JEE Mathematics Solutions of Equations and Inequations MCQs Set 04

Answer: (b) 11

Explanation:
1. \( |t|=t \) holds exactly when \( t\ge0 \), so \( x^2+6x+p\ge0 \) for every real \( x \).
2. This needs discriminant \( 36-4p\le0 \), that is \( p\ge9 \).
3. The smallest prime that is at least 9 is 11.

Teacher's Note:
1. An equation of the form |t| = t is the same as the inequality t >= 0.
2. A quadratic with positive leading coefficient is non-negative for all x exactly when its discriminant is at most zero.

Answer: (a) \( (-\infty,4] \)

Explanation:
1. \( |x-1|+a=\pm4 \) gives \( |x-1|=4-a \) or \( |x-1|=-4-a \).
2. A real solution exists if at least one right side is non-negative.
3. \( 4-a\ge0 \) gives \( a\le4 \); and \( -4-a\ge0 \) gives \( a\le-4 \), which is already included.
4. So \( a\in(-\infty,4] \).

Teacher's Note:
1. |x - 1| = k is solvable exactly when k >= 0.
2. Take the union of the conditions from the two branches.

Answer: (a) 1

Explanation:
1. For \( x\ge\tfrac32 \): left side \( 4x \), so \( 4x=4x+6 \), impossible.
2. For \( -\tfrac32\le x\lt \tfrac32 \): left side \( 6 \), so \( 6=4x+6 \) gives \( x=0 \), which is valid.
3. For \( x\lt -\tfrac32 \): left side \( -4x \), so \( -4x=4x+6 \) gives \( x=-\tfrac34 \), which is not in the range.
4. Exactly one value, \( x=0 \).

Teacher's Note:
1. Check each root against the interval of its own case.
2. The middle interval is the one where the two absolute values partly cancel.

Answer: (c) 3

Explanation:
1. If \( y\ge0 \): \( x-y=1 \) and \( |x+2|+y=5 \) give \( |x+2|+x=6 \).
2. For \( x\ge-2 \): \( 2x+2=6 \), so \( x=2 \) and \( y=1 \) (valid). For \( x\lt -2 \) we get \( -2=6 \), impossible.
3. If \( y\lt 0 \): \( x+y=1 \) gives \( |x+2|-x=4 \); \( x\ge-2 \) gives \( 2=4 \), impossible; \( x\lt -2 \) gives \( x=-3 \), \( y=4 \), which contradicts \( y\lt 0 \).
4. The only solution is \( (2,1) \), so \( x+y=3 \).

Teacher's Note:
1. Handle the sign of y first because |y| appears in the second equation.
2. Reject any pair that breaks the sign assumed for the case.

Answer: (a) 1

Explanation:
1. Let \( t=\log_3x \), so \( \log_{\sqrt3}x=2t \) and the equation is \( |2t-2|-|t-2|=2 \).
2. \( t\ge2 \): \( (2t-2)-(t-2)=t=2 \), valid, so \( x=9 \).
3. \( 1\le t\lt 2 \): \( (2t-2)-(2-t)=3t-4=2 \) gives \( t=2 \), not in range.
4. \( t\lt 1 \): \( (2-2t)-(2-t)=-t=2 \) gives \( t=-2 \), so \( x=\tfrac19 \).
5. The real solutions are \( 9 \) and \( \tfrac19 \), so exactly one is an integer.

Teacher's Note:
1. Convert both logarithms to the same base before splitting into cases.
2. The question asks only for integral values, so discard x = 1/9.

Answer: (d) 15

Explanation:
1. Factorise: \( x^2-6x+5=(x-1)(x-5) \), \( 2x^2-3x+1=(x-1)(2x-1) \), \( x^2-3x+2=(x-1)(x-2) \).
2. Dividing by \( |x-1| \) (note \( x=1 \) is itself a solution): \( \big||x-5|-|2x-1|\big|=3|x-2| \).
3. Put \( a=x-5,\ b=2x-1 \); then \( a+b=3(x-2) \) and \( \big||a|-|b|\big|=\min(|a+b|,|a-b|) \).
4. The equation says this minimum equals \( |a+b| \), i.e. \( |a+b|\le|a-b| \), i.e. \( ab\le0 \), so \( (x-5)(2x-1)\le0 \) and \( \tfrac12\le x\le5 \).
5. The solution set is \( [\tfrac12,5] \); its integers are 1, 2, 3, 4, 5 with sum 15.

