CBSE Class 11 Mathematics Limits And Derivatives Worksheet Set 12

INTEGER QUESTIONS

Question. Equation of the normal to the curve \( y = (1 + x)^y + \sin^{-1}(\sin^2 x) \) at \( x = 0 \) is \( x + y = k \), then \( k \) is
Answer: At \( x = 0 \), \( y = (1+0)^y + 0 \implies y = 1 \).
Differentiating: \( y' = y(1+x)^{y-1} \cdot 1 + (1+x)^y \ln(1+x) y' + \dots \) At \( (0, 1) \), \( y' = 1 \cdot 1 + 0 + 0 = 1 \). Slope of normal \( = -1 \).
Normal equation: \( y - 1 = -1(x - 0) \implies x + y = 1 \).
\( \implies \) \( k = 1 \).

Question. If the acute angles between the curves \( y = |x^2 - 1| \) and \( y = |x^2 - 3| \) at their points of intersection be \( \theta \) such that \( \tan \theta = \frac{4\sqrt{2}}{k} \), then \( k \) is equal to
Answer: Angle of intersection for \( x > 1 \) and \( x < -1 \) are
similar because of similarity about y-axis. for \( x > 1 \),
point of intersection is \( x = \sqrt{2} \).
 

Question. If the function \( f(x) = x^3 + e^{x/2} \) and \( g(x) = f^{-1}(x) \), then the value of \( g'(1) \) is 
Answer: \( f(0) = 0 + e^0 = 1 \implies g(1) = 0 \).
\( g'(1) = \frac{1}{f'(g(1))} = \frac{1}{f'(0)} \).
\( f'(x) = 3x^2 + \frac{1}{2} e^{x/2} \implies f'(0) = 0 + \frac{1}{2} = \frac{1}{2} \).
\( g'(1) = \frac{1}{1/2} = 2 \).
\( \implies \) Answer is 2.

Question. \( f(x) \) is cubic polynomial with \( f(2) = 18 \) and \( f(1) = -1 \). Also \( f(x) \) has local maxima at \( x = -1 \) and \( f'(x) \) has local minima at \( x = 0 \), then
(a) the distance between \( (-1, 2) \), and \( (a, f(a)) \), where \( x = a \) is the point of local minima is \( 2\sqrt{5} \)
(b) \( f(x) \) is increasing for \( x \in [1, 2\sqrt{5}] \)
(c) \( f(x) \) has local minima at \( x = 1 \)
(d) the value of \( f(0) = 15 \)
Answer: (a) the distance between \( (-1, 2) \), and \( (a, f(a)) \), where \( x = a \) is the point of local minima is \( 2\sqrt{5} \)

Question. Let \[ f(x) = \begin{cases} e^x, & 0 \le x \le 1 \\ 2 - e^{x-1}, & 1 < x \le 2 \\ x - e, & 2 < x \le 3 \end{cases} \] , \( g(x) = \int_0^x f(t) dt, x \in [1, 3] \) then \( g(x) \) has
(a) local maxima at \( x = 1 + \ln 2 \) and local minima at \( x = e \)
(b) local maxima at \( x = 1 \) and local minima at \( x = 2 \)
(c) no local maxima
(d) no local minima
Answer: (a) local maxima at \( x = 1 + \ln 2 \) and local minima at \( x = e \)

Question. For function \( f(x) = x \cos \frac{1}{x}, x \ge 1 \),
(a) for atleast one \( x \) in interval \( [1, \infty), f(x + 2) - f(x) < 2 \)
(b) \( \lim_{x \to 0} f'(x) = 1 \)
(c) \( [1, \infty), f(x + 2) - f(x) > 2 \)
(d) \( f'(x) \) is strictly decreasing in the interval \( [1, \infty) \)
Answer: (c) \( [1, \infty), f(x + 2) - f(x) > 2 \)

Question. If \( f(x) = \int_0^x e^{t^2} (t-2)(t-3) dt \) for all \( x \in (0, \infty) \), then
(a) \( f \) has a local maximum at \( x = 2 \)
(b) \( f \) is decreasing on \( (2, 3) \)
(c) there exists some \( c \in (0, \infty) \) such that \( f''(c) = 0 \)
(d) \( f \) has a local minimum at \( x = 3 \)
Answer: (a) \( f \) has a local maximum at \( x = 2 \)

Question. Statement-1: Both \( \sin x \) and \( \cos x \) are decreasing functions in the interval \( (\frac{\pi}{2}, \pi) \)
Statement-2: If a differentiable function decreases in an interval \( (a, b) \), then its derivative also decreasing in \( (a, b) \).
which of the following is true?
(a) both st1 and st2 are wrong
(b) both st1 and st2 are correct but st2 is not correct explanation for st1
(c) both st1 and st2 are correct and st2 is correct explanation for st1
(d) st1 is correct and st2 is wrong
Answer: (d) st1 is correct and st2 is wrong

COMPREHENSION QUESTIONS

PASSAGE - 1
Consider the curve \( x = 1 - 3t^2 \), \( y = t - 3t^3 \). If a tangent at point \( (1-3t^2, t-3t^3) \) inclined at an angel \( \theta \) to the positive \( x \)-axis and another tangent at point \( P(-2, 2) \) cuts the curve again at \( Q \).

