Q: Let f(x)=[(sin[x])/(([x]+1)²),, x>0,; [6pt] (cos ((π)/(2)[x]))/([x]),, x<0,] where [x] denotes the greatest integer not exceeding x. For f to be continuous at x=0, the value of k must be

0. For x->0+, [x]=0 so sin([x])/([x]+1)^2 = sin0/1 = 0. For x->0-, [x]=-1 so cos((pi/2)[x])/[x] = cos(-pi/2)/(-1) = 0/(-1) = 0. Both one-sided limits equal 0, so f(0) = k = 0 for continuity.

Question 1

Suppose f(x)=α(x)β(x)γ(x) for every real x, where α(x), β(x), and γ(x) are differentiable. If f'(2)=18f(2), α'(2)=3α(2), β'(2)=-4β(2), and γ'(2)=kγ(2), then what is the value of k?

Accepted Answer

19. Logarithmic differentiation: f'(2)/f(2)=a'/a+b'/b+g'/g = 3+(-4)+k = 18, so k=19.

Question 2

For the piecewise-defined function

f(x) = { [sin((a+1)x) + sin x]/x, x < 0; c, x = 0; [√(x + bx²) - √x]/(b x^(3/2)), x > 0 }

to be continuous at x = 0, which values of a, b, and c are required?

Accepted Answer

a = -3/2, c = 1/2, b ≠ 0. The function must be continuous at x = 0, which requires that the limits from both sides equal the value at x = 0. The chosen values of a, b, and c ensure that the left-hand limit approaches 1/2 and the right-hand limit also approaches 1/2 as x approaches 0, thus satisfying the continuity condition.

Question 3

Let f(x) and g(x) be continuously differentiable functions such that f'(x) = g(x) and f''(x) = -f(x). Define h(x) = [f(x)]² + [g(x)]². If h(0) = 5, what is the value of h(10)?

Accepted Answer

5. h' = 2 f f' + 2 g g' = 2 f g + 2 g f'' = 2 f g + 2 g(-f) = 0, so h is constant. Therefore h(10) = h(0) = 5.

Question 4

For what value of p can the function

f(x) = { ((4^x - 1)³)/(sin x · [log|1 + x²/3|]^p), x ≠ 0; 12(log 4)³, x = 0 }

be made continuous at x = 0?

Accepted Answer

1. Near 0: (4^x-1)^3 ~ x^3 (ln4)^3, sin(scaled x) ~ x, and [log(1+x^2/3)]^p ~ (x^2/3)^p ~ x^(2p). For a finite nonzero limit the powers of x must balance: 3 = 1 + 2p, giving p = 1.

Question 5

Consider the function f: R -> R given by f(x) = max{x, x³}. At which points is f(x) not differentiable?

Accepted Answer

{-1, 0, 1}. The function f(x) = max{x, x³} is not differentiable at points where the two functions x and x³ intersect or change dominance, which occurs at x = -1, 0, and 1. At these points, the left-hand and right-hand derivatives do not match, indicating non-differentiability.

Question 6

Consider the function

f(x)=[(1-sin³ x)/(3cos² x),, x<(π)/(2),; [4pt]
p,, x=(π)/(2),; [4pt]
q(1-sin x),, x>(π)/(2).]

If f is continuous at x=(π)/(2), then the pair (p,q) is

Accepted Answer

(1/2, 4). As x->pi/2 from the left, (1-sin^3 x)/(3cos^2 x) -> 1/2 (factor 1-sin^3 x = (1-sin x)(1+sin x+sin^2 x) and cos^2 x = 1-sin^2 x), so p = 1/2. Matching the right-hand piece q(1-sin x)/(pi-2x)^2 -> q/8 = 1/2 gives q = 4. Hence (p,q) = (1/2, 4).

Question 7

At x = 0, which of the following functions is differentiable?

Accepted Answer

sin(|x|) - |x|. cos|x| is smooth but |x| is not, so options 0 and 1 fail. For sin|x|+|x| the one-sided derivatives at 0 are +2 and -2 (fails). For sin|x|-|x| they are 1-1=0 and -1+1=0, equal, so it is differentiable at x=0.

Question 8

Define
f(x)=((e^x-1)²)/(sin(x/a) log(1+x/4)) for x≠ 0, and set f(0)=12. If f is continuous at x=0, what is the value of a?

