Q.By using the properties of definite integrals, evaluate the integral ∫0π/2sin3/2x+cos3/2xsin3/2xdx
Concept understanding — Property Of Symmetry
Symmetry Property of Definite Integrals
Suppose you must find ∫−22x3dx. You could integrate directly — or you could notice the graph of x3 is anti-symmetric about the origin, so every positive bit of area on the right is cancelled by an equal negative bit on the left, and the answer is simply 0. When the interval is symmetric about zero, the symmetry of the function does the work for you.
Even and odd functions
- An even function satisfies f(−x)=f(x) (e.g. x2, cosx, ∣x∣). Its graph is a mirror image across the y-axis, so the area on [−a,0] equals the area on [0,a].
- An odd function satisfies f(−x)=−f(x) (e.g. x3, sinx). Its left half is the negative mirror of its right half, so the two areas cancel.
∫−aaf(x)dx=⎩⎨⎧2∫0af(x)dx0if f is evenif f is odd
Why it works
Split at zero and substitute u=−x in the left piece:
∫−aafdx=∫−a0f(x)dx+∫0af(x)dx=∫0af(−u)du+∫0af(x)dx.
If f is even, f(−u)=f(u) and the two integrals add to 2∫0af. If f is odd, f(−u)=−f(u) and they cancel to 0.
Using it
∫−33x4dx=2∫03x4dx=2[5x5]03=5486,∫−ππsinxdx=0.
Two conditions must both hold: the interval must be [−a,a], and the function must actually be even or odd. Something like x2+x is neither, so the shortcut does not apply — check by replacing x with −x before you use it.
Quick test: substitute −x. Same expression back ⇒ even; the negative of it ⇒ odd; anything else ⇒ no symmetry shortcut.
The even/odd symmetry property of definite integrals is one of the core 'properties of definite integrals' listed in the NCERT Class 12 Integrals chapter, and it's a fast, guaranteed-marks CBSE board technique whenever the limits are symmetric about zero. Students searching 'definite integral of odd and even function' or 'properties of definite integrals class 12 examples' will find this double-if-even, zero-if-odd rule is exactly the shortcut those board solutions rely on.
Concept: Property of Symmetry — using the substitution x→2π−x to exploit the complementary relationship between sinx and cosx.
Let
I=∫0π/2sin3/2x+cos3/2xsin3/2xdx.
Step 1: Substitute x=2π−t, so dx=−dt. When x=0, t=2π; when x=2π, t=0.
I=∫π/20sin3/2(2π−t)+cos3/2(2π−t)sin3/2(2π−t)(−dt)=∫0π/2cos3/2t+sin3/2tcos3/2tdt.
Step 2: Renaming the dummy variable t back to x, we have
I=∫0π/2sin3/2x+cos3/2xcos3/2xdx.
Step 3: Add the two expressions for I:
2I=∫0π/2sin3/2x+cos3/2xsin3/2x+cos3/2xdx=∫0π/21dx=2π.
Thus I=4π.
The value is 4π.
Using the property ∫0af(x)dx=∫0af(a−x)dx, the given integral simplifies to 4π.
The trick here is symmetry. When you see an integral from 0 to π/2 with a ratio of sines and cosines, the substitution x→π/2−x often turns the denominator into a mirror image of itself. This lets you add the original and transformed integrals, giving a simple result.
Let’s work through it.
- Define the integral. Let
I=∫0π/2sin3/2x+cos3/2xsin3/2xdx.
- Apply the symmetry substitution. Use the property ∫0af(x)dx=∫0af(a−x)dx. Here a=π/2, so replace x by π/2−x:
I=∫0π/2sin3/2(π/2−x)+cos3/2(π/2−x)sin3/2(π/2−x)dx.
Recall the co-function identities:
sin(π/2−x)=cosx and cos(π/2−x)=sinx.
So the integral becomes
I=∫0π/2cos3/2x+sin3/2xcos3/2xdx.
- Add the two forms. Now we have two expressions for I:
I=∫0π/2sin3/2x+cos3/2xsin3/2xdx
and
I=∫0π/2sin3/2x+cos3/2xcos3/2xdx.
Add them:
2I=∫0π/2sin3/2x+cos3/2xsin3/2x+cos3/2xdx=∫0π/21dx.
