JEE Mathematics Application of Derivatives MCQs Set 21

Answer: (d) (1), (3) and (4) are correct

Explanation:
1. (1) is true: sin x is increasing on the closed interval.
2. (2) is false: take x₁ = 1.5 (sin x₁ ≈ 0.997) and x₂ slightly more than 3π/2 (sin x₂ ≈ −1). Then x₁ < x₂ but sin x₁ > sin x₂, so sin x is not increasing on the union.
3. (3) is true: at each point of the union the function is increasing in a neighbourhood.
4. (4) is true: it is increasing in each of the two intervals separately.

Teacher's Note:
1. Increasing in a set (comparing any two points) is different from increasing at each point.
2. A union of two separate intervals is not an interval of monotonicity.

Answer: (c) only one minimum

Explanation:
1. \( f'(x)=4x+16x^3+36x^5+\cdots+200\cdot50\,x^{99}=x\,(4+16x^2+\cdots) \), where the bracket is positive for every real x.
2. So f' < 0 for x < 0 and f' > 0 for x > 0: f has exactly one turning point, a minimum at x = 0.

Teacher's Note:
1. Every term of f'(x) has an odd power of x with a positive coefficient.
2. The bracket is a sum of positive terms with even powers.

Answer: (d) \( \frac{11}4 \)

Explanation:
1. The circle has centre (7/4, 0) and radius 1, so the shortest distance between the curves is the shortest distance from the centre to the ellipse minus 1.
2. On the ellipse \( y^2=18-\frac{9x^2}{16} \), so the squared distance from the centre is \( d^2=\left(x-\frac74\right)^2+18-\frac{9x^2}{16}=\frac{7x^2}{16}-\frac{7x}{2}+\frac{49}{16}+18 \).
3. This is minimum at \( x=\frac{7/2}{7/8}=4 \) (inside the range of x), where \( d^2=7-14+18+\frac{49}{16}=\frac{225}{16} \), so d = 15/4.
4. Shortest distance = 15/4 − 1 = 11/4.

Teacher's Note:
1. The centre (7/4, 0) lies on the major axis, so the nearest point of the ellipse is found by minimising the distance with x as the variable.
2. The normal to the circle always passes through its centre.

Answer: (b) \( y-\frac12=-\frac1{16}(x-8) \)

Explanation:
1. Point of contact \( (t,4/t) \), slope \( -4/t^2 \). Tangent: \( y=1-\frac{4}{t^2}x \) if it passes through (0, 1) with that slope.
2. The point on the curve must satisfy it: \( \frac4t=1-\frac4t \), so t = 8.
3. Point (8, 1/2), slope −1/16: y − 1/2 = −(1/16)(x − 8).

Teacher's Note:
1. The tangent line through (0, 1) has equation y = mx + 1.
2. Use the fact that the point of contact lies on both the curve and the line.

Answer: (c) \( \frac1{18\pi} \) cm/min

Explanation:
1. The outer radius is R = 10 + 5 = 15 cm and the volume of ice is \( V=\frac43\pi(R^3-10^3) \).
2. So \( \frac{dV}{dt}=4\pi R^2\frac{dR}{dt} \) and \( -50=4\pi(225)\frac{dR}{dt} \).
3. dR/dt = −50/(900π) = −1/(18π): the thickness decreases at 1/(18π) cm/min.

Teacher's Note:
1. The inner ball does not change, so only the outer radius R varies.
2. The negative sign shows that the thickness is decreasing.

Answer: (c) 2, 0

Explanation:
1. \( f(x)=2-|x+1| \). On [−2, 1] the maximum is 2 at x = −1.
2. f(−2) = 2 − 1 = 1 and f(1) = 2 − 2 = 0, so the least value is 0.

Teacher's Note:
1. The square root of a perfect square is a modulus.
2. Compare end-point values with the value at the corner x = −1.

Answer: (b) \( \frac12(a+b)^2 \)

Explanation:
1. If the sides of the circumscribed rectangle make an angle θ with the sides of the given one, its sides are \( a\cos\theta+b\sin\theta \) and \( a\sin\theta+b\cos\theta \).
2. Area \( =ab+\frac{a^2+b^2}2\sin2\theta \).
3. Maximum at θ = 45°: \( ab+\frac{a^2+b^2}2=\frac{(a+b)^2}2 \).

Teacher's Note:
1. Express both sides of the outer rectangle with one angle.
2. sin 2θ has maximum value 1.

Answer: (a) x = 3

Explanation:
1. Near x = 3, \( f(x)=\frac{x-1}{x^2} \) and \( f'(x)=\frac{x^2-2x(x-1)}{x^4}=\frac{2-x}{x^3} \), which is −1/27 < 0 at x = 3: decreasing.
2. At x = 2, f'(2) = 0 (stationary, not decreasing). At x = 1 the function falls to 0 and then rises, so it is a minimum, neither increasing nor decreasing there.