Accepted Answer

3. As x->0, (e^x-1)^2 ~ x^2 and sin(x/a)log(1+x/4) ~ (x/a)(x/4) = x^2/(4a), so the limit is 4a. Continuity needs 4a = 12, hence a = 3.

Question 9

Let

f(x)=[(sin[x])/(([x]+1)²),, x>0,; [6pt]
(cos ((π)/(2)[x]))/([x]),, x<0,]

where [x] denotes the greatest integer not exceeding x. For f to be continuous at x=0, the value of k must be

Accepted Answer

0. For x->0+, [x]=0 so sin([x])/([x]+1)^2 = sin0/1 = 0. For x->0-, [x]=-1 so cos((pi/2)[x])/[x] = cos(-pi/2)/(-1) = 0/(-1) = 0. Both one-sided limits equal 0, so f(0) = k = 0 for continuity.

Question 10

Select the statement(s) that are true:
(a) If the derivative f'(a) exists and is finite at a point, then f must be continuous at x = a.
(b) The function f(x) = 3tan(5x) - 7 is differentiable at every point where it is defined.

Accepted Answer

Both (a) and (b). (a) If f'(a) exists and is finite, differentiability implies continuity, so f is continuous at a. (b) f(x)=3tan(5x)-7 is built from tan, which is differentiable everywhere it is defined; so it is differentiable on its whole domain. Both true -> 'Both (a) and (b)'.

Question 11

Consider the following statements:

Statement-1: Let g(x) be differentiable with g(1) ≠ 0 and g(-1) ≠ 0. If Rolle's theorem cannot be applied to f(x) = (x² - 1)/g(x) on the interval [-1,1], then g(x) must have at least one zero in (-1,1).

Statement-2: Whenever f(a) = f(b), Rolle's theorem

Accepted Answer

Statement-1 is true, Statement-2 is false. Since f(-1)=f(1)=0 already, Rolle's on f=(x^2-1)/g(x) fails only if f is not continuous/differentiable, i.e. g has a zero in (-1,1); so Statement-1 is true. Statement-2 is false because f(a)=f(b) alone is not sufficient (continuity and differentiability are also required). Hence Statement-1 true, Statement-2 false.

Question 12

Consider the following statements:

Statement I: The function f(x)=|x| sin x is differentiable at x=0.
Statement II: Even if f(x) is not differentiable at x=a and g(x) is differentiable at x=a, the product f(x)g(x) may still be differentiable at x=a.

Which of the following is correct?

Accepted Answer

Statement I is true, Statement II is true, and Statement II correctly explains Statement I. f'(0)=lim h->0 (|h|sin h)/h = lim (sin h)(|h|/h)*... evaluating both sides gives 0, so f is differentiable at 0: Statement I is TRUE. Statement II is a true general fact (e.g. this very function), and it correctly explains why I holds. So both statements are true and II explains I.

Question 13

For the function f(x)=3 cos⁴ x + 10 cos³ x + 6 cos² x - 3, defined on 0≤x≤π, which statement is true?

Accepted Answer

It increases on (π/2,2π/3).. f'(x) = -6 sinx cosx (2cosx+1)(cosx+2). On (0,pi) sinx>0 and cosx+2>0, so the sign is set by -cosx(2cosx+1). On (pi/2,2pi/3): cosx<0 but 2cosx+1>0, giving f'>0, so f increases there. Hence the true statement is 'It increases on (pi/2,2pi/3)'.

Question 14

Consider the following statements:

Statement I: If f(0)=0 and f'(x)=ln l(x+√(1+x²) r), then f(x) remains positive for every real x.

Statement II: The function f(x) increases for x>0 and decreases for x<0.

Accepted Answer

Statement I is true, Statement II is true, and Statement II correctly explains Statement I. f'(x)=sinh^-1(x), which is positive for x>0 and negative for x<0, so f increases for x>0 and decreases for x<0 (Statement II true). Thus f has its minimum at x=0 where f(0)=0, so f(x)>=0 for all x (Statement I true). Statement II (the monotonic behaviour) is exactly why f never goes negative, so II correctly explains I.

Question 15

For the equation sin x + xcos x = 0, in which interval is there at least one solution?

Accepted Answer

(0, π). In the interval (0, ext{π}), the function ext{sin} x is positive and ext{x cos} x is negative, indicating that the sum can equal zero at least once due to the Intermediate Value Theorem, confirming the existence of a solution.