Question. The value of \( \tan \theta + \sec \theta \) is equal to
(a) 3t
(b) t
(c) \( t - t^2 \)
(d) \( t^2 - 2t \)
Answer: (a) 3t

Question. The point \( Q \) will be
(a) \( (1, -2) \)
(b) \( \left( -\frac{1}{3}, -\frac{2}{9} \right) \)
(c) \( (-2, 1) \)
(d) \( (1, 0) \)
Answer: (b) \( \left( -\frac{1}{3}, -\frac{2}{9} \right) \)

Question. The angle between the tangents at \( P \) and \( Q \) will be
(a) \( \frac{\pi}{4} \)
(b) \( \frac{\pi}{6} \)
(c) \( \frac{\pi}{2} \)
(d) \( \frac{\pi}{3} \)
Answer: (c) \( \frac{\pi}{2} \)

PASSAGE - 2
Let \( f'(\sin x) < 0 \) and \( f''(\sin x) > 0 \) \( \forall x \in (0, \frac{\pi}{2}) \) and \( g(x) = f(\sin x) + f(\cos x) \).

Question. Which of the following is true in \( (0, \frac{\pi}{2}) \)?
(a) \( g' \) is increasing
(b) \( g' \) is decreasing
(c) \( g' \) has a point of minima
(d) \( g' \) has a point of maxima
Answer: (a) \( g' \) is increasing

Question. Which of the following is true?
(a) \( g(x) \) is decreasing in \( (\frac{\pi}{4}, \frac{\pi}{2}) \)
(b) \( g(x) \) increasing in \( (0, \frac{\pi}{4}) \)
(c) \( g(x) \) is monotonically increasing
(d) None of the options
Answer: (d) None of the options

PASSAGE - 3
Let \( f(x) = (1-x)^2 \sin^2 x + x^2 \) for all \( x \in R \), and let \( g(x) = \int_1^x \left( \frac{2(t-1)}{t+1} - \ln t \right) f(t) dt \) for all \( x \in (1, \infty) \).

Question. Which of the following is true?
(a) \( g \) is increasing on \( (1, \infty) \)
(b) \( g \) is decreasing on \( (1, \infty) \)
(c) \( g \) is increasing on \( (1, 2) \) and decreasing on \( (2, \infty) \)
(d) \( g \) is decreasing on \( (1, 2) \) and increasing on \( (2, \infty) \)
Answer: (b) \( g \) is decreasing on \( (1, \infty) \)

Question. Consider the statements:
P: There exists some \( x \in R \) such that \( f(x) + 2x = 2(1+x^2) \)
Q: There exists some \( x \in R \) such that \( 2f(x) + 1 = 2x(1+x) \) then
(a) both P and Q are true
(b) P is true and Q is false
(c) P is false and Q is true
(d) both P and Q are false
Answer: (c) P is false and Q is true

PASSAGE - 4
The relation between the distance of any point on the curve from the origin & length of the perpendicular from the origin to the tangent at the point is called pedal equation of the curve equation of tangent at a point \( (h, k) \), \( y - k = f'(h)(x - h) \).
\( p = \text{length of perpendicular from } (0, 0) \text{ to the tangent} = \frac{|k - hf'(h)|}{\sqrt{1+(f')^2}} \)
\( k = f(h), r^2 = h^2 + k^2 \)
\( p \)-perpendicular distance \( (0, 0) \) to the tangent at \( (h, k) \), \( r \) - distance from origin.