Teacher's Note:
1. For any two numbers a and b, the quantity ||a| - |b|| equals the smaller of |a + b| and |a - b|.
2. This turns a nested modulus equation into the simple condition ab <= 0.

Answer: (a) -1

Explanation:
1. The expression is undefined at \( x=\pm1 \), so these two integers are not in the solution set.
2. Rewrite \( \dfrac{3|x|-2}{|x|-1}=3+\dfrac{1}{|x|-1} \). For \( |x|\gt 1 \) this exceeds 3, so the inequality holds for every such \( x \).
3. For \( 0\le u=|x|\lt 1 \): if \( u\le\tfrac23 \) the condition \( 2-3u\ge2(1-u) \) gives \( u=0 \); if \( u\gt \tfrac23 \) it gives \( u\ge\tfrac45 \). So no integer other than 0 lies in this range, and \( x=0 \) gives exactly 2, which satisfies the inequality.
4. Hence the only integers outside the solution set are \( 1 \) and \( -1 \), whose product is \( -1 \).

Teacher's Note:
1. Integers where the expression is undefined must be counted as not belonging to the solution set.
2. Check the integer 0 directly: the value is |(-2)/(-1)| = 2, which satisfies the inequality.

Answer: (a) (1), (2) and (3) only

Explanation:
1. \( f(x)=x+1-2(1-x)=3x-1 \) for \( -1\le x\lt 1 \); \( f(x)=x+1-2(x-1)=3-x \) for \( x\ge1 \); \( f(x)=x-3 \) for \( x\lt -1 \).
2. The function rises to \( f(1)=2 \) and falls on both sides, so the maximum is 2: statement (1) is true.
3. \( f(x)=1 \): \( 3x-1=1\Rightarrow x=\tfrac23 \) and \( 3-x=1\Rightarrow x=2 \): two solutions, statement (2) is true.
4. \( f(x)=2 \) only at the peak \( x=1 \): statement (3) is true. \( f(x)=3 \) is above the maximum, so there is no solution: statement (4) is false.

Teacher's Note:
1. Write the function piecewise first; the maximum of a piecewise linear function occurs at a breakpoint.
2. A horizontal line at the maximum value touches the graph only once.

Answer: (a) (1), (2) and (3) only

Explanation:
1. For \( x\gt 0 \): \( |x|\lt a/x \) means \( x^2\lt a \), possible only for \( a\gt 0 \), giving \( 0\lt x\lt \sqrt a \).
2. For \( x\lt 0 \): multiplying by negative \( x \) reverses the sign: \( -x^2\gt a \), i.e. \( x^2\lt -a \), possible only for \( a\lt 0 \), giving \( -\sqrt{-a}\lt x\lt 0 \).
3. For \( a=0 \) the right side is \( 0 \) and \( |x|\lt 0 \) is impossible, so the set is empty.
4. Statements (1), (2), (3) are correct; (4) is wrong because the upper limit is \( \sqrt a \), not \( a \).

Teacher's Note:
1. Multiplying an inequality by a negative number reverses it, so treat x > 0 and x < 0 separately.
2. Test with a = 4: the solution is (0, 2), not (0, 4).

Answer: (d) 6

Explanation:
1. Put \( u=\ln|x-1| \) and multiply the equation by \( \ln3\ln4\ln5 \): \( u^3=u\ln3\ln4+u^2\ln5 \).
2. So \( u\,(u^2-u\ln5-\ln3\ln4)=0 \).
3. The quadratic has discriminant \( \ln^25+4\ln3\ln4\gt 0 \) and a negative product of roots, so it has two distinct non-zero real roots.
4. Thus \( u \) takes 3 distinct values (0 and the two roots). Each value gives \( |x-1|=e^u \), that is 2 values of \( x \) (for \( u=0 \): \( x=0,2 \), where both sides equal 0).
5. Total solutions: 6.

Teacher's Note:
1. Substituting u = ln|x - 1| turns all three logarithms into multiples of u.
2. Each positive value of |x - 1| gives two values of x.

