JEE Mathematics Application of Derivatives MCQs Set 12

Answer: (d) \( \frac1{y^2}-\frac1{x^2}=1 \)

Explanation:
1. Let the point of contact be (x, y). The tangent passes through the origin, so \( \frac yx=\cos x \).
2. Since \( \sin^2x+\cos^2x=1 \): \( y^2+\frac{y^2}{x^2}=1 \).
3. Divide by \( y^2 \): \( 1+\frac1{x^2}=\frac1{y^2}\Rightarrow\frac1{y^2}-\frac1{x^2}=1 \).

Teacher's Note:
1. A tangent through the origin has slope y/x, which must equal the derivative cos x.
2. Then eliminate x using sin²x + cos²x = 1.

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

Explanation:
1. \( y'=2xe^{-x}-x^2e^{-x}=xe^{-x}(2-x) \).
2. Since \( e^{-x}>0 \), y′ > 0 when x(2 − x) > 0, i.e. 0 < x < 2.

Teacher's Note:
1. The sign of the derivative decides the interval of increase.
2. e^(−x) is always positive, so only x(2 − x) matters.

Answer: (d) None of these

Explanation:
1. \( f'(x)=100x^{99}+\cos x \).
2. On \( (0,\pi/2) \) and \( (0,1) \) both terms are positive, so f′ > 0.
3. On \( (\pi/2,\pi) \), \( 100x^{99}>100(1.5)^{99} \) far exceeds |cos x| ≤ 1, so f′ > 0 again.
4. The function is not decreasing on any of the intervals.

Teacher's Note:
1. Estimate sizes: 100x⁹⁹ is huge compared with cos x once x > 1.
2. A function decreases only where its derivative is negative.

Answer: (a) x + 2y = 0

Explanation:
1. Put x = 0: y = 1 − 1 = 0, so the point is (0, 0).
2. \( y'=-\frac12e^{x/2}=-\frac12 \) at x = 0.
3. Tangent: \( y=-\frac x2 \), i.e. x + 2y = 0.

Teacher's Note:
1. Find the point first, then the slope.
2. Write the tangent in the form given in the options.

Answer: (b) \( \tan^{-1}(4\sqrt2/7) \)

Explanation:
1. Equate: \( x^2-1=3-x^2 \) gives \( x^2=2 \), so the intersection points are \( (\pm\sqrt2,1) \).
2. At \( x=\sqrt2 \): for the first curve \( y=x^2-1 \) so \( m_1=2x=2\sqrt2 \); for the second \( y=3-x^2 \) so \( m_2=-2x=-2\sqrt2 \).
3. \( \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|=\left|\frac{4\sqrt2}{1-8}\right|=\frac{4\sqrt2}{7} \).

Teacher's Note:
1. Decide the sign inside each modulus at x² = 2 before differentiating.
2. By symmetry the other intersection point gives the same angle.

Answer: (a) \( \frac{19}{3}a+1+(\sin3-\sin2) \)

Explanation:
1. \( f'(x)=ax^2+1+\cos x>0 \) for a > 0 (since a x² ≥ 4a > 0 and 1 + cos x ≥ 0), so f is increasing on [2, 3].
2. Difference \( =f(3)-f(2)=\int_2^3(at^2+1+\cos t)dt=\frac{19a}{3}+1+\sin3-\sin2 \).

Teacher's Note:
1. First show monotonicity, so the greatest and least values occur at the end points.
2. Integrate between 2 and 3 directly.

Answer: (d) (1, 0)

Explanation:
1. Differentiate: \( y'-e^{xy}(y+xy')+1=0\Rightarrow y'=\frac{ye^{xy}-1}{1-xe^{xy}} \).
2. The tangent is vertical when the denominator is zero: \( xe^{xy}=1 \).
3. Check (1, 0): it lies on the curve (\( 0-e^0+1=0 \)) and \( 1\cdot e^{0}=1 \), with numerator \( 0-1\ne0 \). The tangent is vertical there; (1, 1) and (0, 1) do not lie on the curve.

Teacher's Note:
1. Vertical tangent means dy/dx is infinite, so set the denominator of y′ to zero.
2. Always check that the point lies on the curve.