The integrand simplifies to 1 (provided the denominator is never zero on [0,π/2], which it isn’t — both terms are non-negative and only vanish at the endpoints, but the sum is positive in between).
- Evaluate the simple integral.
∫0π/21dx=2π.
Hence 2I=π/2, so
I=4π.
A common mistake is to forget that the substitution x→a−x changes the limits but the property handles that automatically — you don’t need to recompute them. Also, be careful: the exponent 3/2 is fine here because the functions are well-defined and positive on (0,π/2).
This trick works for any integral of the form ∫0π/2f(sinx)+f(cosx)f(sinx)dx where f is any function for which the substitution works — the answer is always π/4, as long as the denominator never vanishes.
The value of the integral is 4π.
Method: The f+gf complementary-integral trick
For ∫0af(x)+f(a−x)f(x)dx (with f,g swapping under x→a−x), reflection produces the complementary fraction; the two add to 1.
Steps
Step 1: Set I and reflect with x→a−x.
I=∫0af(x)+g(x)f(x)dx⇒I=∫0ag(x)+f(x)g(x)dx,
where the reflection swaps f↔g (e.g. sin↔cos).
Step 2: Add the two forms.
The denominators are identical, so
2I=∫0af(x)+g(x)f(x)+g(x)dx=∫0a1dx=a.
Step 3: Solve for I.
I=2a.
This pattern always gives half the interval length, independent of the specific f.
Common Mistakes
Mistake 1: Thinking the exponent 23 changes the answer.
Why it's wrong: the f+gf reflection trick gives 4π for any power on sin/cos, since the two complementary fractions always sum to 1. Correct approach: reflect x→2π−x and add — the exponent is irrelevant.
Mistake 2: Sign or limit errors when substituting x=2π−t.
Why it's wrong: the −dt and swapped limits must cancel; mishandling them corrupts the reflected integral. Correct approach: reverse the limits to absorb the minus sign, then rename t back to x.
- AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQQ.Let 'a' be a non-zero real number. If the equation whose roots are the squares of the roots of the cubic equation x3−ax2+ax−1=0 is identical with this cubic equation, then 'a' = (A) 31 (B) 3 (C) 21 (D) 2
›Reveal solutionSolution
Use Vieta's formulas on the cubic, form the sum of squares of the roots via (∑p)2−2∑pq, and match it to the original cubic's coefficient since the two cubics are identical. Answer: a=3.
Concept and Intuition
If a new cubic (built from squares of the old roots) is identical to the old cubic, then its elementary symmetric functions of roots must match term-by-term with the old cubic's. This lets us bypass explicitly finding the roots — we only need symmetric-function identities.
Step-by-Step Solution
- Let p,q,r be the roots of x3−ax2+ax−1=0. By Vieta's: p+q+r=a, pq+qr+rp=a, pqr=1.
- The cubic with roots p2,q2,r2 is x3−(∑p2)x2+(∑p2q2)x−(pqr)2=0.
- Compute ∑p2=(p+q+r)2−2(pq+qr+rp)=a2−2a.
- Compute (pqr)2=12=1, matching the original's constant term −1 automatically — consistent for any a, so it isn't restrictive.
- For the new cubic to be identical to x3−ax2+ax−1=0, we need ∑p2=a (matching the x2-coefficient): a2−2a=a⇒a2−3a=0⇒a(a−3)=0.
- Since a=0 is given, a=3.
- Cross-check with the x-coefficient: ∑p2q2=(pq+qr+rp)2−2pqr(p+q+r)=a2−2a, same expression, so it also forces a=3 consistently.
Common Mistakes
- Forgetting to reject the extraneous root a=0 (explicitly excluded by the problem).
- Mixing up which symmetric function corresponds to which coefficient (sign errors on the x2 term).
✓Final answerThe correct option is (B) — 3.
ANSWER: B
- AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQQ.If α,β and γ are the roots of the equation x3−13x2+kx+189=0 such that β−γ=2, then β+γ:k+α= (A) 4:3 (B) 2:1 (C) 6:5 (D) 3:4
›Reveal solutionSolution
Vieta's formulas plus the given β−γ=2 pin down all three roots exactly (−3,7,9); the requested ratio works out to 4:3.
Concept and Intuition
Vieta's formulas relate a cubic's coefficients directly to symmetric functions of its roots (sum, sum of pairwise products, product). Combined with the extra condition β−γ=2, this is enough information to solve for all three roots explicitly, after which any requested expression can simply be evaluated numerically.