Question. The pedal equation of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
(a) \( \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} - \frac{r^2}{a^2 b^2} \)
(b) \( \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{r^2}{a^2 b^2} \)
(c) \( \frac{2}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} - \frac{r^2}{a^2 b^2} \)
(d) \( \frac{2}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{r^2}{a^2 b^2} \)
Answer: (a) \( \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} - \frac{r^2}{a^2 b^2} \)

Question. The equation of a curve is given in parametric form \( x = a \cos^3 \theta, y = a \sin^3 \theta \), the pedal equation of this curve is
(a) \( r^2 = a^2 + 3p^2 \)
(b) \( r^2 = a^2 - 3p^2 \)
(c) \( r^2 = 3a^2 - p^2 \)
(d) \( r^2 = 3a^2 + p^2 \)
Answer: (b) \( r^2 = a^2 - 3p^2 \)

Question. The cartesian equation of curve is given by \( c^2(x^2 + y^2) = x^2 y^2 \). the pedal equation of this curve is
(a) \( \frac{1}{p^2} + \frac{3}{r^2} = \frac{1}{c^2} \)
(b) \( \frac{3}{p^2} + \frac{1}{r^2} = \frac{1}{c^2} \)
(c) \( \frac{1}{p^2} - \frac{3}{r^2} = \frac{1}{c^2} \)
(d) \( \frac{3}{p^2} - \frac{1}{r^2} = \frac{1}{c^2} \)
Answer: (a) \( \frac{1}{p^2} + \frac{3}{r^2} = \frac{1}{c^2} \)

PASSAGE - 5
Tangent at any point \( P_1 \) (other than \( (0, 0) \)) on the curve \( y = x^3 \) meets the curve agent at \( P_2 \). Tangent at \( P_2 \) meets curve agent at \( P_3 \) and so on.

Question. Abcissae of \( P_1, P_2, P_3, ..., P_n \) are in
(a) A.P
(b) G.P
(c) H.P
(d) None of the options
Answer: (b) G.P

Question. ordinates of \( P_1, P_2, P_3, ..., P_n \) are in
(a) A.P
(b) G.P
(c) H.P
(d) None of the options
Answer: (d) None of the options

Question. Ratio of areas of triangles \( P_2 P_3 P_4 \) and \( P_1 P_2 P_3 \) is
(a) 4
(b) 8
(c) 16
(d) 32
Answer: (c) 16

Question. Let \( f(x) \) be a real valued function defined by \( f(x) = x^2 - 2|x| \) and \[ g(x) = \begin{cases} \min \{f(t) : -2 \le t \le x\}, & x \in [-2, 0) \\ \max \{f(t) : 0 \le t \le x\}, & x \in [0, 3] \end{cases} \]
Column-I
(A) \( f(x) \) is not continuous at \( x \) equal to
(B) \( g(x) \) is not derivable at \( x \) equal to
(C) The points of local extremum of \( g(x) \) is/are
(D) Absolute maximum value of \( g(x) \) is equal to
Column-II
(P) -2
(Q) 0
(R) 1
(S) 2
(T) 3
Answer: (A) \( \to \) (Q), (B) \( \to \) (Q), (S), (C) \( \to \) (Q), (R), (S), (D) \( \to \) (T)

SUBJECTIVE QUESTIONS

Question. Show that \( 2\sin x + 2\tan x \ge 3x \) where \( 0 \le x < \frac{\pi}{2} \).
Answer: Let \( f(x) = 2 \sin x + 2 \tan x - 3x \). Determine if \( f(x) \) is increasing.
\( \implies f'(x) = 2 \cos x + 2 \sec^2 x - 3 \)
since, \( 0 \leq x < \pi/2, f'(x) > 0 \)
Thus, \( f(x) \) is increasing, when \( x \geq 0 \),
\( f(x) \geq f(0) \).

Question. Let \( f(x) = \begin{cases} -x^3 + \frac{b^3 - b^2 + b - 1}{b^2 + 3b + 2}, & 0 \le x < 1 \\ 2x - 3, & 1 \le x \le 3 \end{cases} \) Find all possible real values of \( b \) such that \( f(x) \) has the smallest value at \( x = 1 \).
Answer: Given \( f(x) = \begin{cases} -x^3 + \frac{b^3 - b^2 + b - 1}{b^2 + 3b + 2} & 0 \leq x \leq 1 \\ 2x - 3 & 1 \leq x \leq 3 \end{cases} \). Find \( b \) such that the smallest value of \( f(x) \) occurs at \( x = 1 \).
\( f'(x) = \begin{cases} -3x^2 & 0 < x < 1 \\ 2 & 1 < x < 3 \end{cases} \)
So, \( f(x) \) is decreasing on \( [0, 1) \) and increasing on \( (1, 3] \).
Here, \( f(1) = -1 \) is the smallest value at \( x = 1 \).
\( \therefore \) its smallest value occurs as:
\( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (-x^3) + \frac{b^3 - b^2 + b - 1}{b^2 + 3b + 2} \geq -1 \)
\( \frac{b^3 - b^2 + b - 1}{b^2 + 3b + 2} \geq 0 \implies \frac{(b^2 + 1)(b - 1)}{(b + 1)(b + 2)} \geq 0 \)
\( \therefore b \in (-2, -1) \cup [1, \infty) \).
\( b \in (-2, -1) \cup [1, \infty) \)

