JEE Mathematics Application of Derivatives MCQs Set 31

Answer: (d) \( 1 \)

Explanation:
1. The slope of the normal is \( \tan\frac{3\pi}4=-1 \).
2. The slope of the tangent is the negative reciprocal, so \( f'(3)=1 \).

Teacher's Note:
1. Tangent slope and normal slope multiply to \( -1 \).

Answer: (b) \( \frac{\log\frac ab}{1+(\log a)(\log b)} \)

Explanation:
1. \( a^x=b^x \) gives \( x=0 \), where both curves pass through \( (0,1) \).
2. The slopes there are \( m_1=\log a \) and \( m_2=\log b \).
3. \( \tan\alpha=\frac{m_1-m_2}{1+m_1m_2}=\frac{\log a-\log b}{1+\log a\log b}=\frac{\log\frac ab}{1+(\log a)(\log b)} \).

Teacher's Note:
1. Curves meet where their values are equal; then apply the angle formula for slopes.

Answer: (b) \( a^2-b^2=0 \)

Explanation:
1. For the hyperbola, \( \frac{dy}{dx}=\frac{b^2x}{a^2y} \); for \( xy=c^2 \), \( \frac{dy}{dx}=-\frac yx \).
2. Orthogonality at a common point needs \( \frac{b^2x}{a^2y}\cdot\left(-\frac yx\right)=-1 \), i.e. \( \frac{b^2}{a^2}=1 \).
3. So \( a^2-b^2=0 \).

Teacher's Note:
1. Product of slopes is \( -1 \) for perpendicular curves; the point coordinates cancel.

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

Explanation:
1. On \( \left(\frac\pi2,\pi\right) \), \( (\sin x)'=\cos x<0 \) and \( (\cos x)'=-\sin x<0 \), so both functions decrease. The assertion is true.
2. A decreasing function need not have a decreasing derivative. For example \( f(x)=\frac1x \) decreases on \( (1,2) \), but \( f'(x)=-\frac1{x^2} \) increases there. So the reason is false.

Teacher's Note:
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Answer: (a) \( \left(\frac1{2n+1},\frac1{2n}\right) \)

Explanation:
1. \( f'(x)=\frac{\pi}{x^2}\sin\frac\pi x \), which is positive when \( \sin\frac\pi x>0 \).
2. This holds when \( 2n\pi<\frac\pi x<(2n+1)\pi \), i.e. \( \frac1{2n+1}<x<\frac1{2n} \).

Teacher's Note:
1. The factor \( \frac\pi{x^2} \) is always positive, so only the sign of the sine matters.

Answer: (b) \( \frac\pi4<x<\frac{3\pi}8 \)

Explanation:
1. \( f(x)=1-\frac12\sin^22x \), so \( f'(x)=-\sin4x \).
2. \( f'>0 \) when \( \sin4x<0 \), i.e. \( 4x\in(\pi,2\pi) \), so \( x\in\left(\frac\pi4,\frac\pi2\right) \) in the first period.
3. Only \( \frac\pi4<x<\frac{3\pi}8 \) lies inside this interval; the other options include points where \( \sin4x\ge0 \).

Teacher's Note:
1. Rewrite \( \sin^4x+\cos^4x \) using \( \sin^2x+\cos^2x=1 \) to get a single trigonometric term.

Answer: (c) \( [-1,1]\cup[3,\infty) \)

Explanation:
1. \( g'(x)=(2x-2)\left[f'(x^2-2x+8)-f'(14+2x-x^2)\right] \).
2. Since \( f''\ge0 \), \( f' \) is increasing, so the bracket is \( \ge0 \) iff \( x^2-2x+8\ge14+2x-x^2 \), i.e. \( (x-3)(x+1)\ge0 \).
3. \( g'\ge0 \) when \( x>1 \) and the bracket is \( \ge0 \) (so \( x\ge3 \)), or when \( x<1 \) and the bracket is \( \le0 \) (so \( -1\le x<1 \)). Hence g increases on \( [-1,1]\cup[3,\infty) \).