Answer: (d) \( [-2-2\sqrt6,\,-1)\cup[2\sqrt6-2,\,3] \)

Explanation:
1. Domain: \( 21-4x-x^2\ge0 \) gives \( -7\le x\le3 \); also \( x\ne-1 \).
2. If \( x\gt -1 \): need \( \sqrt{21-4x-x^2}\le1 \), i.e. \( x^2+4x-20\ge0 \), so \( x\ge2\sqrt6-2 \) (the other root is below -1). Combined with the domain: \( [2\sqrt6-2,3] \).
3. If \( x\lt -1 \): need \( \sqrt{21-4x-x^2}\ge1 \), i.e. \( x^2+4x-20\le0 \), so \( -2-2\sqrt6\le x\lt -1 \).
4. Union: \( [-2-2\sqrt6,-1)\cup[2\sqrt6-2,3] \).

Teacher's Note:
1. Treat the sign of the denominator separately and keep the square-root domain in view.
2. x = -1 must be excluded, but the endpoints -2 - 2 sqrt 6 and 3 can be included.

Answer: (a) \( [-1,\sqrt[3]{4}) \)

Explanation:
1. The denominator is positive exactly when \( 4-x^3\gt 0 \), i.e. \( x\lt \sqrt[3]4 \).
2. The numerator is non-negative when \( |x+2|\ge|x| \), i.e. \( (x+2)^2\ge x^2 \), i.e. \( 4x+4\ge0 \), so \( x\ge-1 \).
3. Hence \( -1\le x\lt \sqrt[3]4 \).

Teacher's Note:
1. For |a| >= |b|, square both sides: a^2 >= b^2.
2. A square root in the denominator must be strictly positive.

Answer: (c) 2017

Explanation:
1. \( f_1(x)=2 \) gives \( |x|=4 \) or \( |x|=0 \): \( x=0,\pm4 \), that is 3 solutions.
2. \( f_2(x)=2 \) needs \( f_1(x)=0 \) or \( 4 \): \( x=\pm2,\pm6 \), that is 4 solutions.
3. \( f_3(x)=2 \) has solutions \( 0,\pm4,\pm8 \): 5 solutions. Each further step adds exactly one more solution.
4. So \( f_n(x)=2 \) has \( n+2 \) solutions, and for \( n=2015 \) there are 2017.

Teacher's Note:
1. Work out the first three or four cases and look for the pattern.
2. Check the pattern: n = 1, 2, 3 give 3, 4, 5 solutions.

Answer: (a) (1) and (2) only

Explanation:
1. \( |x-a|+|x+a|=2|a| \) when \( |x|\le|a| \) and \( =2|x| \) when \( |x|\gt |a| \); so it is always at least \( 2|a| \).
2. If \( b\le2|a| \) the inequality \( \lt b \) can never hold: statement (1) is true.
3. If \( b\gt 2|a| \): we need \( 2|x|\lt b \) (the case \( |x|\le|a| \) is automatically included), so \( -\tfrac b2\lt x\lt \tfrac b2 \): statement (2) is true.
4. Statement (3) is false, because for \( b\lt 2|a| \) there is no solution at all.

Teacher's Note:
1. The sum |x - a| + |x + a| has minimum value 2|a|.
2. Combine this with the observation that it equals 2 max(|x|, |a|).

Answer: (d) \( b\gt c\gt 0 \)

Explanation:
1. \( |x-a|-b=\pm c \) gives \( |x-a|=b+c \) or \( |x-a|=b-c \).
2. Four distinct roots need two different positive values of \( |x-a| \): \( b+c\gt b-c\gt 0 \), which requires \( c\gt 0 \) and \( b\gt c \).
3. So \( b\gt c\gt 0 \) (no condition on \( a \) is needed).

Teacher's Note:
1. Each positive value of |x - a| gives two roots, x = a plus or minus that value.
2. The root count does not depend on the position a of the centre.

