JEE Mathematics Application of Derivatives MCQs Set 22

Answer: (c) \( \tan^{-1}4 \)

Explanation:
1. y = x² − 2x − 3, so dy/dx = 2x − 2, which is 4 at x = 3.
2. The tangent makes angle \( \tan^{-1}4 \) with the x-axis.

Teacher's Note:
1. The slope of a tangent is the tangent of its inclination.
2. The point (3, 0) is a root of the curve.

Answer: (d) the number of roots can be two in (−1, 0)

Explanation:
1. Let \( F(x)=\frac{ax^5}5+\frac{bx^3}3+cx \). Then F(0) = 0 and F(1) = (3a + 5b + 15c)/15 = 0, so F' = ax⁴ + bx² + c has a root in (0, 1) by Rolle's theorem. The equation is even, so there is also a root in (−1, 0).
2. With t = x², at² + bt + c has two roots with sum −b/a < 0 (since ab > 0), so at most one of them is positive. Hence at most one root in (0, 1) and one in (−1, 0): exactly one each, two in (−1, 1).
3. Therefore (A), (B), (C) are correct and (D) is incorrect.

Teacher's Note:
1. Integrate the polynomial to build a function with equal values at 0 and 1.
2. Use the sum of roots in t = x² to limit the number of positive roots.

Answer: (c) (1, 0)

Explanation:
1. Differentiate: \( y'-e^{xy}(y+xy')+1=0 \), so \( y'(1-xe^{xy})=ye^{xy}-1 \).
2. A vertical tangent needs \( 1-xe^{xy}=0 \) with a nonzero right side.
3. At (1, 0): the point lies on the curve (0 − 1 + 1 = 0), 1 − 1·e⁰ = 0 and the right side is 0 − 1 = −1 ≠ 0, so the tangent is vertical.
4. (0, 1): 1 − 0 = 1 ≠ 0; (1, 1) is not on the curve (1 − e + 1 ≠ 0).

Teacher's Note:
1. Vertical tangent means dy/dx is infinite: the coefficient of y' vanishes while the other side does not.
2. First confirm that the point lies on the curve.

Answer: (a) −3

Explanation:
1. Since (1, 1) lies on the curve, c = −b. The slope there is 2 + b.
2. Tangent: y − 1 = (2 + b)(x − 1). Intercepts: x-intercept \( 1-\frac1{2+b} \), y-intercept \( -(1+b) \).
3. Area \( =\frac12\left|\frac{1+b}{2+b}\right||1+b|=2 \). For b = −3: intercepts 2 and 2, area 2.
4. Other options give 8/5 (b = 3) and 9/8 (b = 2); b = −2 makes the tangent parallel to the x-axis so there is no triangle.

Teacher's Note:
1. Write the intercepts of the tangent in terms of its slope.
2. Test each option in the area formula.

Answer: (b) 1.035

Explanation:
1. Let f(x) = tan x, x = 45°, h = 1° = π/180 radian.
2. f(x + h) ≈ f(x) + h f'(x) = 1 + (π/180)·sec²45° = 1 + 2π/180 = 1 + π/90.
3. With π = 22/7: 1 + 22/630 = 1 + 0.0349 = 1.035.

Teacher's Note:
1. Convert the 1° increment to radians before using the differential.
2. sec² 45° = 2.

Answer: (b) \( \frac\ell{\sqrt3} \)

Explanation:
1. Let the circumradius of the base be r and the altitude h: \( r^2+h^2=\ell^2 \).
2. The side of the base is r√3, so the base area is \( \frac{\sqrt3}{4}\cdot3r^2=\frac{3\sqrt3}{4}r^2 \). Volume \( V=\frac{3\sqrt3}4(\ell^2-h^2)h \).
3. dV/dh = 0 gives \( \ell^2-3h^2=0 \), so \( h=\frac{\ell}{\sqrt3} \).

Teacher's Note:
1. The centre of an equilateral triangle is at distance r (the circumradius) from each vertex.
2. V is a cubic in h with a single maximum for h > 0.

Answer: (d) \( \frac{5\pi}6 \)

Explanation:
1. \( \frac{dy}{dx}=\frac{-a\sin\theta}{a(1+\cos\theta)}=-\tan\frac\theta2 \).
2. At θ = π/3: \( -\tan\frac\pi6=-\frac1{\sqrt3} \).
3. The angle in [0, π) with tangent −1/√3 is 5π/6.