Teacher's Note:
1. Differentiate with the chain rule and use that \( f' \) is increasing to compare its two values.

Answer: (a) \( \operatorname{cosec}x \)

Explanation:
1. \( \operatorname{cosec}x \) increases on \( \left(\frac\pi2,\pi\right) \) and decreases on \( \left(\pi,\frac{3\pi}2\right) \) (it is not even defined at \( \pi \)).
2. \( \tan x \) has derivative \( \sec^2x>0 \), \( x^2 \) and \( |x-1| \) have positive derivative for \( x>1 \), so they are monotonic increasing.

Teacher's Note:
1. Check the sign of the derivative on the whole interval; cosec x changes its behaviour at the middle.

Answer: (c) \( \left(-\frac\pi3,\frac\pi3\right) \)

Explanation:
1. \( f'(x)=\sec^2x-4<0 \) iff \( \sec^2x<4 \), i.e. \( |\cos x|>\frac12 \).
2. On \( \left(-\frac\pi2,\frac\pi2\right) \) this gives \( -\frac\pi3<x<\frac\pi3 \), so the largest interval offered is \( \left(-\frac\pi3,\frac\pi3\right) \).

Teacher's Note:
1. Solve \( f'(x)<0 \) and then pick the widest option that fits.

Answer: (d) \( \lambda>4 \)

Explanation:
1. The numerator of \( f'(x) \) is \( (\lambda\cos x-6\sin x)(2\sin x+3\cos x)-(\lambda\sin x+6\cos x)(2\cos x-3\sin x) \).
2. This simplifies to \( (3\lambda-12)(\sin^2x+\cos^2x)=3\lambda-12 \).
3. \( f'>0 \) iff \( \lambda>4 \).

Teacher's Note:
1. For a ratio of two linear forms in sin and cos, the derivative reduces to a constant over a square.

Answer: (d) \( a=11 \)

Explanation:
1. \( f'(x)=3x^2-12x+a \) has roots \( 2\pm\frac1{\sqrt3} \) (one of them is \( \frac{2\sqrt3+1}{\sqrt3}=2+\frac1{\sqrt3} \) and the sum of roots is 4).
2. Product of roots \( =\frac a3=\left(2+\frac1{\sqrt3}\right)\left(2-\frac1{\sqrt3}\right)=\frac{11}3 \), so \( a=11 \).

Teacher's Note:
1. The sum and product of the roots of \( f' \) fix the unknown coefficient.

Answer: (b) 0

Explanation:
1. \( \int_1^2f'(x)\,dx=f(2)-f(1) \).
2. Rolle's theorem requires \( f(1)=f(2) \), so the integral is 0.

Teacher's Note:
1. Use the fundamental theorem of calculus together with the equal end values.

Answer: (c) \( 1-\frac{\sqrt{21}}6 \)

Explanation:
1. \( f(x)=x^3-3x^2+2x \), \( f(0)=0 \) and \( f\!\left(\frac12\right)=\frac38 \), so the chord slope is \( \frac{3/8}{1/2}=\frac34 \).
2. \( f'(c)=3c^2-6c+2=\frac34 \) gives \( 12c^2-24c+5=0 \), so \( c=1\pm\frac{\sqrt{21}}6 \).
3. Only \( c=1-\frac{\sqrt{21}}6\approx0.236 \) lies in \( \left(0,\frac12\right) \).

Teacher's Note:
1. Always check that the c you find lies inside the open interval.

Answer: (d) \( \frac12 \)

Explanation:
1. \( f(1)=0=f(0) \), so the end values are equal.
2. For continuity at 0 we need \( \lim_{x\to0^+}x^\alpha\log x=0 \), which needs \( \alpha>0 \).
3. Among the options only \( \alpha=\frac12 \) is positive.