Teacher's Note:
1. Check the sign of f' at each point after removing the modulus.
2. A point with f' = 0 and a sign change is a turning point.

Answer: (a) 4

Explanation:
1. Let the horizontal distance be x and the cord length s. Then \( s^2=x^2+300^2 \), with s = 500 giving x = 400.
2. Differentiate: \( s\frac{ds}{dt}=x\frac{dx}{dt} \), so \( \frac{ds}{dt}=\frac{400}{500}\times5=4 \) km/h.

Teacher's Note:
1. The height is constant, so only the horizontal distance changes.
2. Find x using the Pythagoras relation first.

Answer: (a) \( (-\infty,-3] \)

Explanation:
1. \( f'(x)=3(a+2)x^2-6ax+9a\le0 \) for all x requires a + 2 < 0 and discriminant ≤ 0.
2. Discriminant: \( 36a^2-108a(a+2)\le0 \), i.e. \( a(a+3)\ge0 \), so a ≤ −3 or a ≥ 0.
3. With a < −2 the common set is a ≤ −3. (a = −2 gives f'(x) = 12x − 18, which changes sign.)

Teacher's Note:
1. A quadratic stays non-positive when its leading coefficient is negative and its discriminant is non-positive.
2. Check the boundary a = −2 separately.

Answer: (d) g(x) has no extremum

Explanation:
1. The reflection is the inverse function, \( g(x)=\sinh^{-1}x=\ln\left(x+\sqrt{x^2+1}\right) \).
2. Its derivative \( g'(x)=\frac1{\sqrt{x^2+1}} \) is positive and never zero, so g is strictly increasing with no extremum and no horizontal tangent.
3. It is also finite everywhere, so there is no vertical tangent. At the origin g'(0) = 1, so the tangent is y = x, not y = −x.

Teacher's Note:
1. f(x) = sinh x; its inverse is the inverse hyperbolic sine.
2. A strictly increasing function has no local extremum.

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

Explanation:
1. The shortest distance lies along a common normal. Take the point \( (2s^2,4s) \) on the first parabola; its normal has slope −s.
2. The point \( (3+u^2,2u) \) on the second parabola has normal slope −u. A common normal needs s = u.
3. Then the distance is \( \sqrt{(3+s^2-2s^2)^2+(2s-4s)^2}=\sqrt{(3-s^2)^2+4s^2}=\sqrt{s^4-2s^2+9} \), minimised at s² = 1 giving \( \sqrt8=2\sqrt2 \).

Teacher's Note:
1. The line joining the nearest points is normal to both curves.
2. Minimise the expression in s² (a perfect square plus a constant).

Answer: (a) \( a=11,\ b\in\mathbb R \)

Explanation:
1. f is a polynomial, hence continuous and differentiable. We need f(1) = f(3).
2. f(1) = −5 + a + b and f(3) = −27 + 3a + b. Equating: −5 + a = −27 + 3a, so a = 11.
3. b does not appear in the condition, so b is any real number.

Teacher's Note:
1. Only f(1) = f(3) needs to be imposed for a polynomial.
2. The constant b cancels.

Answer: (c) (4, 4)

Explanation:
1. Circle: centre (0, 12), radius 4. The nearest point of the parabola to the circle is the nearest point to the centre, found from a normal passing through (0, 12).
2. Normal at \( (t^2,2t) \): \( y+tx=2t+t^3 \). Through (0, 12): \( t^3+2t-12=0 \), which has the single real root t = 2.
3. The point is (4, 4).

Teacher's Note:
1. The line joining the centre of the circle to the nearest point is normal to the parabola.
2. t³ + 2t − 12 is increasing, so there is only one real root.

Answer: (b) \( \frac{-16}{9t^3} \)

Explanation:
1. For y = 12/x, dy/dx = −12/x² = −4/(3t²) at the point, so the normal slope is 3t²/4.
2. The slope of the chord to the second point is \( \frac{4/t_1-4/t}{3t_1-3t}=-\frac{4}{3tt_1} \).
3. Equate: \( -\frac4{3tt_1}=\frac{3t^2}4 \), so \( t_1=-\frac{16}{9t^3} \).

Teacher's Note:
1. The second point lies on the normal, so the chord slope equals the normal slope.
2. Simplify (4/t₁ − 4/t) = 4(t − t₁)/(t t₁).

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

Explanation:
1. Tangent to \( y=-x^2 \) at x = p (p < 0): \( y=-2px+p^2 \). Tangent to \( y=x^2+8 \) at x = q (q ≥ 0): \( y=2qx+8-q^2 \).
2. Equal slopes: −2p = 2q, so p = −q. Equal intercepts: p² = 8 − q², so 2q² = 8, q = 2.
3. Slope 4 and intercept 8 − 4 = 4: y = 4x + 4.