Answer: (a) t = 0

Explanation:
1. \( \frac{dx}{dt}=2t,\ \frac{dy}{dt}=2t-1 \).
2. A tangent perpendicular to the x-axis is vertical, so \( \frac{dx}{dt}=0 \) with \( \frac{dy}{dt}\ne0 \).
3. This gives t = 0, where dy/dt = −1 ≠ 0.

Teacher's Note:
1. A vertical tangent has infinite slope: dx/dt = 0 and dy/dt ≠ 0.
2. A horizontal tangent would instead be dy/dt = 0.

Answer: (d) None of these

Explanation:
1. Rewrite as \( (y-2)^2=2x^3-4 \). Slope: \( y'=\frac{3x^2}{y-2} \).
2. The tangent at (x₀, y₀) passes through (1, 2): \( (y_0-2)^2=3x_0^2(x_0-1) \). Combined with the curve: \( 2x_0^3-4=3x_0^3-3x_0^2 \), i.e. \( (x_0+1)(x_0-2)^2=0 \).
3. x₀ = −1 gives \( (y_0-2)^2=-6 \) (impossible); x₀ = 2 gives \( (y_0-2)^2=12 \), so the points are \( (2,2\pm2\sqrt3) \), which is not among (a)-(c).

Teacher's Note:
1. Use the condition that the tangent line at the contact point passes through the external point.
2. Check the original options against the curve equation: (2, 2 ± √3) does not lie on it.

Answer: (c) 15

Explanation:
1. Differentiate: \( 2x+y+xy'+2yy'=0\Rightarrow y'=-\frac{2x+y}{x+2y} \). At (1, −3): \( y'=-\frac{-1}{-5}=-\frac15 \).
2. Subtangent length \( =\left|\frac{y}{y'}\right|=\frac{3}{1/5}=15 \).

Teacher's Note:
1. Subtangent = |y / (dy/dx)|, the projection of the tangent segment on the x-axis.
2. Check that the given point is on the curve: 1 − 3 + 9 = 7.

Answer: (a) always 0

Explanation:
1. An increasing function of a decreasing function is decreasing, so h is decreasing on \( [0,\infty) \).
2. Then for x ≥ 0, \( h(x)\le h(0)=0 \). But h maps into \( [0,\infty) \), so h(x) ≥ 0.
3. Hence h(x) = 0 for every x ≥ 0, and h(x) − h(1) = 0 always.

Teacher's Note:
1. Combine the monotonicity of the composition with the range condition.
2. h ≥ 0 and h ≤ 0 forces h ≡ 0.

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

Explanation:
1. \( g'(x)=-f'(2-x)+f'(4+x) \).
2. Since f″ > 0, f′ is increasing, so \( f'(4+x)>f'(2-x) \) exactly when \( 4+x>2-x \), i.e. x > −1.
3. Hence g′ > 0 on \( (-1,\infty) \).

Teacher's Note:
1. f″ > 0 means f′ is strictly increasing.
2. Compare the two arguments 4 + x and 2 − x.

Answer: (b) 3%

Explanation:
1. \( V=\frac43\pi r^3\Rightarrow\frac{\Delta V}{V}=3\frac{\Delta r}{r} \).
2. So the percentage error in V is 3 × 1% = 3%.

Teacher's Note:
1. For V ∝ rⁿ, the relative error in V is n times the relative error in r.
2. Take logarithms and differentiate: dV/V = 3 dr/r.

Answer: (d) \( 4\sqrt{65} \)

Explanation:
1. P′ has a minimum at x = 1, so P″(1) = 0. Write \( P=ax^3+bx^2+cx+d \): b = −3a. P′(−1) = 0 gives c = −9a.
2. P(−1) = 5a + d = 10 and P(1) = −11a + d = −6, so a = 1, d = 5 and \( P=x^3-3x^2-9x+5 \).
3. Maximum at x₁ = −1 (y₁ = 10), minimum at x₂ = 3 (y₂ = −22). \( AB=\sqrt{4^2+32^2}=\sqrt{1040}=4\sqrt{65} \).