Question 16

Let f be defined on the interval [-5, 5] by
f(x) = x when x is rational,
and f(x) = -x when x is irrational.
Which of the following is true about the continuity of f?

Accepted Answer

f is discontinuous at every point except x = 0. The function f is defined differently for rational and irrational numbers, leading to discontinuity at all points except x = 0, where both definitions converge to the same value. At x = 0, both f(0) = 0 and the limit as x approaches 0 from both rational and irrational sides is also 0, making it continuous only at that point.

Question 17

Suppose f(x) is differentiable at x = 1 and lim(h→0) [1/h] f(1 + h) = 5, then f'(1) equals

Accepted Answer

5. For lim_{h->0} f(1+h)/h to be finite, f(1)=0. Then f(1+h)/h = [f(1+h)-f(1)]/h -> f'(1). Hence f'(1)=5.

Question 18

Let f be differentiable for all x. If f(1) = -2 and f'(x) ≥ 2 for x ∈ [1, 6], then

Accepted Answer

f(6) ≥ 8. Since the derivative f'(x) is always greater than or equal to 2 on the interval [1, 6], this indicates that the function f is increasing at a rate of at least 2 units for every unit increase in x. Starting from f(1) = -2, after 5 units of increase (from x=1 to x=6), the minimum value of f(6) can be calculated as f(1) + 5 * 2 = -2 + 10 = 8, thus f(6) must be at least 8.

Question 19

If f is a real valued differentiable function satisfying |f(x) - f(y)| ≤ (x - y)², x, y ∈ R and f(0) = 0, then f(1) equals

Accepted Answer

0. Dividing by |x-y| gives |f(x)-f(y)|/|x-y| <= |x-y| -> 0, so f'(x)=0 for all x and f is constant. With f(0)=0, f(1)=0.

Question 20

Let f(x) = {(x - 1) sin(1/(x - 1)), x ≠ 1; 0, x = 1}. Then which one of the following is true?

Accepted Answer

f is differentiable at x = 0 but not at x = 1. The function is continuous at x = 1, but its derivative does not exist there due to the oscillatory behavior of the sine term as x approaches 1. However, at x = 0, the function is smooth and differentiable.

Question 21

Let y be an implicit function of x defined by x² - 2x² cot y - 1 = 0. Then y'(1) equals

Accepted Answer

-1. At x=1 the relation x^2-2x^2 cot y-1=0 gives -2 cot y=0, so cot y=0 and y=pi/2. Differentiating: 2x-4x cot y+2x^2 csc^2(y) y'=0. At x=1, cot y=0, csc^2 y=1: 2+2y'=0, so y'(1)=-1.

Question 22

The values of p and q for which the function f(x) = { [sin(p + 1)x + sin x]/x, x < 0; q, x = 0; [√(x + x²) - √x]/x^(3/2), x > 0 } is continuous for all x in R, are

Accepted Answer

p = -3/2, q = 1/2. As x->0-, [sin(p+1)x+sin x]/x -> (p+1)+1 = p+2. As x->0+, [sqrt(x+x^2)-sqrt(x)]/x^(3/2) = (sqrt(1+x)-1)/x -> 1/2, so q=1/2. Continuity needs p+2=1/2, giving p=-3/2, q=1/2.

Question 23

For the function f(x)=|x-2|+|x-5|, where x∈ R, consider the following statements:
Statement-1: f'(4)=0
Statement-2: f is continuous on [2,5], differentiable on (2,5), and f(2)=f(5).

Accepted Answer

Both Statement-1 and Statement-2 are true, but Statement-2 does not correctly explain Statement-1.. Statement-1 is true because the function has a minimum at x=4, where the derivative is zero. Statement-2 is also true as the function is continuous and differentiable in the specified intervals, but it does not provide a reason for the derivative being zero at that specific point.

Question 24

Let f and g be differentiable on [0,1] such that f(0)=2, g(0)=0, g(1)=2, and f(1)=6. Then there exists some c in (0,1) for which

Accepted Answer

f'(c)=2g'(c). The correct option is based on the application of the Mean Value Theorem, which states that if two functions are continuous and differentiable on a closed interval, then there exists at least one point in the interval where the derivatives are proportional to the changes in the function values. Here, the change in f from 2 to 6 and the change in g from 0 to 2 suggests that the derivative of f at some point is twice that of g, leading to the conclusion that f'(c) = 2g'(c).

Question 25

For the piecewise function g(x) = { k√(x+1), 0 ≤ x ≤ 3; mx + 2, 3 < x ≤ 5 }, if g is differentiable, what is the value of k + m?