Teacher's Note:
1. Continuity at the left end is the real restriction here.

Answer: (b) \( b=-5,\ c=8 \)

Explanation:
1. \( f(1)=f(2) \) gives \( 1+b+c=8+4b+2c \), so \( 3b+c+7=0 \).
2. \( f'\!\left(\frac43\right)=\frac{16}3+\frac{8b}3+c=0 \) gives \( 8b+3c+16=0 \).
3. Solving: \( b=-5 \), \( c=8 \).

Teacher's Note:
1. Two conditions, two unknowns: equal end values and zero derivative at the given point.

Answer: (c) The cubic has minima at \( \sqrt{\frac p3} \) and maxima at \( -\sqrt{\frac p3} \)

Explanation:
1. \( f'(x)=3x^2-p=0 \) gives \( x=\pm\sqrt{\frac p3} \) and \( f''(x)=6x \).
2. \( f''>0 \) at \( x=\sqrt{\frac p3} \) (minimum) and \( f''<0 \) at \( x=-\sqrt{\frac p3} \) (maximum).

Teacher's Note:
1. A cubic with positive leading coefficient has its local maximum on the left and local minimum on the right.

Answer: (a) \( (0,1) \)

Explanation:
1. Let \( F(x)=\frac{ax^3}3+\frac{bx^2}2+cx \), so \( F(0)=0 \) and \( F(1)=\frac{2a+3b+6c}6=0 \).
2. By Rolle's theorem \( F'(x)=ax^2+bx+c \) vanishes somewhere in \( (0,1) \).

Teacher's Note:
1. Build the antiderivative whose end values are equal and apply Rolle.

Answer: (b) \( a \) has no real value

Explanation:
1. The line is \( y=3-x \). Tangency needs \( \frac{ax}{1+x}=3-x \) to have a double root.
2. \( ax=(3-x)(1+x) \) gives \( x^2+(a-2)x-3=0 \), with discriminant \( (a-2)^2+12>0 \) for every real a.
3. So the equation always has two distinct roots and tangency is impossible.

Teacher's Note:
1. Tangency means a repeated root when the line is substituted in the curve.

Answer: (a) \( f(x) \) is strictly increasing

Explanation:
1. \( f'(x)=3x^2+2bx+c \) has discriminant \( 4b^2-12c<4c-12c<0 \) because \( b^2<c \).
2. So \( f'(x)>0 \) for all x and f is strictly increasing.

Teacher's Note:
1. A quadratic derivative with negative discriminant never changes sign.

Answer: (d) \( a\in(5,\infty) \)

Explanation:
1. \( f'(x)=\left|\log_2[\ldots]\right|\ge0 \), so f increases wherever it is defined for all x.
2. For \( \log_2[\log_3(\log_4(\cos t+a))] \) to exist for every t we need \( \log_3(\log_4(\cos t+a))>0 \), i.e. \( \log_4(\cos t+a)>1 \), i.e. \( \cos t+a>4 \).
3. The smallest value of \( \cos t+a \) is \( a-1 \), so \( a-1>4 \) and \( a>5 \).

Teacher's Note:
1. The real work is making the nested logarithms defined for every t.

Answer: (d) none of these

Explanation:
1. The slopes at P are \( \frac{2a}{y_1} \) and \( \frac{x_1}{2b} \); orthogonality needs \( ax_1=-by_1 \).
2. From \( y_1^2=4ax_1 \) the sign of \( x_1 \) is that of a, and from \( x_1^2=4by_1 \) the sign of \( y_1 \) is that of b, so \( ax_1>0 \) and \( by_1>0 \) and the product condition \( ax_1=-by_1 \) can never hold.
3. So the parabolas never intersect orthogonally at a point other than the origin, and none of the given relations is possible.

Teacher's Note:
1. Check the signs of the coordinates before solving for a relation between the constants.