Teacher's Note:
1. Write the tangent to each branch at its own point of contact and equate slope and intercept.
2. Check that p = −2 is negative and q = 2 is non-negative.

Answer: (c) \( 2\sqrt8,\ 3\sqrt2 \)

Explanation:
1. Take a vertex at \( (4\cos\theta,3\sin\theta) \). The rectangle has sides 8cos θ and 6 sin θ, area \( 24\sin2\theta \).
2. Maximum when θ = π/4: sides \( 8\cdot\frac1{\sqrt2}=4\sqrt2=2\sqrt8 \) and \( 6\cdot\frac1{\sqrt2}=3\sqrt2 \).

Teacher's Note:
1. Use the parametric form of the ellipse.
2. A = 24 sin 2θ is largest at 2θ = 90 degrees.

Answer: (a) \( 2x-y-1=0 \)

Explanation:
1. Differentiate: \( \frac{dy}{dx}=-\frac1{2\sqrt x} \), which equals −1/2 at x = 1.
2. Slope of the normal = 2.
3. Normal: y − 1 = 2(x − 1), i.e. 2x − y − 1 = 0.

Teacher's Note:
1. Normal slope = −1 / (tangent slope).
2. Check that (1, 1) satisfies the final equation.

Answer: (c) \( <\frac1{2N} \)

Explanation:
1. For n > N²: \( \sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n} \).
2. Since √n > N and √(n+1) > N, the denominator exceeds 2N.
3. So the difference is less than 1/(2N).

Teacher's Note:
1. Rationalise the difference of square roots.
2. Use √n > N to bound the denominator.

Answer: (b) 20 cm/min

Explanation:
1. Radius and depth are in the ratio 1 : 2, so at depth h the radius is h/2 and \( V=\frac13\pi\frac{h^2}{4}h=\frac{\pi h^3}{12} \).
2. Then \( \frac{dV}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt} \). With 77 litre = 77000 cm³ and h = 70: \( 77000=\frac{22}{7}\cdot\frac{4900}{4}\frac{dh}{dt}=3850\frac{dh}{dt} \).
3. dh/dt = 20 cm/min.

Teacher's Note:
1. Convert all lengths to cm and litres to cm³ before substituting.
2. Express the radius through the depth using similar triangles.

Answer: (d) one half that of the cone

Explanation:
1. Let the cone have radius R and height H, and the cylinder radius r and height h. By similar triangles \( h=H\left(1-\frac rR\right) \).
2. Curved surface \( S=2\pi rh=2\pi H\left(r-\frac{r^2}{R}\right) \).
3. dS/dr = 0 gives \( 1-\frac{2r}R=0 \), so r = R/2 (and S'' < 0, a maximum).

Teacher's Note:
1. Use similar triangles to eliminate the height of the cylinder.
2. The expression r − r²/R is maximum at r = R/2.

Answer: (a) \( x+y=0 \)

Explanation:
1. For small x, \( f(x)=\frac{\sin x^2}{x}\approx\frac{x^2}{x}=x \), so \( f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x}=\lim_{x\to0}\frac{\sin x^2}{x^2}=1 \).
2. The tangent at the origin has slope 1, so the normal has slope −1.
3. Normal: y = −x, i.e. x + y = 0.

Teacher's Note:
1. Use the definition of the derivative at 0 because the formula for f is not valid at x = 0.
2. The normal slope is the negative reciprocal of the tangent slope.

Answer: (b) \( \tan\theta=\left|\frac{\log(a/b)}{1+\log a\log b}\right| \)

Explanation:
1. The curves meet where a^x = b^x, i.e. at x = 0 (the point (0, 1)).
2. Slopes there: \( m_1=\ln a,\ m_2=\ln b \).
3. Angle between them: \( \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|=\left|\frac{\ln(a/b)}{1+\ln a\ln b}\right| \).

Teacher's Note:
1. Find the intersection point first and take the derivatives there.
2. Use tan θ = |(m₁ − m₂)/(1 + m₁m₂)|.

Answer: (c) is equal to 1

Explanation:
1. A non-decreasing function of a non-increasing function is non-increasing, so h is non-increasing.
2. Hence h(2) ≤ h(1) = 1. Since the range of f is within [1, 10], h(2) ≥ 1.
3. So h(2) = 1.

Teacher's Note:
1. Compose the monotonicities: increasing ∘ decreasing is decreasing.
2. Use the range [1, 10] for the lower bound.

Answer: (d) (0, 1]

Explanation:
1. The minimum of \( Ax^2+Bx+C \) is \( C-\frac{B^2}{4A} \): \( m(b)=1-\frac{b^2}{1+b^2}=\frac1{1+b^2} \).
2. As b varies over the reals, 1 + b² ranges over [1, ∞), so m(b) ranges over (0, 1].

Teacher's Note:
1. The minimum value is attained at x = −b/(1 + b²).
2. The value 0 is not attained because 1/(1 + b²) is always positive.