Question. It is given that \( y = ax^3 + bx^2 + cx + 5 \) touches x-axis at \( P(-2, 0) \) which implies that x-axis is tangent at \( (-2, 0) \) and the curve also passes through \( (-2, 0) \). The curve cuts y-axis at \( (0, 5) \) and gradient at this point is 3. Find \( a, b, c \).
Answer: \( \frac{dy}{dx} = 3ax^2 + 2bx + c \)
Since x-axis is tangent at \( (-2, 0) \):
\( \frac{dy}{dx} \big|_{x=-2} = 0 \implies 0 = 3a(-2)^2 + 2b(-2) + c \implies 0 = 12a - 4b + c \) ... (i)
Slope of tangent at \( (0, 5) \) is 3:
\( \frac{dy}{dx} \big|_{(0,5)} = 3 \implies 3 = 3a(0)^2 + 2b(0) + c \implies 3 = c \) ... (ii)
Curve passes through \( (-2, 0) \):
\( 0 = a(-2)^3 + b(-2)^2 + c(-2) + 5 \implies 0 = -8a + 4b - 2c + 5 \) ... (iii)
Hence, \( a = -1/2, b = -3/4, c = 3 \).

Question. Prove that \( \sin(\tan x) \geq x \).
Answer: Let \( f(x) = \sin(\tan x) - x \)
\( f'(x) = \cos(\tan x) \cdot \sec^2 x - 1 = \cos(\tan x) \{1 + \tan^2 x\} - 1 \)
\( = \tan^2 x (\cos(\tan x)) + \cos(\tan x) - 1 > \tan^2 x \cos(\tan x) - \frac{\tan^2 x}{2} \)
\( f'(x) > \tan^2 x \left[ \cos(\tan x) - \frac{1}{2} \right] \)
\( > \tan^2 x \{ \cos(\tan x) - \cos(\pi/3) \} > 0 \)
\( \implies f(x) \) is increasing function \( \forall x \in [0, \pi/4] \)
as \( f(0) = 0 \implies f(x) \geq 0 \forall x \in [0, \pi/4] \implies \sin(\tan x) \geq x \).
\( \sin(\tan x) \geq x \) for \( x \in [0, \pi/4] \).

Question. Consider the function \( f(x) = \tan x \), defined on \( [ a , b ] \) such that \( a , b \in \left( 0, \frac{\pi}{2} \right) \).
Answer: Applying Lagrange’s mean value theorem, we have
\( f'(c) = \frac{f(b) - f(a)}{b - a} \) for some \( c \in ( a , b ) \)
\( \implies \sec^2 C = \frac{\tan b - \tan a}{b - a} \)
\( \implies f(a,b) = \sec^2 C \)
\( \implies f(a,b) > 1 \) [\( \because \sec^2 C > 1 \) as \( C \in ( 0, \pi/2 ) \)]

Question. Given \( f'(x) = 12x^2 - 2x - 2 = 2[6x^2 - x - 1] = 2(3x + 1)(2x - 1) \). Define \( g(x) = \begin{cases} f(x) & \text{if } 0 \le x < \frac{1}{2} \\ f\left(\frac{1}{2}\right) & \text{if } \frac{1}{2} \le x \le 1 \\ 3 - x & \text{if } 1 < x \le 2 \end{cases} \). Find \( g\left(\frac{1}{4}\right) + g\left(\frac{3}{4}\right) + g\left(\frac{5}{4}\right) \).
Answer: Hence \( g\left(\frac{1}{4}\right) + g\left(\frac{3}{4}\right) + g\left(\frac{5}{4}\right) = f\left(\frac{1}{4}\right) + f\left(\frac{1}{2}\right) + g\left(\frac{5}{4}\right) = \frac{5}{2} \)

Question. Consider the function \( f(x) = \frac{x^2}{(x^3 + 200)} \). Find the greatest term of the sequence.
Answer: \( f'(x) = x \frac{(400 - x^3)}{(x^3 + 200)^2} = 0 \) When \( x = (400)^{1/3} \) \( x = (400)^{1/3} - h \implies f'(x) > 0 \) \( x = (400)^{1/3} + h \implies f'(x) < 0 \therefore f(x) \) has maxima at \( x = (400)^{1/3} \) Since \( 7 < (400)^{1/3} < 8 \), either \( a_7 \) or \( a_8 \) is the greatest term of the sequence. \( \because a_7 = \frac{49}{543} \) and \( a_8 = \frac{8}{89} \) and \( \frac{49}{543} > \frac{8}{89} \)
\( \implies a_7 = \frac{49}{543} \) is the greatest term.