Step-by-Step Solution
- From x3−13x2+kx+189=0: α+β+γ=13, αβ+βγ+γα=k, αβγ=−189.
- Let s=β+γ. Since β−γ=2: β=2s+2, γ=2s−2, so βγ=4s2−4. Also α=13−s.
- Substitute into αβγ=−189: (13−s)⋅4s2−4=−189⇒(13−s)(s2−4)=−756.
- Expand: 13s2−52−s3+4s=−756⇒−s3+13s2+4s+704=0⇒s3−13s2−4s−704=0.
- Test s=16: 163−13(16)2−4(16)−704=4096−3328−64−704=0. ✓ So s=16 (the cubic's other factor s2+3s+44 has negative discriminant, so s=16 is the only real solution).
- With s=16: β=216+2=9, γ=216−2=7, α=13−16=−3. Check product: (−3)(9)(7)=−189 ✓.
- Compute k=αβ+βγ+γα=(−3)(9)+(9)(7)+(7)(−3)=−27+63−21=15.
- So β+γ=16 and k+α=15+(−3)=12. Ratio =16:12=4:3.
Common Mistakes
- Trying to solve for β,γ before eliminating α via the sum condition — introduces unnecessary extra variables.
- Sign error substituting α=−3 into k+α (getting 15+3=18 instead of 15−3=12).
- Not checking that the cubic in s has a unique real root, and mistakenly picking an extraneous complex solution.
✓Final answerThe correct option is (A) — 4:3.
ANSWER: A
- AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQQ.If the sum of the cubes of the roots of the equation x3−ax2+bx−c=0 is zero, then a3+3c= (A) −2ab (B) 2ab (C) −3ab (D) 3ab
›Reveal solutionSolution
Using Newton's identity for the sum of cubes of the roots of a cubic in terms of its elementary symmetric functions gives a3+3c=3ab directly.
Concept and Intuition
For roots α,β,γ of x3−e1x2+e2x−e3=0, Newton's identities give the power sums in terms of e1,e2,e3. In particular ∑α3=e13−3e1e2+3e3. Here matching the given cubic to this standard form: e1=a (sum of roots), e2=b (sum of pairwise products), e3=c (product of roots).
Step-by-Step Solution
- Compare x3−ax2+bx−c=0 with x3−e1x2+e2x−e3=0: so α+β+γ=a, αβ+βγ+γα=b, αβγ=c.
- Newton's identity: α3+β3+γ3=(α+β+γ)3−3(α+β+γ)(αβ+βγ+γα)+3αβγ=a3−3ab+3c.
- Given the sum of cubes is 0: a3−3ab+3c=0⇒a3+3c=3ab.
Common Mistakes
- Sign errors when matching e2,e3 to b,c from the given cubic's alternating-sign form.
- Misremembering the Newton's-identity coefficient (it's −3e1e2+3e3, not +3e1e2).
✓Final answerThe correct option is (D) — 3ab.
ANSWER: D
- AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQQ.If (h, k) is the image of the point (2, −3) with respect to the line 5x−3y=2, then h+k= (A) −3 (B) −343 (C) −341 (D) 5
›Reveal solutionSolution
Direct application of the point-reflection-about-a-line formula gives the image (−3,0), so h+k=−3 — option (A).
Concept and Intuition
The image of a point (x1,y1) reflected across a line ax+by+c=0 lies such that the line is the perpendicular bisector of the segment joining the point and its image. This gives the standard formula
h=x1−a2+b22a(ax1+by1+c),k=y1−a2+b22b(ax1+by1+c).
Step-by-Step Solution
- Write the line as 5x−3y−2=0, so a=5, b=−3, c=−2.
- Compute D=ax1+by1+c=5(2)+(−3)(−3)+(−2)=10+9−2=17.
- Compute a2+b2=25+9=34.
- Compute the scale factor a2+b22D=3434=1.
- h=x1−a(1)=2−5=−3.
- k=y1−b(1)=−3−(−3)=0.
- So (h,k)=(−3,0) and h+k=−3.
Common Mistakes
- Sign error writing the line in the form ax+by+c=0 (here c=−2, not +2).
- Forgetting the factor of 2 in the reflection formula (that version instead gives the foot of the perpendicular, not the full reflection).