Teacher's Note:
1. P′ minimum at 1 gives P″(1) = 0; P maximum at −1 gives P′(−1) = 0.
2. Then use the two function values to find the constants.

Answer: (c) \( (0,1/e) \)

Explanation:
1. \( f'(x)=x^x(1+\ln x) \).
2. f′ < 0 when ln x < −1, i.e. 0 < x < 1/e.

Teacher's Note:
1. Use logarithmic differentiation for x^x.
2. Decreasing means f′ < 0.

Answer: (c) \( -17-(x-2)^4 \)

Explanation:
1. We need f(2) = −17, f′(2) = f″(2) = 0 and a maximum at 2.
2. For \( -17-(x-2)^4 \): f(2) = −17, \( f'=-4(x-2)^3 \) and \( f''=-12(x-2)^2 \) both vanish at 2, and f ≤ −17 near 2, so it is a local maximum.
3. (a) has a minimum of 0 and (b) has value 3 at x = 2, so both fail.

Teacher's Note:
1. Check the value at x = 2 first, then the type of extremum.
2. The fourth-power term makes the first three derivatives behave as required.

Answer: (a) \( a=-\tfrac34,\ b=-\tfrac18 \)

Explanation:
1. \( f'(x)=\frac ax+2bx+1 \) must vanish at x = 1 and x = 3.
2. \( a+2b+1=0 \) and \( \frac a3+6b+1=0\Rightarrow a+18b+3=0 \).
3. Subtract: 16b + 2 = 0, so \( b=-\frac18 \) and \( a=-1+\frac14=-\frac34 \).

Teacher's Note:
1. Extrema occur where f′ = 0; set up two equations.
2. Subtract the equations to eliminate a.

Answer: (d) A is false, but R is true.

Explanation:
1. \( f(x)=(x-1)|x-1| \) equals \( (x-1)^2 \) for x ≥ 1 and \( -(x-1)^2 \) for x < 1, with \( f'(x)=2|x-1| \) existing everywhere (including 0 at x = 1).
2. So f is continuous and differentiable on every interval and the LMVT conditions hold: A is false.
3. R is true: |x − 1| has a corner at x = 1.

Teacher's Note:
1. Check differentiability of the product, not of its factors separately.
2. A product can be smoother than one of its factors.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Let \( F(x)=ax^4+bx^3+cx^2+dx \). Then F′(x) = f(x), F(0) = 0 and \( F(3)=3(27a+9b+3c+d)=0 \).
2. By Rolle's theorem (R) on [0, 3], F′(x) = f(x) = 0 for some x in (0, 3).
3. So both statements are true and R is the reason.

Teacher's Note:
1. Integrate f(x) to find a function whose derivative is f.
2. Check that F(0) = F(3).

Answer: (d) A is false, but R is true.

Explanation:
1. A is false: Rolle's theorem also needs f to be continuous on [a, b] and differentiable on (a, b).
2. For R: y′ = 3x² − 2x − 1 = 0 at x = 1 and y(1) = 1, so the tangent is y = 1. Then \( x^3-x^2-x+1=(x-1)^2(x+1)=0 \) gives x = −1 again. R is true.

Teacher's Note:
1. Rolle's theorem has three conditions, not just equal end values.
2. For the tangent, substitute y = 1 into the cubic.

Answer: (d) A is false, but R is true.

Explanation:
1. The sum of distances from 1, 2, 3 is least at the median x = 2 (value 2), not at 3 (value 3): A is false.
2. f has corners at 1, 2 and 3, so it is not differentiable at x = 3: R is true.

Teacher's Note:
1. The sum of absolute deviations is minimised at the median.
2. Each |x − k| has a corner at x = k.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. \( f'(x)=1-\sin x\ge0 \) and equals 0 only at isolated points, so f is strictly increasing: A is true.
2. R is also a true general statement (a strictly increasing function may have f′ = 0 at isolated points), but it does not explain why x + cos x is increasing, since here f′ vanishes at infinitely many points.

Teacher's Note:
1. Check f′ ≥ 0 and that its zeros are isolated.
2. Decide whether R actually explains A.