Accepted Answer

2. For the function g(x) to be differentiable at x = 3, both the function values and their derivatives must match at that point. Solving the equations derived from the continuity and differentiability conditions leads to the conclusion that k + m equals 2.

Question 26

Let x be a real number, and define f(x)=|log 2-sin x| and g(x)=f(f(x)). Then:

Accepted Answer

g'(0)=cos(log 2). Near x=0, log2-sin x>0, so f(x)=log2-sin x and f'(x)=-cos x. Also log2-sin(f(x))>0 near 0, so g(x)=log2-sin(f(x)), giving g'(x)=-cos(f(x)) f'(x)=cos(f(x)) cos x. At x=0, f(0)=log2, so g'(0)=cos(log2).

Question 27

Consider the function f: R → R given by
f(x) = 5, for x ≤ 1;
a + bx, for 1 < x < 3;
b + 5x, for 3 ≤ x < 5;
30, for x ≥ 5.
Which of the following is true about f?

Accepted Answer

f is never continuous, whatever the values of a and b. Continuity needs a+b=5 (at x=1), a+2b=15 (at x=3) and b=5 (at x=5). The last two give a=5, but then a+b=10 != 5, a contradiction. No a,b make f continuous, so f is never continuous.

Question 28

Let f(x) = 15 - |x - 10| for x ∈ R. The function g(x) = f(f(x)) fails to be differentiable at which of the following sets of x-values?

Accepted Answer

{5, 10, 15}. g(x)=f(f(x)) with f(x)=15-|x-10|. Non-differentiability occurs at the inner kink x=10 and where f(x)=10, i.e. |x-10|=5 -> x=5,15. So the set is {5,10,15}.

Question 29

Consider a function f: [−7, 0] → R that is continuous on [−7, 0] and differentiable on (−7, 0). Given that f(−7) = −3 and f'(x) ≤ 2 for every x in (−7, 0), the quantity f'(−1) + f(0) must belong to which interval for all such functions?

Accepted Answer

(−∞, 20]. By MVT, f(0) <= f(-7)+2*7 = 11, and f'(-1) <= 2, so f'(-1)+f(0) can reach 13 (take f'=2 everywhere, f(x)=2x+11). Since f' has no lower bound, the sum is unbounded below, so it must lie in (-inf, 20].

Question 30

Value of c for which conclusion of Mean Value Theorem holds for the function f(x) = logₑ x on the interval [1, 3] is

Accepted Answer

2 log₃ e. For f(x)=ln x on [1,3], f'(c) = (f(3)-f(1))/(3-1) = (ln3)/2. Since f'(c)=1/c, c = 2/ln3 = 2 log_3 e.

Question 31

Let f: R → R be given by
f(x) = { k - 2x, for x ≤ -1
 { 2x + 3, for x > -1
If x = -1 is a point where f attains a local minimum, then one possible value of k is

Accepted Answer

-1. For f to have a local minimum at x = -1, the left-hand limit (from the piece k - 2x) must equal the right-hand limit (from the piece 2x + 3) at that point. Setting these equal gives k - 2(-1) = 2(-1) + 3, which simplifies to k + 2 = 1, leading to k = -1.

Question 32

Consider the continuous function f: R → R given by
f(x) = 1 / (e^x + 2e^(-x)).

Statement-1: There exists some c ∈ R such that f(c) = 1/3.
Statement-2: For every x ∈ R, 0 < f(x) ≤ 1/(2√2).

Choose the correct option.

Accepted Answer

Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.. Statement-1 is true because the function f(x) is continuous and takes values between 0 and 1/(2√2), which includes 1/3. Statement-2 is true as it correctly describes the range of f(x), thus supporting the existence of some c such that f(c) = 1/3.

Question 33

Consider the function
f(x) = { tan x / x, for x ≠ 0
 { 1, for x = 0
Statement 1: x = 0 is a point of minimum of f.
Statement 2: f'(0) = 0.
Which of the following is correct?

Accepted Answer

Statement 1 is true, Statement 2 is true, but Statement 2 does not correctly explain Statement 1.. At x = 0, the function f(x) achieves a minimum value of 1, making Statement 1 true. Statement 2 is also true because the derivative at that point is zero, indicating a critical point, but this does not directly confirm it as a minimum since further analysis is needed.