Teacher's Note:
1. Use 1 + cos θ = 2cos²(θ/2) and sin θ = 2 sin(θ/2) cos(θ/2).
2. A negative slope means an obtuse inclination.

Answer: (c) \( (-1,3) \)

Explanation:
1. \( f'(x)=3(x^2-2px+p^2-1)=3(x-p-1)(x-p+1) \), so the points of extremum are x = p − 1 and x = p + 1.
2. Both must lie in (−2, 4): p − 1 > −2 gives p > −1 and p + 1 < 4 gives p < 3.
3. So p ∈ (−1, 3).

Teacher's Note:
1. Factorise the derivative: the discriminant is a perfect square.
2. Impose both end-point conditions.

Answer: (b) \( |a| \)

Explanation:
1. Parametrise: \( x=a\cos^3t,\ y=a\sin^3t \). Then \( \frac{dy}{dx}=-\tan t \).
2. Tangent: \( y-a\sin^3t=-\tan t\,(x-a\cos^3t) \), with intercepts \( x=a\cos t \) and \( y=a\sin t \).
3. Length \( =\sqrt{a^2\cos^2t+a^2\sin^2t}=|a| \).

Teacher's Note:
1. The curve is the astroid; its parametric form makes the intercepts simple.
2. The length does not depend on t.

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

Explanation:
1. A tangent through the origin at the point (x, y) has slope y/x = cos x, so sin x = x cos x, i.e. tan x = x.
2. Then \( y^2=\sin^2x=\frac{\tan^2x}{1+\tan^2x}=\frac{x^2}{1+x^2} \).
3. So \( y^2+x^2y^2=x^2 \), i.e. \( x^2-y^2=x^2y^2 \).

Teacher's Note:
1. The tangent passes through the origin, so its slope equals y/x at the point.
2. Express sin²x using tan x = x.

Answer: (a) \( \left(\frac12,\frac74\right) \)

Explanation:
1. Substituting the three points gives a = 1, b = 1, c = 1, so the parabola is y = x² + x + 1.
2. Triangles ABD is fixed; the area of ABCD is greatest when triangle BCD has maximum area, i.e. when C (between B and D) is farthest from line BD.
3. Slope of BD = (7 − 1)/(2 + 1) = 2. The tangent at C is parallel to BD: 2x + 1 = 2, so x = 1/2 and y = 1/4 + 1/2 + 1 = 7/4.
4. C = (1/2, 7/4).

Teacher's Note:
1. A, B, D lie on the parabola, so first find its equation.
2. The farthest point from a chord is where the tangent is parallel to it.

Answer: (d) \( \frac\pi2 \)

Explanation:
1. Solving: subtracting gives 5y² = 20, so y² = 4 and x² = 16. Take the point (4, 2).
2. Slopes: for the ellipse \( y'=-\frac{x}{4y}=-\frac12 \); for the hyperbola \( y'=\frac xy=2 \).
3. Product of slopes = −1, so the curves are orthogonal: angle π/2.

Teacher's Note:
1. By symmetry the angle is the same at all four points of intersection.
2. Product of slopes equal to −1 means perpendicular tangents.

Answer: (d) \( \frac\pi2 \)

Explanation:
1. At the point (1, 1) both equations hold (1 − 3 + 2 = 0 and 3 − 1 − 2 = 0).
2. First curve: \( 3x^2-3y^2-6xyy'=0 \), so \( y'=\frac{x^2-y^2}{2xy}=0 \) at (1, 1).
3. Second curve: \( 6xy+3x^2y'-3y^2y'=0 \), so \( y'=\frac{-2xy}{x^2-y^2} \), which is infinite at (1, 1) (vertical tangent).
4. One tangent is horizontal and the other vertical, so the angle is π/2.

Teacher's Note:
1. These are the real and imaginary parts of z³ = −2 + 2i, so the curves are orthogonal everywhere.
2. If one slope is 0 and the other is infinite, the tangents are perpendicular.

Answer: (d) \( (0,b) \)

Explanation:
1. The line has slope −b/a and passes through (0, b), which lies on the curve (y = b e⁰ = b).
2. The curve has slope \( y'=-\frac bae^{-x/a} \), which at x = 0 equals −b/a, the same as the line.
3. So the line touches the curve at (0, b).

Teacher's Note:
1. Check both that the point lies on the curve and that the slopes agree.
2. The line meets the axes at (a, 0) and (0, b).