Answer: (a) (1), (2) and (3) only

Explanation:
1. The domain is \( x\gt 0 \), so every root is positive.
2. For \( x\ge1 \): \( \ln x+x=2 \); the left side increases from 1 (at \( x=1 \)) to infinity, so exactly one root exists, \( x\approx1.557 \).
3. For \( 0\lt x\lt 1 \): \( x-\ln x=2 \); the left side decreases from infinity to 1, so exactly one root exists, \( x\approx0.159 \).
4. Hence two solutions, both positive, one above 1 and one below 1: (1), (2), (3) are correct and (4) is false.

Teacher's Note:
1. Split at x = 1 where ln x changes sign, and use monotonic behaviour to count roots.
2. Evaluate the left side at the end points of each interval to confirm a root exists.

Answer: (c) 2

Explanation:
1. \( |x-1|+|x+1|=2 \) for \( |x|\le1 \) and \( 2|x| \) for \( |x|\gt 1 \).
2. For \( |x|\le1 \): \( 2^x=2 \) gives \( x=1 \).
3. For \( x\gt 1 \): \( 2^x=2x \); the curves meet at \( x=2 \) (and at \( x=1 \), already counted).
4. For \( x\lt -1 \): \( 2^x\lt \tfrac12 \) while \( -2x\gt 2 \), so there is no solution.
5. Solutions: \( x=1 \) and \( x=2 \), that is 2.

Teacher's Note:
1. Write the right side piecewise and compare with the exponential on each piece.
2. 2^x = 2x has the roots x = 1 and x = 2 because the exponential is convex.

Answer: (c) 2

Explanation:
1. For \( |x|\le2 \): \( |x-2|+|x+2|=4 \), so \( x^2=3 \) and \( x=\pm\sqrt3 \), both within the range.
2. For \( |x|\gt 2 \): the right side is \( 2|x|-1 \), so \( x^2-2|x|+1=0 \), i.e. \( (|x|-1)^2=0 \) and \( |x|=1 \), which is not in this range.
3. Hence 2 solutions.

Teacher's Note:
1. For |x| <= 2 the two distances to 2 and -2 add up to the constant 4.
2. Always verify each root against the range of its case.

Answer: (c) \( \dfrac{15}{2} \)

Explanation:
1. The lower boundary is the V-shape with vertex \( (2,0) \); the upper boundary is an inverted V with vertex \( (3,4) \).
2. They meet where \( 2-x=1+x \) i.e. \( (0.5,1.5) \), and where \( x-2=7-x \) i.e. \( (4.5,2.5) \).
3. The region is the quadrilateral with vertices \( (0.5,1.5),(2,0),(4.5,2.5),(3,4) \).
4. Shoelace formula: \( \tfrac12|(-3)+5+10.5+2.5|=\tfrac12\cdot15=\tfrac{15}{2} \).

Teacher's Note:
1. Find the intersection points first; the two graphs are made of straight segments so the region is a polygon.
2. The sides are perpendicular (slopes 1 and -1), so it is a rectangle with sides 3 sqrt2 / 2 and 5 sqrt2 / 2.

Answer: (b) \( \min(p,q)=\tfrac12(p+q-|p-q|) \)

Explanation:
1. Since \( \max(p,q)=\tfrac12(p+q+|p-q|) \) and \( \min(p,q)=\tfrac12(p+q-|p-q|) \), option (b) is an identity.
2. Option (a) fails when \( r\le\max(p,q) \) (then both sides are equal). Option (c) fails whenever \( p=q \) or \( r \) is small: for example \( p=1,q=2,r=5 \) gives \( 2\lt 1 \), which is false.

Teacher's Note:
1. Test each option with simple numbers such as p = 1, q = 2, r = 0.
2. The max and min formulas with |p - q| are worth memorising.

Answer: (d) None of these

Explanation:
1. (a): at \( x=2 \): \( f(4)=3 \) but \( (f(2))^2=1 \), so it is false.
2. (b): at \( x=y=1 \): \( f(2)=1 \) but \( f(1)+f(1)=0 \), so it is false.
3. (c): at \( x=-2 \): \( f(|-2|)=f(2)=1 \) but \( |f(-2)|=3 \), so it is false.
4. So none of (a), (b), (c) holds.

Teacher's Note:
1. One counter-example is enough to reject an identity.
2. Choose values that make the modulus act differently on the two sides.

