) the vertex \(c=a+1>1\), so \(f'\) is smallest at \(c=1\): \(-4a^2-16a+8\).
3. We need \(-4a^2-16a+8>0\), i.e. \(a^2+4a-2<0\), giving \(0<a<\sqrt6-2\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (a) \(f(x)\) is increasing and \(g(x)\) is decreasing

Explanation:
1. \(f'(x)=\dfrac{\sin x-x\cos x}{\sin^2x}>0\) because \(\tan x>x\) on \((0,\frac\pi2)\).
2. \(g'(x)=\dfrac{\tan x-x\sec^2x}{\tan^2x}\) has numerator \(\sin x\cos x-x\) over \(\cos^2x\), which is negative because \(\sin2x<2x\). So \(g\) decreases.

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (b) \(a_7\)

Explanation:
1. For \(h(x)=\dfrac{x^2}{x^3+200}\): \(h'(x)=\dfrac{x(400-x^3)}{(x^3+200)^2}\), zero at \(x=\sqrt[3]{400}\approx7.37\).
2. Compare \(a_7=\dfrac{49}{543}\approx0.0902\) with \(a_8=\dfrac{64}{712}\approx0.0899\): \(a_7\) is larger.

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (d) increasing on \([1,2)\)

Explanation:
1. On \([1,2)\), \(f=x^2+1\), which is increasing.
2. At \(x=2\) the value drops from nearly \(5\) to \(\dfrac52\), so \(f\) is not increasing on \([1,4)\); it is also not decreasing there.

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (a) \(\dfrac13\)

Explanation:
1. \(\tan A\tan B=\dfrac{\cos(A-B)-\cos(A+B)}{\cos(A-B)+\cos(A+B)}=\dfrac{c-\frac12}{c+\frac12}\) with \(c=\cos(A-B)\le1\).
2. This increases with \(c\), so the maximum is at \(c=1\): \(\dfrac{1/2}{3/2}=\dfrac13\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (b) \(\left(-\infty,-\dfrac43\right)\cup(2,\infty)\)

Explanation:
1. \(\cot\frac x2\sin^2\frac x2=\sin\frac x2\cos\frac x2=\frac12\sin x\), so \(f'(x)=(4a-3)+(a-7)\cos x\).
2. No critical point means \(|4a-3|>|a-7|\), i.e. \((3a+4)(5a-10)>0\), giving \(a<-\dfrac43\) or \(a>2\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (d) \(|a|\ge1\)

Explanation:
1. Near \(-2\), \(x^2+5x+6=(x+2)(x+3)\to0\), so \(f(x)\) approaches \(2\) from below.
2. For a maximum at \(-2\) we need \(f(-2)=a^2+1\ge2\), i.e. \(|a|\ge1\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (a) is continuous on \(\left(0,\dfrac\pi2\right]\)

Explanation:
1. For \(0<x\le\frac\pi2\), \(0\le\cos x<1\), so \([\cos x]=0\).
2. Hence \(f(x)=1\), a constant: continuous, neither strictly monotonic, with maximum \(1\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (d) \(v\sqrt{2-\sqrt2}\)

Explanation:
1. After time \(t\) each is at distance \(vt\); the separation is \(d=2vt\sin\dfrac{45^\circ}2=2vt\sin22.5^\circ\).
2. Rate \(=2v\sin22.5^\circ=v\sqrt{2-2\cos45^\circ}=v\sqrt{2-\sqrt2}\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (c) \(2y=x\)

Explanation:
1. The nearest point has tangent parallel to \(y=2x\): \(\dfrac{dy}{dx}=\dfrac xy=2\).
2. So \(x=2y\) at \(P\), i.e. \(2y=x\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (b) \(H=\varnothing\) and \(V=\{(1,1)\}\)

Explanation:
1. \(\dfrac{dy}{dx}=\dfrac y{y^2-x}\). Horizontal tangent needs \(y=0\), but then \(2=0\): no such point, so \(H=\varnothing\).
2. Vertical tangent needs \(x=y^2\): \(y^3-3y^3+2=0\) gives \(y=1\), \(x=1\). So \(V=\{(1,1)\}\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (b) \(0\)

Explanation:
1. Let \(q(x)=p(x)-x^2\): \(q(0)=0\) and \(q(x)>0\) for \(x\ne0\), so \(q\) has a minimum at \(0\), giving \(q''(0)\ge0\).
2. But \(q''(0)=\dfrac12-2<0\), a contradiction. No such polynomial exists.

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (c) \(40\)

Explanation:
1. Fuel cost per hour \(=kv^2\) with \(k=\dfrac{48}{256}=\dfrac3{16}\); cost per mile \(=\dfrac3{16}v+\dfrac{300}v\).
2. Minimum when \(\dfrac3{16}=\dfrac{300}{v^2}\), so \(v^2=1600\) and \(v=40\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (d) \(1\)

Explanation:
1. With \(s=x+y\): \((1+y')\dfrac{s-1}s=2\), so \(1+y'=\dfrac{2s}{s-1}\), which is infinite when \(s=1\).
2. Then \(\alpha+\beta=1\) (indeed the curve gives \(1-0=2\alpha+5\), \(\alpha=-2\), \(\beta=3\)).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (c) \(\dfrac{16}{\sqrt3}\)

Explanation:
1. By symmetry take the rectangle \(|x|\le t\); height \(=\left(2-\dfrac{t^2}2\right)-(t^2-4)=6-\dfrac{3t^2}2\).
2. Area \(=12t-3t^3\), maximum at \(t^2=\dfrac43\), giving \(\dfrac{16}{\sqrt3}\approx9.24\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (b) \(2\)

Explanation:
1. The symmetry about \(x=1\) makes \(P\) a polynomial in \((x-1)^2\): \(P(x)=Q\left((x-1)^2\right)\).
2. \(P(1)=0\) gives \(Q(0)=0\), so \((x-1)^2\) divides \(P\); a polynomial like \((x-1)^2\) itself shows that \(m=3\) is not guaranteed.

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (d) \(2\)

Explanation:
1. \(f'(x)=(b-1)\left[1-2(b-2)\sin2x\right]\).
2. For \(b=1\) the derivative is zero everywhere. Otherwise we need \(|2(b-2)|<1\), i.e. \(\dfrac32<b<\dfrac52\), so the only integer is \(b=2\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (a) \(1\)

Explanation:
1. Orthogonality at a common point: \(aa_1x^2+bb_1y^2=0\); subtracting the two equations and combining gives \(\left(\dfrac1a-\dfrac1b\right)=\left(\dfrac1{a_1}-\dfrac1{b_1}\right)\) (the conics are confocal).
2. Rearranging, \(\dfrac1a-\dfrac1{a_1}=\dfrac1b-\dfrac1{b_1}\), so \(\lambda=1\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (c) \(9\)

Explanation:
1. The slant height is \(1\) and the base radius is \(r\); \(V=\dfrac\pi3r^2\sqrt{1-r^2}\) is maximum when \(r^2=\dfrac23\).
2. The remaining sector has arc \(2\pi r\), so its area is \(\pi r\) and \(\dfrac A{A_1}=\dfrac{\pi}{\pi-\pi r}=\dfrac1{1-\sqrt{2/3}}=3+\sqrt6\).
3. So \(m=3\), \(n=6\), \(m+n=9\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.