Answer: (a) \( f''(1)+f'(0)=1 \)

Explanation:
1. Let \( f(x)=Ax^2+Bx+C \), so \( f''=2A \) and \( f'(0)=B \).
2. Touching \( y=x \) at \( x=1 \) gives \( f'(1)=2A+B=1 \).
3. Hence \( f''(1)+f'(0)=2A+B=1 \).

Teacher's Note:
1. For a parabola the second derivative is constant, which makes this relation immediate.

Answer: (b) \( f\left(\frac{x_1+x_2}2\right)<\frac{f(x_1)+f(x_2)}2 \)

Explanation:
1. \( f''>0 \) means f is convex (its graph lies below every chord).
2. Therefore the value at the midpoint is less than the average of the end values.

Teacher's Note:
1. Convexity gives a midpoint inequality; nothing is known about the convexity of \( f' \).

Answer: (a) \( \left(\frac\pi4,\frac\pi2\right) \)

Explanation:
1. \( g'(x)=\cos x\,f'(\sin x)-\sin x\,f'(\cos x) \), and \( f'<0 \), \( f''>0 \) on \( (0,1) \) so \( f' \) is negative and increasing.
2. For \( x>\frac\pi4 \): \( \sin x>\cos x \), so \( f'(\sin x)>f'(\cos x) \), i.e. \( |f'(\sin x)|<|f'(\cos x)| \). Then \( g'(x)=\sin x\,|f'(\cos x)|-\cos x\,|f'(\sin x)|>0 \).
3. For \( x<\frac\pi4 \) the inequalities reverse and \( g'<0 \). So g increases on \( \left(\frac\pi4,\frac\pi2\right) \).

Teacher's Note:
1. Compare the two terms of \( g' \) using that \( f' \) is increasing and \( \sin x \) versus \( \cos x \).

Answer: (b) \( \frac67 \)

Explanation:
1. \( t^2+3t-8=2 \) gives \( t=2 \) or \( t=-5 \); \( y=-1 \) only for \( t=2 \).
2. \( \frac{dy}{dx}=\frac{4t-2}{2t+3}=\frac67 \) at \( t=2 \).

Teacher's Note:
1. First find the parameter value that gives the point on both coordinates.

Answer: (d) \( -\infty<a\le-3 \)

Explanation:
1. \( f'(x)=3(a+2)x^2-6ax+9a\le0 \) for all x needs \( a+2<0 \) and discriminant \( \le0 \).
2. \( 36a^2-108a(a+2)\le0 \) gives \( -72a^2-216a\le0 \), so \( a(a+3)\ge0 \), i.e. \( a\le-3 \) or \( a\ge0 \).
3. With \( a<-2 \) the common range is \( a\le-3 \).

Teacher's Note:
1. A quadratic stays non-positive only if its leading coefficient is negative and its discriminant is not positive.

Answer: (c) \( f(x) \) is an increasing function

Explanation:
1. \( f'(x)=\frac{\sin x-x\cos x}{\sin^2x}>0 \) because \( \tan x>x \) for \( 0<x<\frac\pi2 \).
2. \( g'(x)=\frac{\tan x-x\sec^2x}{\tan^2x}=\frac{\sin x\cos x-x}{\sin^2x}<0 \) because \( \sin x\cos x=\frac12\sin2x<x \).
3. So f increases and g decreases.

Teacher's Note:
1. Use the inequality \( \tan x>x>\sin x \) on \( \left(0,\frac\pi2\right) \).

Answer: (b) \( ad-bc<0 \)

Explanation:
1. The numerator of the derivative is \( (a\cos x-b\sin x)(c\sin x+d\cos x)-(a\sin x+b\cos x)(c\cos x-d\sin x) \).
2. This simplifies to \( (ad-bc)(\sin^2x+\cos^2x)=ad-bc \).
3. The function decreases when \( ad-bc<0 \).

Teacher's Note:
1. The derivative reduces to the determinant \( ad-bc \) over a square.