Answer: (a) \( \left|\dfrac rp-7\right|\ge4\sqrt3 \)

Explanation:
1. Since \( p,q,r \) are in A.P., \( q=\dfrac{p+r}2 \).
2. Real roots need \( q^2\ge4pr \): \( (p+r)^2\ge16pr \). Dividing by \( p^2 \) and writing \( k=r/p \): \( k^2-14k+1\ge0 \).
3. This gives \( k\le7-4\sqrt3 \) or \( k\ge7+4\sqrt3 \), i.e. \( |k-7|\ge4\sqrt3 \).

Teacher's Note:
1. The condition for real roots is discriminant >= 0.
2. Dividing by p^2 reduces the condition to a quadratic in the ratio r/p.

Answer: (b) 4

Explanation:
1. Put \( t=|x-2|\ge0 \): \( t^2+t-2=0 \) gives \( (t+2)(t-1)=0 \), so \( t=1 \) (since \( t=-2 \) is rejected).
2. \( |x-2|=1 \) gives \( x=1 \) or \( x=3 \).
3. Sum of the roots \( =1+3=4 \).

Teacher's Note:
1. The substitution t = |x - 2| must give a non-negative t.
2. The two roots are symmetric about x = 2, so their sum is 2 x 2 = 4.

Answer: (b) \( \alpha\lt 0\lt \beta\lt |\alpha| \)

Explanation:
1. The product of the roots is \( c\lt 0 \), so the roots have opposite signs: \( \alpha\lt 0\lt \beta \).
2. The sum of the roots is \( -b\lt 0 \), so \( \beta\lt -\alpha=|\alpha| \).
3. Hence \( \alpha\lt 0\lt \beta\lt |\alpha| \).

Teacher's Note:
1. Use the sum and product of the roots to get the signs without solving.
2. A negative sum with opposite signs means the negative root is larger in magnitude.

Answer: (c) does not exist

Explanation:
1. For every real \( x \): \( t^2x^2\ge0 \), \( |x|\ge0 \) and \( 9\gt 0 \), so the left side is at least 9.
2. The left side never equals 0, so the equation has no real roots and the product of real roots does not exist.

Teacher's Note:
1. Sum of non-negative terms plus a positive constant can never be zero.
2. If there are no real roots, any statement about their product is meaningless.

Answer: (d) no solution

Explanation:
1. A square root is non-negative, so we need \( x-3\ge0 \), i.e. \( x\ge3 \).
2. Squaring: \( 3x^2+x+5=x^2-6x+9 \) gives \( 2x^2+7x-4=0 \), so \( (2x-1)(x+4)=0 \) and \( x=\tfrac12 \) or \( x=-4 \).
3. Neither value satisfies \( x\ge3 \), so there is no solution.

Teacher's Note:
1. Always impose the condition (right side >= 0) before squaring a radical equation.
2. Both roots of the squared equation are extraneous here.

Answer: (c) \( (-\infty,0) \)

Explanation:
1. We need \( |x|-x\gt 0 \) (positive because it is under a root in the denominator).
2. For \( x\ge0 \): \( |x|-x=0 \), not allowed. For \( x\lt 0 \): \( |x|-x=-2x\gt 0 \), allowed.
3. So the domain is \( (-\infty,0) \).

Teacher's Note:
1. |x| - x is zero for all non-negative x and equals -2x for negative x.
2. The expression under a root in a denominator must be strictly positive.

Answer: (b) \( \dfrac{2\sqrt{13}}9 \)

Explanation:
1. \( \dfrac1\alpha+\dfrac1\beta=\dfrac{\alpha+\beta}{\alpha\beta}=-\dfrac qr=4 \), so \( q=-4r \).
2. A.P.: \( 2q=p+r \) gives \( p=2q-r=-9r \).
3. The equation becomes \( -9rx^2-4rx+r=0 \), i.e. \( 9x^2+4x-1=0 \).
4. \( |\alpha-\beta|=\dfrac{\sqrt{16+36}}{9}=\dfrac{\sqrt{52}}9=\dfrac{2\sqrt{13}}9 \).

Teacher's Note:
1. Express p and q in terms of r using the two given conditions.
2. |alpha - beta| equals the square root of the discriminant divided by |p|.