Expand: (1+x)5=1+5x+…(1+x)5=1+5x+…, (1−x)5=1−5x+…(1−x)5=1−5x+…. Subtract gives about 10x10x. Dividing by x gives 10. Higher terms vanish in limit.
If f(x)={sinxx,x≠0k,x=0f(x)={xsinx,k,x=0x=0 is continuous at 0, k equals
A k = 0
B k = 2
C k = −1
D k = 1
limx→0sinxx=1limx→0xsinx=1. To make the piecewise function continuous at 0, define f(0)=kf(0)=k equal to this limit.
If f(x)={x2,x≤13x+b,x>1f(x)={x2,3x+b,x≤1x>1 is continuous at x=1, then b equals
A −1
B 0
C −2
D 1
Continuity at 1 needs 12=3(1)+b12=3(1)+b. So 1=3+b⇒b=−21=3+b⇒b=−2.
If f(x)={x2,x≤1ax+b,x>1f(x)={x2,ax+b,x≤1x>1 is differentiable at x=1, then a equals
A 2
B 1
C 3
D 0
Differentiability at 1 requires left derivative 2x2x at 1 equals right derivative a. So a=2a=2. (Continuity would then decide b, but a comes from slopes.)
If f(x)=∣x−1∣+∣x+1∣f(x)=∣x−1∣+∣x+1∣, f is not differentiable at
A x = 0
B x = 2
C x = 3
D x = 1
Absolute value creates corners at points where inside becomes zero. Here corners occur at x=1 and x=−1. So derivative fails at both those points due to slope change.
Sum of non-differentiable points of f(x)=∣x−1∣+∣x+1∣f(x)=∣x−1∣+∣x+1∣ is
A 1
B −1
C 0
D 2
The corners are at x=1 and x=−1. Their sum is 1+(−1)=01+(−1)=0. These are exactly the points where the derivative fails due to slope change.
If g(x)=x3g(x)=3x, g is not differentiable at
A x = 0
B x = 1
C x = 2
D x = −2
g′(x)=13x−2/3g′(x)=31x−2/3, which becomes unbounded at x=0. The graph has a vertical tangent there, so derivative does not exist as a finite number.
Evaluate limx→01+x−1−x−xx3limx→0x31+x−1−x−x
A 0
B 1/8
C −1/8
D 1/4
Use expansions: 1±x=1±x2−x28±x316+…1±x=1±2x−8×2±16×3+…. Difference becomes x+x38+…x+8×3+…. Subtract x leaves x388x3. Divide gives 1/8.
Evaluate limx→0arctanx−xx3limx→0x3arctanx−x
A 1/3
B 0
C −1/6
D −1/3
Series: arctanx=x−x33+…arctanx=x−3×3+…. Then arctanx−x≈−x33arctanx−x≈−3×3. Dividing by x3x3 gives −1/3.
Evaluate limx→0sin−1x−xx3limx→0x3sin−1x−x
A 1/6
B −1/6
C 0
D 1/3
Expansion: sin−1x=x+x36+…sin−1x=x+6×3+…. So sin−1x−x≈x36sin−1x−x≈6×3. Divide by x3x3 to get 1/6.
Evaluate limx→01−cosxxsinxlimx→0xsinx1−cosx
A 1
B 2
C 1/2
D 0
Use 1−cosx∼x221−cosx∼2×2 and sinx∼xsinx∼x. Then denominator xsinx∼x2xsinx∼x2. Ratio ∼x2/2×2=1/2∼x2x2/2=1/2.
Both sinxx→1xsinx→1 and ln(1+x)x→1xln(1+x)→1. The ratio of two expressions each tending to 1 also tends to 1.
If f(x)={x+k,x<0x2,x≥0f(x)={x+k,x2,x<0x≥0 is continuous at 0, k equals
A k = 1
B k = −1
C k = 0
D k = 2
Continuity at 0 needs LHL = f(0). For x<0, limit is 0+k=k0+k=k. For x≥0, f(0)=0²=0. So k must be 0.
For f(x)={x+k,x<0x2,x≥0f(x)={x+k,x2,x<0x≥0, differentiability at x=0 is
A always true
B true if k=0
C true if k=1
D never true
Even if k=0 makes it continuous, left derivative is 1 (from x+k) while right derivative at 0 is 0 (from x²). Since slopes differ, derivative at 0 never exists.
If h(x)=∣x∣xh(x)=∣x∣x, then h′(0) equals
A 1
B 0
C −1
D does not exist
For x≥0, h=x2h=x2 so derivative at 0 from right is 0. For x<0, h=−x2h=−x2 so derivative from left is 0. Both match, so h′(0)=0.