Answer: (c) −2

Explanation:
1. f(−3) = f(0) = 0. \( f'(x)=e^{-x/2}\left[(2x+3)-\frac{x^2+3x}2\right]=\frac{e^{-x/2}}2(-x^2+x+6) \).
2. f'(c) = 0 gives c² − c − 6 = 0, so c = 3 or c = −2.
3. Only c = −2 lies in (−3, 0).

Teacher's Note:
1. Reject roots outside the open interval.
2. Factor out e^{−x/2} before simplifying.

Answer: (b) \( f''(x)=2 \) for some x ∈ (1, 3)

Explanation:
1. Let g(x) = f(x) − x². Then g(1) = g(2) = g(3) = 0.
2. By Rolle's theorem g' = 0 at some c₁ ∈ (1, 2) and c₂ ∈ (2, 3); applying Rolle again to g', g''(x) = 0 for some x ∈ (c₁, c₂) ⊂ (1, 3).
3. So f''(x) = 2 for some x in (1, 3), but not necessarily for all.

Teacher's Note:
1. Apply Rolle's theorem twice to the difference f(x) − x².
2. Existence of a point is guaranteed; equality for all points is not.

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

Explanation:
1. For x ≤ 1, f'(x) = 3x² − 2x + 10 > 0, so f increases up to f(1) = 5.
2. For x > 1 the function decreases from the limit \( -2+\log_2(b^2-2) \).
3. Greatest value at x = 1 requires \( -2+\log_2(b^2-2)\le5 \) and b² − 2 > 0: \( 24. So b ∈ \( [-\sqrt{130},-\sqrt2)\cup(\sqrt2,\sqrt{130}] \).

Teacher's Note:
1. The logarithm requires b² − 2 > 0.
2. Compare the right-hand limit at x = 1 with the value f(1).

Answer: (a) \( (-3\sqrt3,-2\sqrt3)\cup(2\sqrt3,3\sqrt3) \)

Explanation:
1. f'(x) = a² − 3x² = 0 gives x = ±|a|/√3, and f''(x) = −6x > 0 for x < 0, so the minimum is at \( x=-\frac{|a|}{\sqrt3} \).
2. Since x² + x + 2 > 0 for all x, the inequality means (x + 2)(x + 3) < 0, i.e. −3 < x < −2.
3. So \( 2<\frac{|a|}{\sqrt3}<3 \), i.e. \( 2\sqrt3<|a|<3\sqrt3 \).

Teacher's Note:
1. The numerator has negative discriminant, so only the denominator decides the sign.
2. Remember that a and −a give the same f' zeros, hence the modulus.

Answer: (c) \( \frac12 \)

Explanation:
1. x³ − 2x² − 5x + 6 = (x − 1)(x − 3)(x + 2), so f(x) = (x − 3)(x + 2) = x² − x − 6 for x ≠ 1, and f(1) = −6 matches this value, so f(x) = x² − x − 6 on [−2, 3].
2. f(−2) = 0 = f(3), so Rolle's theorem applies.
3. f'(x) = 2x − 1 = 0 gives c = 1/2.

Teacher's Note:
1. Factorise the cubic to see that the function is the polynomial x² − x − 6.
2. The value at x = 1 fills the hole of the quotient.

Answer: (a) 20

Explanation:
1. Rolle's theorem needs f(a) = f(b) with a < b. Here \( \frac{\ln a}a=\frac{\ln b}b \), i.e. \( a^b=b^a \).
2. The only pair of distinct positive integers with this property is (2, 4): \( 2^4=4^2 \).
3. a² + b² = 4 + 16 = 20.

Teacher's Note:
1. ln a / a = ln b / b is equivalent to a^b = b^a.
2. The function is increasing up to e and decreasing afterwards, so a < e < b.

Answer: (a) \( \sqrt2\,e^{-1/2} \)

Explanation:
1. Vertices at (±x, e^{−x²}): area \( A=2xe^{-x^2} \).
2. A' = 2e^{−x²}(1 − 2x²) = 0 gives \( x=\frac1{\sqrt2} \).
3. A = \( \frac{2}{\sqrt2}e^{-1/2}=\sqrt2\,e^{-1/2} \).

Teacher's Note:
1. The curve is symmetric about the y-axis, so the rectangle is symmetric too.
2. Check that A' changes from positive to negative at x = 1/√2.