Question 34

Let p(x) be a function on the real line satisfying p'(x)=p'(1-x) for every x in the interval [0,1], with p(0)=1 and p(1)=41. Then the value of ∫₀¹ p(x) dx is

Accepted Answer

21. The condition p'(x) = p'(1-x) implies that the derivative of p is symmetric about x = 0.5, which suggests that p(x) can be expressed in a form that reflects this symmetry. Given the boundary conditions p(0) = 1 and p(1) = 41, we can deduce that the average value of p over the interval [0, 1] is 21, leading to the conclusion that the integral of p from 0 to 1 equals 21.

Question 35

If f(x) is a non-zero polynomial of degree four, having local extreme points at x = -1, 0, 1; then the set S = {x ∈ R: f(x) = f(0)} contains exactly:

Accepted Answer

two irrational and one rational number.. Since f(x) is a degree four polynomial, it can have at most four real roots. Given that it has local extrema at x = -1, 0, and 1, the function must cross the horizontal line y = f(0) at least three times, leading to two additional roots. This results in a total of three distinct points where f(x) equals f(0), which can include two irrational and one rational number.

Question 36

A function y=f(x) has a second order derivative f''(x)=6(x-1). If its graph passes through the point (2,1) and at that point the tangent to the graph is y=3x-5, then the function is

Accepted Answer

(x-1)³. The second derivative, when integrated, gives the first derivative, which must match the slope of the tangent line at the given point. Integrating twice and applying the conditions of the point and tangent line leads to the function being (x) = (x-1)³, which satisfies both the second derivative and the initial conditions.

Question 37

If f and g are differentiable functions in [0, 1] satisfying f(0) = 2 = g(1), g(0) = 0 and f(1) = 6, then for some c ∈ ]0, 1[:

Accepted Answer

f'(c) = 2g'(c). The correct option is based on the application of the Mean Value Theorem, which implies that there exists a point c in the interval (0, 1) where the rates of change of the functions f and g are related by their respective changes over the interval. Given the values of f and g at the endpoints, the relationship f'(c) = 2g'(c) reflects the proportional change between the two functions.

Question 38

For x ∈ R, f(x) = |log 2 − sinx| and g(x) = f(f(x)), then:

Accepted Answer

g'(0) = cos(log 2). The correct option states that g'(0) = cos(log 2) because the composition of functions preserves differentiability, and at x = 0, the derivative of f(x) evaluated at that point leads to this result based on the chain rule.

Question 39

If the function f defined as f(x) = 1/x − (k − 1)/(e^(2x) − 1), x ≠ 0, is continuous at x = 0, then the ordered pair (k, f(0)) is equal to ?

Accepted Answer

(3, 1). The function is continuous at x = 0 if the limit as x approaches 0 equals the function value at that point. By evaluating the limit of f(x) as x approaches 0, we find that it converges to 1 when k is set to 3, thus making the ordered pair (3, 1) the correct choice.

Question 40

Let f(x) be a polynomial of degree 4 having extreme values at x = 1 and x = 2. If lim x→0 ((f(x)/x²) + 1) = 3 then f(-1) is equal to -

Accepted Answer

9/2. From lim(f/x^2+1)=3 we get f(0)=0, f'(0)=0 and the x^2 coefficient = 2. Extrema at x=1,2 fix the quartic as f(x)=(1/2)x^4-2x^3+2x^2. Then f(-1)=1/2+2+2=9/2.

Question 41

Let S = {(λ, μ) ∈ R × R: f(t) = (|λ| e^(|t|) − μ) · sin(2|t|), t ∈ R, is a differentiable function}. Then S is a subset of

Accepted Answer

R × [0, ∞). The correct option is R × [0, ∞) because for the function f(t) to be differentiable, the term |λ| e^(|t|) must be non-negative, which requires λ to be any real number and μ to be non-negative, thus forming the set of all pairs (λ, μ) where μ is in [0, ∞).

Question 42

Let f(x) = x/√(a² + x²) - (d - x)/√(b² + (d - x)²), x ∈ R, where a, b and d are non-zero real constants. Then:
(1) f is an increasing function of x
(2) f is neither increasing nor decreasing function of x
(3) f' is not a continuous function of x
(4) f is a decreasing function of x

Accepted Answer

f is an increasing function of x. The function f(x) is increasing because its derivative f'(x) is positive for all x in the real numbers, indicating that as x increases, f(x) also increases.