If p(x)=∣x∣p(x)=∣x∣, then left derivative at 0 is
A −1
B 1
C 0
D undefined
For x<0, |x|=−x, so slope is −1. Approaching 0 from left keeps the same slope. Hence left derivative at 0 equals −1.
Evaluate limx→0∣x∣xlimx→0x∣x∣ from the right side
A −1
B 0
C does not exist
D 1
For x>0, ∣x∣=x∣x∣=x, so ∣x∣/x=1∣x∣/x=1. Hence the right-hand limit at 0 equals 1, even though the two-sided limit fails.
Evaluate limx→0∣x∣xlimx→0x∣x∣ from the left side
A 0
B −1
C 1
D does not exist
For x<0, ∣x∣=−x∣x∣=−x, so ∣x∣/x=−1∣x∣/x=−1. Therefore the left-hand limit at 0 is −1, showing a jump between sides.
If f(x)=x2∣x∣f(x)=∣x∣x2 for x≠0, then limx→0f(x)limx→0f(x) is
A 1
B does not exist
C ∞
D 0
For x≠0, x2/∣x∣=∣x∣x2/∣x∣=∣x∣. As x→0, |x|→0 from both sides. Hence the two-sided limit exists and equals 0.
If y=x2e2xy=x2e2x, then y(n)y(n) contains factor
A e2xe2x always
B exex always
C e0xe0x always
D no exponential
Every derivative of e2xe2x keeps e2xe2x as a factor (with powers of 2). Product with x2x2 still results in each higher derivative having e2xe2x factor.
In Leibnitz theorem, (fg)(3)(fg)(3) has coefficient of f′g′′f′g′′ equal
A 1
B 6
C 3
D 0
Leibnitz gives (fg)(3)=(30)fg(3)+(31)f′g′′+(32)f′′g′+(33)f(3)g(fg)(3)=(03)fg(3)+(13)f′g′′+(23)f′′g′+(33)f(3)g. Here (31)=3(13)=3, so coefficient is 3.
If f(x)=x4f(x)=x4, then f(5)(x)f(5)(x) equals
A 24
B 120
C 24x24x
D 0
A polynomial of degree 4 becomes zero after differentiating 5 times. Each differentiation lowers degree by 1, so the fifth derivative of x4x4 is 0.
If f(x)=x4f(x)=x4, then f(4)(x)f(4)(x) equals
A 24x24x
B 24
C 4
D 0
Differentiate: x4→4×3→12×2→24x→24×4→4×3→12×2→24x→24. The fourth derivative is the constant 24. After that, derivatives become 0.
Evaluate limx→0sinx−x+x36x5limx→0x5sinx−x+6×3
A 1/120
B −1/120
C 0
D 1/24.
Using series sinx=x−x36+x5120+…sinx=x−6×3+120×5+…. After adding x366x3 and subtracting x, the leading term is x5120120x5. Divide by x5x5 gives 1/120.
From cosx=1−x22+x424−x6720+…cosx=1−2×2+24×4−720×6+…, we get 1−cosx−x22+x424≈x67201−cosx−2×2+24×4≈720×6. Dividing by x6x6 gives 1/720.
If ff is continuous and one-to-one near a, then inverse exists near
A a only
B 0 only
C f(a) only
D ∞ only
The inverse function is defined around y-values near f(a)f(a). If f is one-to-one and continuous locally, it maps a neighborhood of a to a neighborhood of f(a)f(a), where inverse works.
If f(x)=x3+xf(x)=x3+x, then (f−1)′(0)(f−1)′(0) equals
A 1
B 0
C 1/3
D 3
Solve f(x)=0⇒x3+x=0⇒x=0f(x)=0⇒x3+x=0⇒x=0. Then f′(x)=3×2+1f′(x)=3×2+1, so f′(0)=1f′(0)=1. Hence ((f^{-1})'(0)=1/f'(0)=1.
Evaluate limx→0x21−cosxlimx→01−cosxx2
A 1
B 1/2
C 0
D 2
Use 1−cosx∼x221−cosx∼2×2. Then x21−cosx∼x2x2/2=21−cosxx2∼x2/2×2=2. This is a common rearranged standard limit.
Evaluate limx→0sinx1−cosxlimx→01−cosxsinx
A 0
B ∞
C 1
D does not exist
For small x, sinx∼xsinx∼x and 1−cosx∼x2/21−cosx∼x2/2. Ratio ∼xx2/2=2x∼x2/2x=x2, which grows without bound as x→0.
Evaluate limx→0+xlnxlimx→0+xlnx
A 0
B 1
C −1
D −∞
As x→0⁺, lnx→−∞lnx→−∞ and x→0, giving 0·∞ type. Rewrite as lnx1/x1/xlnx and apply L’Hospital to get 0.