JEE Mathematics Maxima and Minima MCQs Set 13

Answer: (c) 2abc

Explanation:
1. Substitute y = a²/x: \( S=b^2x+\frac{a^2c^2}x \).
2. By AM-GM, \( S\ge2\sqrt{b^2x\cdot\frac{a^2c^2}x}=2abc \).

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Answer: (c) 4

Explanation:
1. \( f'(x)=4x^3-2x-2=2(x-1)(2x^2+2x+1) \), and the quadratic factor has no real root.
2. So x = 1 is the only critical point, f″(1) = 10 > 0, and f(1) = 4. As x → ±∞, f → ∞, so 4 is the global minimum.

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Answer: (c) \( 4+\sqrt2 \)

Explanation:
1. Write \( 3\cos^2x+4\sin^2x=3+\sin^2x\le4 \) and \( \cos\frac x2+\sin\frac x2=\sqrt2\sin\left(\frac x2+\frac\pi4\right)\le\sqrt2 \).
2. Both bounds are attained together at x = π/2: sin²x = 1 and sin(π/4 + π/4) = 1.
3. Maximum = 4 + √2.

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Answer: (c) \( \sqrt{a^2-b^2} \)

Explanation:
1. Set the derivative to zero: \( a\sec\theta\tan\theta-b\sec^2\theta=0 \), so sin θ = b/a.
2. Then \( \sec\theta=\frac a{\sqrt{a^2-b^2}} \) and \( \tan\theta=\frac b{\sqrt{a^2-b^2}} \).
3. Value \( =\frac{a^2-b^2}{\sqrt{a^2-b^2}}=\sqrt{a^2-b^2} \).

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Answer: (b) 1

Explanation:
1. The maximum possible value is 2, needing cos x = 1 and cos(√2x) = 1.
2. This needs x = 2πm and √2x = 2πn, so √2 = n/m, impossible for non-zero x.
3. Only x = 0 works, so there is exactly 1 point.

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Answer: (b) one point of maximum in the interval (0, π/2)

Explanation:
1. \( f'(x)=\frac{1-x^2\sec^2x}{(1+x\tan x)^2} \), which vanishes when \( x\sec x=1 \), i.e. \( x=\cos x \).
2. The equation x = cos x has exactly one root in (0, π/2), near 0.739.
3. f′ is positive before it and negative after it, and f → 0 at both ends, so it is a maximum.

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

Explanation:
1. Slope \( g(x)=f'(x)=e^x(\sin x+\cos x) \).
2. g′(x) = 2eˣ cos x = 0 gives x = π/2 or 3π/2.
3. g″(x) = 2eˣ(cos x − sin x): negative at π/2 (local maximum) and positive at 3π/2 (local minimum). So a = π/2.

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Answer: (a) a maximum

Explanation:
1. f′(a) = 0 and f′ changes sign from positive to negative at x = a.
2. By the first derivative test, f has a local maximum at x = a.

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Answer: (c) f(x) has a minimum at x = 1

Explanation:
1. \( f'(x)=3(x+3)(x-1) \), positive for x < −3, negative between −3 and 1, positive for x > 1.
2. So x = −3 is a maximum and x = 1 is a minimum (f″(1) = 12 > 0).

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Answer: (a) \( a=-\frac34,\ b=-\frac18 \)

Explanation:
1. \( f'(x)=\frac ax+2bx+1=0 \) at x = 1 and 3.
2. a + 2b + 1 = 0 and \( \frac a3+6b+1=0 \).
3. Solving gives a = −3/4 and b = −1/8.

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Answer: (b) \( e^{1/e} \)

Explanation:
1. Let \( y=-2x^2\ln x \) (log of the function). Then \( y'=-4x\ln x-2x=-2x(2\ln x+1) \).
2. y′ = 0 at \( x=e^{-1/2} \), where y′ changes from positive to negative, a maximum.
3. Maximum \( y=-2e^{-1}\cdot\left(-\frac12\right)=\frac1e \), so the maximum value is \( e^{1/e} \).

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Answer: (b) \( \frac{3\sqrt3}4 \)

Explanation:
1. \( f'(x)=\cos x+\cos2x=0 \) gives \( 2\cos^2x+\cos x-1=0 \), so cos x = 1/2 (or −1).
2. At x = π/3, \( f=\frac{\sqrt3}2\cdot\frac32=\frac{3\sqrt3}4 \), the maximum.

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Answer: (a) \( \frac\pi6 \)

Explanation:
1. \( \sin\theta\sin\varphi=\frac12[\cos(\theta-\varphi)-\cos(\theta+\varphi)]=\frac12\left[\cos(\theta-\varphi)-\frac12\right] \).
2. It is largest when θ = φ, so θ = π/6.

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Answer: (d) neither a maximum nor a minimum

Explanation:
1. \( f'(x)=3(x-2)^2\ge0 \) everywhere, so f is increasing and f′ does not change sign at x = 2.
2. x = 2 is a point of inflection with no extremum.

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Answer: (a) \( \frac{p+q+r}3 \)

Explanation:
1. \( f'(x)=2[(x-p)+(x-q)+(x-r)]=0 \) gives 3x = p + q + r.
2. f″ = 6 > 0, so λ = (p + q + r)/3.

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Answer: (b) exactly one minimum

Explanation:
1. Every term is a non-negative even power of x, so f(x) ≥ 1 with equality only at x = 0.
2. f′(x) = 4x + ... has the sign of x, so f decreases for x < 0 and increases for x > 0. There is exactly one minimum and no maximum.

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Answer: (d) none of these

Explanation:
1. \( f'(x)=-\frac a{x^2}+2x=0 \) at x = −3 gives \( -\frac a9-6=0 \), so a = −54.
2. Then \( f''(x)=\frac{2a}{x^3}+2=\frac{-108}{-27}+2=6\gt 0 \): a minimum, not a maximum.
3. So no value of a gives a maximum at x = −3.

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69. a
70. l
71.
72. p
73. o
74. i
75. n
76. t
77.
78. t
79. u
80. r
81. n
82. s
83.
84. o
85. u
86. t
87.
88. t
89. o
90.
91. b
92. e
93.
94. a
95.
96. m
97. i
98. n
99. i
100. m
101. u
102. m
103. .