Question 43

Let f: [-1, 3] → R be defined as
f(x) = { |x| + [x], -1 ≤ x < 1
 x + |x|, 1 ≤ x < 2
 x + [x], 2 ≤ x ≤ 3
Where [t] denotes the greatest integer less than or equal to t. Then, f is discontinuous at

Accepted Answer

only two points. On [-1,1): at x=0 the left limit is -1 but f(0)=0 (jump). At x=1 the left limit is 1 while the next piece gives 2 (jump). At x=2 both sides give 4 (continuous). So f is discontinuous at exactly two points, x=0 and x=1.

Question 44

If f(x) = { (sin(p+1)x + sin x)/x, x < 0; x/q, x = 0; (√(x + x²) − √x)/x^(3/2), x > 0 } is continuous at x = 0, then the ordered pair (p, q) is equal to:

Accepted Answer

(-3/2, 1/2). To ensure continuity at x = 0, the left-hand limit as x approaches 0 must equal the right-hand limit and the function value at x = 0. By calculating these limits and setting them equal, we find that p must be -3/2 and q must be 1/2 to satisfy the continuity condition.

Question 45

Let f: R → R be a continuously differentiable function such that f(2) = 6 and f'(2) = 1/48. If ∫₆^(f(x)) 4t³ dt = (x−2)g(x), then Lim x→2 g(x) is equal to:

Accepted Answer

18. To find the limit as x approaches 2 of g(x), we can differentiate both sides of the equation using the Fundamental Theorem of Calculus and the Chain Rule. Evaluating at x = 2 gives us g(2) = 4(6³) / f'(2), which simplifies to 18 when substituting the known values of f(2) and f'(2).

Question 46

Let f:[0,2]→ R be a twice differentiable function such that f''(x)>0, for all x∈(0,2). If φ(x)=f(x)+f(2-x), then φ is -

Accepted Answer

decreasing on (0,1) and increasing on (1,2). phi(x)=f(x)+f(2-x) so phi'(x)=f'(x)-f'(2-x). Since f''>0, f' is increasing: for x in (0,1), x<2-x gives phi'<0 (decreasing); for x in (1,2), phi'>0 (increasing). So phi decreases on (0,1) and increases on (1,2).

Question 47

Let f(x) = { -1, -2 ≤ x < 0; x² - 1, 0 ≤ x ≤ 2 } and g(x) = |f(x)| + f(|x|). Then, in the interval (-2, 2), g is: (1) non continuous (2) differentiable at all points (3) not differentiable at two points (4) not differentiable at one point

Accepted Answer

not differentiable at one point. The function g(x) is composed of f(x) and its absolute value, which introduces a potential point of non-differentiability. Specifically, g(x) is not differentiable at x = 0, where f(x) transitions from a constant to a quadratic function, creating a cusp.

Question 48

Let f be a differentiable function such that f(1) = 2 and f'(x) = f(x) for all x ∈ R. If h(x) = f(f(x)), then h'(1) is equal to:

Accepted Answer

4e. The derivative of h(x) can be found using the chain rule, where h'(x) = f'(f(x)) * f'(x). Given that f'(x) = f(x), we can substitute to find h'(1) = f'(f(1)) * f'(1) = f(2) * f(1). Since f(1) = 2 and f(2) = e² (from the differential equation), we get h'(1) = e² * 2 = 4e.

Question 49

The equation of the normal to the curve y = (1 + x)^(2y) + cos²(sin^(-1) x) at x = 0 is:

Accepted Answer

x + 4y = 8. The correct option represents the equation of the normal line at the given point on the curve, which is derived from the slope of the tangent line at that point. By calculating the derivative and using the negative reciprocal for the normal's slope, we find that the equation simplifies to x + 4y = 8.

Question 50

Let f be a twice differentiable function on (1, 6). If f(2) = 8, f'(2) = 5, f'(x) ≥ 1 and f''(x) ≥ 4, for all x ∈ (1, 6), then:

Accepted Answer

f(5) + f'(5) ≥ 28. The correct option is right because, given that f'(x) is always at least 1 and f''(x) is at least 4, f' is increasing. Starting from f'(2) = 5, we can deduce that f'(5) must be at least 5 + 4*3 = 17. Additionally, using the minimum growth rate of f, we find that f(5) must be at least 8 + 1*3 = 11. Therefore, f(5) + f'(5) is at least 11 + 17 = 28.

JEE Main Maths: Continuity and Differentiability questions with solutions

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