Question. Let there be a value of k for which \( x^3 - 3x + k = 0 \) has two distinct roots between 0 and 1.
Answer: Let \( a, b \) be two distinct roots of \( x^3 - 3x + k = 0 \) lying between 0 and 1 such that \( a < b \). Let \( f(x) = x^3 - 3x + k \). Then \( f(a) = f(b) = 0 \). Since between any two roots of a polynomial \( f(x) \), there exists at least one root of its derivative \( f'(x) \). Therefore, \( f'(x) = 3x^2 - 3 \) has at least one root between \( a \) and \( b \). But \( f'(x) = 0 \) has two roots equal to \( \pm 1 \) which do not lie between \( a, b \). Hence : \( f(x) = 0 \) has no real roots lying between 0 and 1 for any value of k.

Question. Let ABCD be a rectangular sheet whose corner C is folded over along EF so as to reach the edge AB at C’. Find the minimum length of fold \( x = EF \).
Answer: Let \( EF = x \). \( \therefore \angle FEC = \theta = \angle FEC' \) From \( \Delta BEC' \) we have \( BE = C'E \cos(\pi - 2\theta) \)
\( \implies BE = -x \cos \theta \cos 2\theta \) \( \therefore BC = BE + EC \) \( 1 = -x \cos \theta \cos 2\theta + x \cos \theta \)
\( \implies x = \frac{1}{\cos \theta (1 - \cos 2\theta)} \dots \dots \dots (1) \) For \( x \) to be minimum, \( Z = \frac{1}{x} = \cos \theta (1 - \cos 2\theta) \) has to be maximum \( \dots \dots \dots (2) \) Differentiating Eq. ( 2 ) w.r.t. \( \theta \), we get \( \frac{dZ}{d\theta} = \cos \theta (+2 \sin 2\theta) - \sin \theta (1 - \cos 2\theta) \) and \( \frac{d^2Z}{d\theta^2} = \cos \theta (4 \sin 4\theta) - 2 \sin 2\theta \sin \theta - \sin \theta (2 \sin 2\theta) - \cos \theta (1 - \cos 2\theta) \) For maximum/minimum, \( \frac{dZ}{d\theta} = 0 \implies 2 \sin \theta (2 - 3 \sin^2 \theta) = 0 \) \( \therefore \sin \theta = +\sqrt{2/3} \) (\( \because \sin \theta \ne 0 \)) When \( \sin \theta = \sqrt{2/3} \implies \frac{d^2Z}{d\theta^2} = -\frac{5}{5\sqrt{3}} - \frac{16}{3\sqrt{3}} = -\frac{1}{\sqrt{3}} - \frac{8}{\sqrt{3}} < 0 \) Hence : \( Z \) is maximum.
\( \implies x = \frac{1}{Z} \) is minimum. [from Eq.(1)] \( \therefore x \) is minimum.
\( \implies \text{min } x = \frac{1}{Z} = \frac{1}{(1 + \sqrt{3}) \left( 1 + \frac{1}{3} \right)} = \frac{3\sqrt{3}}{4} \text{ unit} \)

Question. Let S be the square of unit area and ABCD be the quadrilateral of sides a, b, c, and d. Prove \( 2 \le a^2 + b^2 + c^2 + d^2 \le 4 \).
Answer: Here \( a^2 = (1 - x)^2 + z^2, b^2 = w^2 + (1 - z)^2 \) \( c^2 = (1 - w)^2 + (1 - y)^2, d^2 = x^2 + y^2 \) Adding all the above, \( a^2 + b^2 + c^2 + d^2 \) \( = \{x^2 + (1 - x)^2\} + \{y^2 + (1 - y)^2\} + \{z^2 + (1 - z)^2\} + \{w^2 + (1 - w)^2\} \) where \( 0 \le x, y, z, w \le 1 \) Let us consider function, \( f(x) = x^2 + (1 - x)^2, 0 \le x \le 1 \) Then \( f'(x) = 2x - 2(1 - x) \) Let \( f'(x) = 0 \) for maximum/minimum,
\( \implies 4x - 2 = 0 \implies x = 1/2 \) Again, \( f''(x) = 4 > 0 \) when \( x = 1/2 \) \( \therefore f(x) \) is minimum at \( x = 1/2 \) and maximum at \( x = 1 \)
\( \implies 2 \le a^2 + b^2 + c^2 + d^2 \le 4 \)

Question.
Answer: [Evaluation pending based on derivatives]