✓Final answerThe correct option is (A) — −3.
ANSWER: A
- AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQQ.If x5+ax4+bx3+cx2+5x+d=0 (b∈R) is an odd order reciprocal equation of second type and 21+3i is a root, then b−c= (A) 0 (B) 5 (C) 12 (D) 18
›Reveal solutionSolution
Use the second-type reciprocal relations to pin a,d and b=−c, factor out the known root's minimal polynomial and the automatic root x=1, then match coefficients to get b−c=18.
Concept and Intuition
A reciprocal equation of the second type (odd degree n) has coefficients that are equal in magnitude but opposite in sign when read from either end: ai=−an−i. A structural consequence for odd degree is that x=1 is always a root (substituting x=1 makes the sum telescope to zero by the antisymmetry). Also, since coefficients are real, a non-real root's conjugate is automatically a root too, and together they satisfy a real quadratic factor.
Step-by-Step Solution
- Write the equation as x5+ax4+bx3+cx2+5x+d=0, i.e. coefficients (from x5 down) are 1,a,b,c,5,d.
- Second-type, degree 5: a0=−a5, a1=−a4, a2=−a3 gives 1=−d⇒d=−1; a=−5; b=−c.
- The given root is 21+3i=cos60∘+isin60∘, modulus 1. Its conjugate is also a root, and together they satisfy x2−x+1=0 (sum of root+conjugate =1, product =1).
- For odd-degree second-type equations x=1 is always a root (check: at x=1, the sum is 1+a+b+c+5+d=1−5+b+c+5−1=b+c=0, true since b=−c).
- So the quintic factors as (x−1)(x2−x+1)(x2+px+q) for real p,q. Expand (x−1)(x2−x+1)=x3−2x2+2x−1, then multiply by (x2+px+q): x5+(p−2)x4+(q−2p+2)x3+(2p−2q−1)x2+(2q−p)x−q.
- Match: x4 coeff p−2=a=−5⇒p=−3. Constant −q=d=−1⇒q=1. Check x-coeff: 2q−p=2+3=5 ✓ (matches given coefficient 5).
- Now b=q−2p+2=1+6+2=9 and c=2p−2q−1=−6−2−1=−9 (consistent with b=−c).
- So b−c=9−(−9)=18.
Common Mistakes
- Confusing first-type (ai=an−i) with second-type (ai=−an−i) — only the second type forces x=1 as an automatic root for odd degree.
- Forgetting to use the complex-conjugate root, which is what supplies the quadratic factor x2−x+1.
✓Final answerThe correct option is (D) — 18.
ANSWER: D
- AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQQ.A ray of light passing through the point (2,3) reflects on Y-axis at a point P. If the reflected ray passes through the point (3,2) and P=(a,b) then 5b= (A) a−5 (B) a−13 (C) a+13 (D) a+5
›Reveal solutionSolution
Reflecting the source point across the mirror (the Y-axis) and joining it to the destination point pins down P=(0,513), so 5b=13=a+13 (since a=0).
Concept and Intuition
For light reflecting off a straight mirror, the reflected ray behaves as if it originated from the mirror-image of the true source. So instead of applying the law of reflection angle-by-angle, we reflect the source point (2,3) across the Y-axis to get its image (−2,3); the straight line from this image through the actual destination point (3,2) crosses the Y-axis exactly at the true reflection point P.
Step-by-Step Solution
- P lies on the Y-axis, so P=(a,b)=(0,b).
- Image of source (2,3) reflected across the Y-axis (x→−x): (−2,3).
- The reflected ray's line is the straight line joining this image (−2,3) to the point it actually passes through, (3,2).
- Slope =3−(−2)2−3=5−1. Line: y−2=−51(x−3).
- At x=0: y=2−51(0−3)=2+53=513. So b=513, a=0.
- 5b=5×513=13. Checking the options with a=0: a+13=13 matches; the others (a−5=−5, a−13=−13, a+5=5) do not.
Common Mistakes
- Reflecting the destination point instead of the source (also valid, but must be paired with the other real point correctly) — mixing the two methods causes sign errors.
- Forgetting a=0 is forced because P is explicitly on the Y-axis, and instead trying to solve for a independently.
✓Final answerThe correct option is (C) — a+13.
ANSWER: C
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